EARTH - RESONANCE PCCP IDEAL for NTSE, IJSO,...

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EARTH�S MAGNETIC FIELD

The earth behaves as a huge bar magnet with its magnetic

field on its surface and upto a height of about 3 × 104 km.

The magnetic field of the earth is attributed to one or

more of the following causes:-

(i) Another point of view attributes the earth�s magnetism

is the presence of iron and nickel in the core of the

earth. It is assumed that the rotation of the earth about

its own axis causes the magnetisation of the iron and

nickel and the earth behaves as a bar magnet.

(ii) Prof. Blackett, studied that the rotation of the earth

about its own axis causes earth�s magnetism. For

example, the presence of ions in the upper atmosphere

constitutes a current due to the rotation of the earth

about its own axis.

Long ago, it has been known that earth has a magnetic

field, which is very similar to the field produced by a

huge magnet which is supposed to be at its centre. Its

north pole lies towards the geographic south and south

pole towards the geographic north. The axis of the

earth�s magnetic field does not coincide with the

geographic axis, the angle between them being about

150. The positions of the magnetic poles are not well

defined on the globe, they are spread over an area.

The magnetic pole in the northern hemisphere is

located on a peninsula in northern Canada. The

opposite magnetic pole in the southern hemisphere

lies on the Antarctic continent, south of Australia. The

earth�s magnetic field is supposed to arise from electric

currents in the molten iron rich outer core surrounding

the solid inner core of the earth.

Figure shows some field lines in the earth�s magnetic

field. The field lines usually dip towards or come out of

the earth�s surface at some angle. This angle is called

the magnetic inclination or the magnetic dip at that

place. At the magnetic poles, the angle of dip is 90º.

The line on the earth�s surface passing through the

places having angle of dip 00 is called the magnetic

equator, it passes through Thumba in south India,

where the space research centre is located.

N

S

N

S

Geographical North

Geographical South

Magnetic South

Magnetic North

Geographical Equator

ELEMENTS OF EARTH�S MAGNETIC FIELD

The following three quantities are called elements ofearth�s magnetic field.

(a) Angle of declination :

(b) Angle of dip (or Inclination)

(c) Horizontal component of earth�s magnetic field

(a) Angle of Declination :

The vertical plane passing through the axis of a freelysuspended magnet is called magneticmeridian. The direction of earth�s magnetic field lies

in the magnetic meridian and may not behorizontal. The vertical plane passing through the truegeographical north and south (or geographical axis ofearth) is called geographical meridian. The anglebetween the magnetic meridian and the geographicmeridian at a place is called angle of declination atthat place. The knowledge of declination at a placehelps in finding the true geographical directions atthat place. In our country, the angle of declination iszero in Pondicherry.

Geographicmeridian Angle of

declination

Magneticmeridian

O

G

MAGNETIC EFFECT OF CURRENT AND E.M.I.

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(b) Angle of Dip (or Angle of Inclination) :

The angle between the axis of a freely suspendedmagnetic needle (which is free to rotate in the verticalplane) and the horizontal plane at a place is called theangle of dip (or inclination) at that place.

M

Magnetic axis

Angle of dip

Horizontal plane H

Magnetic needle

(i) Angle of dip at the poles : The magnetic lines offorce at the poles of earth are vertical due to which themagnetic needle becomes vertical. Thus the angle ofdip at the magnetic poles of the earth is 900.

�

�

�

��

�

��

Dip Circle 00 00

900

900

600 60 0

600

300

300

30 0

300

600

Angle of dip at the equator : The lines of force aroundthe magnetic equator of the earth are perfectlyhorizontal. So, the magnetic needle will becomehorizontal there. Thus, the angle of dip at themagnetic equator of the earth will be 00 as shown infigure. The line on the earth�s surface passing through

the places where the angle of dip is 00 is calledmagnetic equator.

�

�

�

��

�

�

�

Dip Circle

60º

0º 0º

30º

90º

90º60º

30º 30º

30º

60º

60º

(c) Horizontal Component of Earth�s

Magnetic Field :

Let I be the total intensity of earth�s magnetic field. At a

place the total intensity I of earth�s magnetic field can be

resolved into two components i.e. horizontal componentH and vertical component V.

H = I cos .......(i)V = I sin .......(ii)

Where is the angle of dip at a place.

on dividing equation (ii) by (i)

tan = HV

.......(iii)

from equation (i) and (ii)

I = 22 HV .......(iv)

BIOT-SAVART�S LAW ( DUE TO WIRE)B

It is an experimental law. Let current i flows in a wire

(may be straight or curved). Due to d length of the

wire the magnetic field at P is :

iddB ....(i)

dB 2r

1 ....(ii)

dB sin ....(iii)

Combining these equations, we get

dB 2r

sinid

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

P

rd

i

dB = 20

r

sinid4

Vector representation

dB = 30

r

rdi4

Here r = position vector of the test point with respect

to (w.r.t.) d

= angle between d and r

.

Using this fundamental formula we can derive the

expression of B

due to a long wire.


(a) B

due to a straight wire :

Due to a straight wire PQ carrying a current I, the B

at

I

Qr

P

A2

1

by the formula :

PointA is given

B = r4I0

(sin

1 + sin

2) (i.e., direction of B

is inward

perpendicular to the plane of page)

Special case :

(i) If the wire is infinitely long then the mag-

netic field at �P� (as shown in the figure) is given by

(using 1 =

2 = 90º and the formula of �B� due to straight

wire)

r

P

×

×

×

×

×

r2

B 0

rB

The direction of B

at various is as shown in the figure.

The magnetic lines of force will be concentric

circles around the wire (as shown earlier)

(ii) If the wire is infinitely long but �P� is as shown in

the figure. The

direction of B

at various points is as shown in the

figure and magnetic field atpoint P

B = r4

0

r

P

×

×

×

×

×

upto

×

90º

RULE FOR FINDING DIRECTION OF MAGNETIC FIELD

The direction of the magnetic field at a point P due to astraight wire can be found by a slight variation in theright hand thumb rule. If we stretch the thumb of theright hand along the current and curl our fingers topass through the point P, the direction of the fingers atP gives the direction of the magnetic field there.

ILLUSTRATIONS

1. A horizontal overhead power line carries a current of90 A in east to west direction. What is the magnitudeand direction of the magnetic field due to the current1.5 m below the line ?

Sol. I = 90º A, r = 1.5 m

B = r4

2 0

I = T

5.1290104 7

B

S

N

E W

B = 710

5.1180

T = 1.2 × 10�5T

Magnetic field is towards south.

2. Find resultant magnetic field at �C� in the figure shown.

a

R

QP

S

i

C

a3

Sol: It is clear that �B� at �C� due all the wires is directed .

Also B at �C� due PQ and SR is same.

Also due to QR and PS is same B

res = 2(B

PQ + B

SP)


BPQ

=

2a

4

i0

(sin 60º + sin 60º),

Bsp

=

2a3

4

i0

(sin 30º + sin 30º)

Bres

= 2a3

i4

3a2

ia2

i3 000

3. Figure shows a square loop made from a uniform wire.Find the magnetic field at the centre of the square if abattery is connected between the points A and C.

A

D C

B

Sol. The current will be equally divided at A. The fields at thecentre due to the currents in the wires AB and DC willbe equal in magnitude and opposite in direction. Theresultant of these two fields will be zero. Similarly, theresultant of the fields due to the wires AD and BC willbe zero. Hence, the net field at the centre will be zero.

4. In the figure shown there are two parallel long wires(placed in the plane of paper) are carrying currents 2and consider points A, C, D on the line perpendicularto both the wires and also in the plane of the paper.

The distances are mentioned. Find (i) B

at A, C, D

(ii) position of point on line A C D where B

is O.

C

a/2a

1 2

2 D

aa

A

Sol. (i) Let us call B

due to (1) and (2) as 1B

and 2B

respectively. Then

at A : 1B

is and 2B

is ×

B1 =

a2

20

and B

2 =

a220

Bres

= B1� B

2 =

43

a0

at C : 1B

is × and 2B

also ×

Bres

= B1 + B

2 =

2a

22a

2

2 00

=

a2

6 0

=

a

3 0

×

at D : 1B

is × and 2B

is and both are equal in

magnitude. B

res = 0

(ii) It is clear from the above solution that B = 0 atpoint �D�.

(b) B

due to circular loop :

(i) At centre : Due to each d element of the loop, B

at point c is inwards (in this case).

ttanulresB at point c is . (inwards)

B =

R2N0 I

,c R

I

N = No. of the turns in the loop.

= R2

, = length of the loop

N can be fraction

.etc

311

,31

,41

or integer..

Direction of B

: The direction of the magnetic field at

the centre of a circular wire can be obtained using theright-hand thumb rule. If the fingers are curled alongthe current, the stretched thumb will point towards themagnetic field (see figure ).

Another way to find the direction is to look into the loopalong its axis. If the current is in anticlockwise direction,the magnetic field is towards the viewer. If the currentis in clockwise direction, the field is away from theviewer.

Semicircular and Quarter of a circle :

I

I8R

14N=

B =

Note : Magnetic field due to an arc is B =

2R2i0


5. A wire as shown in figure carries a current I ampere.The circular parthas a radius r. The magnetic field at the centre C willbe :

r CI

I

Sol. Magnetic field at the centre due to AC and BC is zero.

the magnetic field at the centre will only due to the

43

th part of the circular loop.

r CI

I

rA

BMagnetic field at C =

r243 0I =

r83 0I

6. A straight wire carrying a current of q2 A is bent ito asemicircular arc of radius 2.0 cm as shown in Figure

(a). What is the direction and magnitude of B at the

centre of the arc? Would your answer change if thewire were bent into a semicircular arc of the sameradius but in the opposite way as shown in Fig. 4.29(b)?

Sol. (i) Magnetic field at he centre of the arc is

B = r4I

Here I= 12 A, r = 2.0cm = 0.02 m,

4 × 10�7 TmA�1

B = 02.04

12104 7

= 1.9 × 10�4 T..

According to right hand rule the direction of the field isnormally into the plane of paper.(ii) The magnetic field will be of same magnitude,B = 1.9 × 10�4 T.The direction of the field is normally out of the plane ofpaper.

7. A long wire is bent as shown in figure. What will be themagnitude and direction of the field at the centre O ofthe circular portion , if a current I is passed through thewire ? Assume that the various portions of the wire donot touch at point P.

I I

r

P

O

I I

Sol. The system consists of a straight conductor and a cir-cular loop Field due to straight conductor at point O is

B1 =

r2I

, up the plane of paper

Field due to circular loop at point O is

B2 =

r2I , up th plane of paper

Total field at O is

B = B1 +B

2 =

r2I

11 , up the plane of paper..

9. Figure shows a current loop having two circular seg-ments and joined by two radial lines Find the mag-netic field at the centre O.

RS

O

P Q

I

I

a

b

Sol. Since the point O lines SP and QR so the magneticfield at O due to these straight portions is zero.The magnetic field at O due to the circular segment PQ

is

B1 =

2a4

I

Here, l = length of arc PQ = a

B1 = a

I4

directed normally upward

Similarly, the magnetic field at O due to the circular

segment SR is

B2 = b

I4

, directed normally downward.

The resultant field at O is

B = B1�B

2 =

4I

b1

a1

or B = ab4

)ab(I


MAGNETIC FORCE ON MOVING CHARGE

When a charge q moves with velocity v in a magnetic

field B

, then the magnetic force experienced by moving

charge is given by following formula :

F

= q( Bv

), (Put value of q with sign), v and

B remain in the same plane

v : Instantaneous velocity

B

: Magnetic field at that point.

NOTE :

(i) F

v

and also F

B

(Here is symbol for perpendicular)

(ii) F

v

power due to magnetic force on a charged particle

is zero.

(use the formula of power P = v.F

for its proof).

(iii) Since F

B

so work done by magnetic force is

zero in every part of the motion. The magnetic force

cannot increase or decrease the speed (or kinetic

energy) of a charged particle. It can only change the

direction of velocity.

(iv) On a stationary charged particle, magnetic force is

zero.

(v) If B||v

, then also magnetic force on charged

particle is zero. It moves along a straight line if only

magnetic field is acting. (Here || is symbol for parallel)

10. An infinitely long straight conductor �XY� is carrying a

current of 5 A. An electron is moving with a speed of 105

m/s parallel to the conductor in air from point A to B, as

shown in figure. The perpendicular distance between

the electron and the conductor �XY� is 20 cm. Calculate

the magnitude of the force experienced by the electron.

Write the direction of the force.

x

Belectron

20cm

A

5A Y

Sol. Magnetic field at a distance of 20 cm from current

carrying conductor XY is

B = r4

2 0

I

B = T1020

51022

7

x

Belectron

20cm

A

5A Y

B = 5 × 10�6 T

Force experienced by the electron is

F = evB ( = 90º)

= 1.6 ×10�19 × 105 × 5 × 10�6 N

= 8 × 10�20 N

According to Fleming�s left hand rule direction of force

will be upwards.

11. A charged particle of mass 5 mg and charge q = +2C

has velocity k�4j�3i�2v

. Find out the magnetic

force on the charged particle and its acceleration at

this instant due to magnetic field k�2j�3B

. v

and

B

are in m/s and Wb/m2 respectively..

Sol. BvqF

= 2 × 10�6 ( k�4j�3i�2 ) × )k�2j�3(

= 2 × 10�6 [�6 i� + 4 j� + 6 k� ] N

By Newton�s Law mF

a

= 6

6

105

102

)k�6j�4i�6(

= 0.8 ( k�3j�2i�3 ) m/s2

MAGNETIC FLUX

If we consider a plane perpendicular to a uniform

magnetic field, then the product of the magnitude of

the field and the area of the plane is called the magnetic

flux () linked with that plane. The magnetic flux linked

with this plane is given by

= BA


If the magnetic field B

, instead of being perpendicular

to the plane, makes an angle with the perpendicular

to the plane as shown in figure, then the magnetic flux

linked with the plane will be equal to the product of the

component of the magnetic field perpendicular to the

plane and the area of the plane.

Thus, = (B cos) A = BA cos .......(i)

Case I : If = 0º, then from equation (i) = BA cos0= BA (outgoing flux)

Case II : If = 90º, then from equation (i) = BA cos90= 0

AB

, Case III : If = 180º, then from equation (i) = BA cos180= � BA (incoming flux)

B

A

is positive if the outward normal to the plane is in the

same direction as B

. It is negative if the outward normal

is opposite to B

.

S.I. unit of B and :

The S.I. unit of magnetic flux is weber (Wb). Since

B = /A, the magnetic field is also expressed in

weber/metre2, (Wb-m�2). That is why the magnetic field

induction B is also called the magnetic flux density.

Definition of magnetic flux density (B) :

B = A

, if A = 1 metre2, then B =

In a magnetic field the number of lines of force (flux)

passing through per metre2 perpendicular to the field

is equal to the magnetic flux density.

12. The plane of a coil of area 1m2 and having 50 turns is

perpendicular to a magnetic field of 3 × 10�5

weber/

m2. Find the magnetic flux linked with it.

Sol. = NBA cos

but N = 50, B = 3 × 10�5 wb/m2,

A = 1m2, = 0 or = NBA

= 50 × 3 ×10�5

× 1

= 150 × 10�5 weber

FARADAY�S LAWS OF ELECTROMAGNETIC

INDUCTION

(i) When magnetic flux passing through a loop changes

with time or magnetic lines of force are cut by a

conducting wire then an e.m.f. is produced in the loop

or in that wire. This e.m.f. is called induced e.m.f.

If the circuit is closed then the current will be called

induced current.

(ii) In case of loop, the magnitude of induced e.m.f. is

equal to the rate of change of flux w.r.t. time . In case of

a wire it is equal to the rate at which magnetic lines of

force are cut by a wire.

E= � dtd

(�) sign indicates that the e.m.f. will be induced in such

a way that it will oppose the change of flux. S.I. unit of

magnetic flux is weber, so S.I. unit of induced e.m.f. is

SecondWb

= Volt


LENZ�S LAW

According to this law, e.m.f. will be induced in such away that it will oppose the cause which has producedit. Figure shows a magnet approaching a ring with itsnorth pole towards the ring.

We know that magnetic field lines come out of the northpole and magnetic field intensity increases as we movetowards the magnet.

If we consider the approach of North pole to be thecause of flux change, the lenz�s law suggests that the

side of the coil towards the magnet will behave asNorth pole and will repel the magnet. We know that acurrent carrying coil will behave like North pole if currentin it flows anticlockwise, as seen in figure.If we consider the approach of magnet as the cause ofthe flux change, Lenz�s law suggest that a force

opposite to the motion of magnet will act on the magnet,whatever be the mechanism. Lenz�s law tells that if the

coil is set free, it will move away from magnet, becausein doing so it will oppose the �approach� of magnet

If the magnet is given some initial velocity towards thecoil and is released, it will slow down. It can beexplained as the following.

The current induced in the coil will produce heat. Fromthe energy conservation, if heat is produced, there mustbe an equal decrease of energy in some other form,here it is the kinetic energy of the moving magnet. Thusthe magnet must slow down. So we can justify that thelenz�s law is conservation of energy principle.

13. A coil is placed in a constant magnetic field .The mag-netic field is parallel to the plane of the coil as shownin figure. Find the emf induced in the coil .

B

Sol. = 0 (always) since area is perpendicular to magneticfield. emf = 0

14. Find the emf induced in the coil shown in figure. Themagnetic field is perpendicular to the plane of the coiland is constant.

Area=A

B

Sol.= BA (always)

= const. emf = 0

15. Find the direction of induced current in the coil shownin figure. Magnetic field is perpendicular to the plane ofcoil and it is increasing with time.

Sol. Inward flux is increasing with time. To opposite it out-ward magnetic field should be induced. Hence cur-rent will flow anticlockwise.

SELF INDUCTANCE

When current flows through a coil or circuit, magneticfield is produced and hence a magnetic flux getsassociated with this coil or circuit. This magnetic fluxis directly proportional to the current flowing in the circuit(If other factors remain constant). If current throughthe coil is changed, the magnetic field is producedand hence the magnetic flux associated with itchanges and as a result of which, an e.m.f. is inducedin the coil or circuit, According to Lenz�s law the direction

of induced e.m.f. is such as that it always opposeschange due to which it is produced. As shown in figure,if current increases in circuit, induced e.m.f. is set upin such a way that it will decrease the current i.e.,induced current produced due to induced e.m.f. flowsopposite to the main current. Similarly, if main currentdecreases in the circuit, induced e.m.f. will increase it.Now, the induced current due to induced e.m.f. willflow in the direction of main current.

In this way the phenomenon, in which, on changingthe current in a coil or circuit an induced e.m.f. is set upin that coil or circuit, is called self induction. This inducede.m.f. is called back e.m.f.

(a) Coefficient of Self Induction or Self

Inductance :

When current I flows in a circuit, associated magneticflux is proportional to the current flowing i.e.,

or = L .....(i)Here, L is a constant of proportionality, called theCoefficient of self-induction or self-inductance.Self-inductance of a coil depends on its area, numberof turns and the medium inside it (material of core). If = 1A, then L = . Hence, self-inductance of a coil isequal to the magnetic flux associated with the coil when


unit current flows through it.

In equation (i), on changing the current , will alsochange and an induced e.m.f. E is produced in thecircuit i.e.,

E = dtd

Ldt

)L(ddt

)(d

or L = |E| if dtd

= 1 A/s

Hence, the coefficient of self induction in a circuit isnumerically equal to the induced e.m.f. produced inthe circuit when the rate of change of current is 1 A/s.When flow of current in the circuit starts, induced e.m.f.opposes this change.

16. An average induced e.m.f. of 0.20 V appears in a coilwhen the current in it is changed from 5.0 A in onedirection to 5.0 A in the opposite direction in 0.20s.Find the self-inductance of the coil.

Sol. Average dtdi

= s20.0)A0.5()A0.5(

= � 50 A/s

Using E = � L dtdi

,

0.2 V = L (50 A/s)

or, L = mH0.4s/A50

V2.0

MUTUAL INDUCTANCE

Consider two arbitrary conducting loops 1 and 2.Suppose that I1 is the instantaneous current flowing inloop 1. This current generates a magnetic field B

1 which

links the second circuit, giving rise to a magnetic flux 2

through the second circuit.

Furthermore, it is obvious that the flux through thesecond circuit is zero whenever the current flowingaround the first circuit is zero. It follows that the flux 2

through the second circuit is directly proportional tothe current I1 flowing around the first circuit. Hence ,wecan write 2 = M21I1 where the constant of proportionalityM21 is called the mutual inductance of circuit 2 withrespect to circuit 1. Similarly, the flux

1 through the first

circuit due to the instantaneous current I2 flowingaround the second circuit is directly proportional to thatcurrent, so we can write 1 =M12I2 where M12 is themutual inductance of circuit 1 with respect to circuit 2.It can be shown that M21 = M12.

Note: M is a purely geometric quantity, depending onlyon the size, number of turns, relative position andrelative orientation of the two circuits. The S.I. unit of

mutual inductance is called Henry (H). One henry isequal to a volt-second per ampere.Suppose that the current flowing around circuit 1changes by an amount I1 in a small time interval t.The flux linking circuit 2 changes by an amount 2 =MI1 in the same time interval. According to Faraday�s

law, an e.m.f. E2 = �

t2

is generated around the

second circuit due to the changing magnetic flux linkingthat circuit, Since,

2 = MI

1, this e.m.f. can also be

written E2 = � M

t1

I

Thus, the emf generated around the second circuitdue to the current flowing around the first circuit isdirectly proportional to the rate at which that currentchanges. Likewise, if the current I2 flowing around thesecond circuit changes by an amount I1 in a timeinterval t then the e.m.f. generated around the first

circuit is E1 = �M

t2

I.

Note that there is no direct physical connection(coupling) between the two circuits the coupling is dueentirely due to the magnetic field generated by thecurrents flowing around the circuits.

17. Two conducting circular loops of radii R1

and R2 are

placed in the same plane with their centres coinciding.Find the mutual inductance between them assumingR

2 << R

1.

Sol. Suppose a current i is established in the outer loop.The magnetic field at the centre will be :

B = 1

0

R2i

As the radius R2 of the inner coil is small compared to R

1,

the flux of magnetic field through it will be approximately

= 22

1

0 RR2

i

So that the mutual inductance is

M = 1

22o

R2R

i

TRANSFORMER

It is a device which raises or lowers the voltage in ACcircuits through mutual induction. It consists of twocoils wound on the same core. The coil which isconnected to the source (i.e. to which input is applied)is called primary coil while the other which is connectedto the load (i.e. from which output is taken) is calledsecondary coil. The alternating current passing throughthe primary coil creates a continuously changing fluxthrough the core. This changing flux induces analternating e.m.f. in the secondary coil. As magneticlines of force are closed curves, the flux per turn of theprimary must be equal to the flux per turn of thesecondary coil.


P and S are the flux passing through the primary and

secondary coils, NP and NS are the number of the turnsin primary and secondary coils respectively,then,

P

P

S

S

NN

The number of turns in each coil is constant, if dP anddS are the changes in flux in time dt in primary andsecondary coils respectively,then,

dtd

N1

dt

d

N1 P

P

S

S

orP

S

P

S

NN

EE

dtd

Eas

And as in an ideal transformer there is no loss ofpower, so P = EI = constant, therefore

P

S

S

P

P

S

NN

II

EE

TYPES OF TRANSFORMER

Transformer can be divided mainly into two types:

(a) Step up transformer (b) Step down transformer

(a) Step up transformer :

If the secondary coil has a greater number of turnsthan the primary (N

s > N

p), the voltage is stepped up

(Vs > V

p). This type of arrangement is called a step-up

transformer. In this arrangement, there is less currentin the secondary than in the primary (N

p/N

s < 1 and

Is< I

p ).

Eg.: If the primary coil of a transformer has 100 turnsan the secondary has 200 turns, then

p

s

NN

= 2 and s

p

N

N =

21

. Thus, a 220 V input at 10 A will

step-up to 440 V output at 5.0 A.

(b) Step down Transformer :If the secondary coil has less number of turns than theprimary (N

s < N

p), the voltage is stepped down

(Vs < V

p). This type of arrangement is called a step-

down transformer. In this arrangement, current in thesecondary coil is more increased than in the primary(N

p/N

s < 1 and I

s> I

p ).

Efficiency of Transformer :Efficiency of transformer,

= inputPoweroutputPower

% = i

o

PP

×100

NOTE :

Regarding a transformer it is worth noting that :

(i) It works on AC only and never on DC

(ii) It can increase or decrease either voltage or currentbut not both simultaneously (as power = constant).

(iii) Some power is always lost due to flux leakage,hysteresis, eddy currents, humming and heating of coils.

18. A step-down transformer converts a supply line voltageof 2200 volt into 220 volt. The primary coil has 5000turns. The efficiency and power transmitted by thetransformer are 90% and 8 kilowatt respectively.Calculate the number of turns in the secondary coil.

Sol.p

s

p

s

NN

ee

p

s

NN

2200220

Ns = 500

19. The primary winding of a transformer has 500 turnswhereas its secondary has 5000 turns. The primary isconnected to an AC supply of 20 V, 50 Hz. What will beoutput of secondary coil ?

Sol. We know that

P

S

P

S

EE

NN

or 5005000

= 20ES

or Es =

500205000

= 200 V

Frequency remains same.

GENERATOR

This is a device which convert mechanical energy into

electrical energy using the principle of electromagnetic

induction. It is of two types :


(a) AC Generator or Dynamo :

When a coil (conductor) is rotated in a magnetic field,the magnetic f lux linked with it changes and therefore an alternating e.m.f. is induced in the coil.

Construction : The main parts of a dynamo are:-

(i) Field magnets :

It is a strong horse shoe permanent magnet. Anelectromagnet run by a DC source can also be usedfor high power generators.

(ii) Armature :

It is a soft iron core on which a coil ABCD having a largenumber of turns of insulated copper wire is wound. Thisarmature (or coil) is rotated rapidly in the magnetic fieldbetween the poles of the magnet.(iii) Slip rings :

The ends of the armature (or the coil) are connected totwo coaxial metallic slip rings S

1 and S

2 which rotate

along with the coil.

(iv) Brushes :

Two brushes B1 and B

2 made of carbon, press against

the slip rings S1 and S

2 respectively. The external circuit

(i.e. load) is connected between the other ends ofbrushes. The brushes B

1 and B

2 do not rotate along

with the coil.

Working of an AC generator :

Suppose that the generator coil ABCD is initially in thehorizontal position. Again suppose that the coilABCD is being rotated in the anticlockwise directionbetween the poles N and S of a horse-shoe typemagnet.

(i) As the coil rotates in the anticlockwise direction, theside AB of the coil moves down cutting the magneticlines of force near the N-pole of the magnet and sideCD moves up, cutting the lines of force near the S-poleof the magnet. Due to this, induced current is producedin the sides AB and DC of the coil. On applyingFleming�s right-hand rule to the sides AB and DC of

the coil, we find that the currents are in the directions Bto A and D to C. Thus, the induced currents in the twosides of the coil are in the same direction and we getan effective induced current in the direction BADC.

(ii) After half revolution, the sides AB and DC of thecoil will interchange their positions. The side AB willcome on the right hand side and side DC will come onthe left hand side. So, after half a revolution, side ABstarts moving up and side DC starts moving down. Asa result of this, the direction of induced current in eachside of the coil is reversed after half a revolution. Sincethe direction of induced current in the coil is reversedafter half revolution so that polarity (positive andnegative) of the two ends of the coil also changes afterhalf revolution. The end of coil which was positive inthe first half of rotation becomes negative in the secondhalf. And the end which was negative in the first-halfrevolution becomes positive in the second half ofrevolution. Thus, in 1 revolution of the coil, the currentchanges its direction 2 times.

The alternating current (AC ) produced in India has afrequency of 50 Hz. That is, the coil is rotated at the rateof 50 revolutions per second. Since in 1 revolution ofcoil, the current changes its direction 2 times, so in 50revolutions of coil, the current changes its direction 2 ×50 = 100 times. Thus, the AC supply in India changesits direction 100 times in 1 second. Another way ofsaying this is that the alternating current produced inIndia changes its direction every 1/100 second. Thatis, each terminal of the coil is positive (+) for 1/100 of asecond and negative (-) for the next 1/100 of a second.

After every half revolution, each side of the generatorcoil starts moving in the opposite direction in themagnetic field. The side of the coil which was initiallymoving upwards, after half revolution, it starts movingdownwards. Due to the change in the direction of motionof the two sides of the coil in the magnetic field afterevery half revolution, the direction of current producedin them also changes after every half revolution.

(b) DC Generator (or DC Dynamo) :

�DC generator� means �Direct Current generator�. That

is, a DC generator produces direct current.Construction of a DC Generator :

A simple DC generator consists of a rectangular coilABCD which can be rotated rapidly between the polesnorth and south of a strong horseshoe type magnet M.

The generator coil is made of a large number of turnsof insulated copper wire. The two ends of the coil areconnected to the two copper half rings (or split rings)R

1 and R

2 of a commutator. There are two carbon

brushes B1 and B

2 which press lightly against the two

half rings. When the coil is rotated, the two half ringsR

1 and R

2 touch the two carbon brushes B

1 and B

2 one

by one. So, the current produced in the rotating coil canbe tapped out through the commutator half rings intothe carbon brushes. From the carbon brushes B

1 and

B2, we can take the current into the various electrical

appliances like radio, T.V., electric iron, bulbs, etc.


Permanentmagnet Rotation of coil anticlockwise

Rectangular Coil

Antclockwise

yLoad

Commutator

Carbonbrushes

D.C. Generator

(Split ring)

VVVVD.C.

+ �

B1 B2

Working of a DC generator :

Suppose that the generator coil ABCD is initially in the

horizontal position. Again suppose that the coil ABCD

is being rotated in the anticlockwise direction between

the poles N and S of a horseshoes type magnet.

(i) As the coil rotates in the anticlockwise direction, the

side AB of the coil moves down cutting the magnetic

lines of force near the N-pole of the magnet and side

DC moves up, cutting the lines of force near the S-pole

of the magnet in figure. Due to this, induced current is

produced in the sides AB and DC of the coil. On applying

Fleming�s right-hand rule to the sides AB and DC of

the coil we find that the currents in them are in the

directions B to A and D to C respectively. Thus, we get

an effective induced current in the direction BADC. Due

to this the brush B1 becomes a positive (+) pole and

brush B2 becomes negative (-) pole of the generator.

(ii) After half revolution the sides AB and DC of the coil

will interchange their positions. The side AB will come

on the right hand side and start moving up whereas

side DC will come on the left-hand side and start

moving down. But when sides of the coil interchange

their positions, then the two commutator half rings R1

and R2 automatically change their contacts from one

carbon brush to the other. Due to this change, the current

keeps flowing in the same direction in the circuit. The

brush B1 will always remain positive terminal and brush

B2 will always remain negative terminal of the generator.

Thus, a DC generator supplies a current in one

direction by the use of a commutator consisting of two

half-rings of copper.

Difference between a DC generator and an AC

generator :

In a DC generator we connect the two ends of the coil

to a commutator consisting of two, half rings of copper.

On the other hand, in an AC generator, we connect the

two ends of the coil to two full rings of copper called

slip rings.

ELECTRIC MOTOR

A motor is a device which converts electrical energy

into mechanical energy. Every motor has a shaft or

spindle which rotates continuously when current is

passed into it. The rotation of its shafts is used to drive

the various types of machines in homes and industry.

Electric motor is used in electric fans, washing

machines, refrigerators, mixer and grinder and many

other appliances. A common electric motor works on

direct current. So, it is also called DC motor, which

means a �Direct Current motor�. The electric motor

which we are going to discuss now is actually a DC

motor.

(a) Principle of a Motor :

An electric motor utilizes the magnetic effect of current.

A motor works on the principle that when a rectangular

coil is placed in a magnetic field and current is passed

through it, a torque acts on the coil which rotates it

continuously. When the coil rotates, the shaft attached

to it also rotates. In this way the electrical energy

supplied to the motor is converted into the mechanical

energy of rotation.

(b) Construction of a Motor :

An electric motor consists of a rectangular coil ABCD

of insulated copper wire, wound on a soft iron core

called armature. The soft iron core has not been shown

in figure to make things simple. The coil is mounted

between the curved poles of a U-shaped permanent

magnet in such a way that it can rotate between the

poles N and S. The two ends of the coil are soldered

(or welded) permanently to the two half rings X and Y of

a commutator.

Permanentmagnet Rotation of coil anticlockwise

Rectangular Coil

FF

Antclockwise

+�

QPy

Battery

Commutator

Carbonbrushes

An electric motor

(Split ring)

A commutator is a copper ring split into two parts X andY, these two parts are insulated from one another andmounted on the shaft of the motor.

End A of the coil is welded to part X of the commutatorand end D of the coil is welded to part Y of thecommutator. The commutator rings are mounted on


the shaft of the coil and they also rotate when the coilrotates.

The function of commutator rings is to reverse thedirection of current flowing through the coil every timethe coil just passes the vertical position during arevolution.

We cannot join the battery wires directly to the twocommutator�s half rings to pass current into the coil

because if we do so, then the connecting wires will gettwisted when the coil rotates. So, to pass the electriccurrent to the coil, we use two carbon strips P and Qknown as brushes. The carbon brushes P and Q arefixed to the base of the motor and they press lightlyagainst the two half rings of the commutator. Thefunction of carbon brushes is to make contact with therotating rings of the commutator and through them tosupply current to the coil. It should be noted that anyone brush touches only one ring at a time, so thatwhen the coil rotates, the two brushes will touch boththe rings one by one.

(c) Working of a Motor :

Suppose that initially the coil ABCD is in the horizontalposition as shown in figure. On pressing the switch,the current enters the coil through carbon brush P andcommutator half ring X. The current flows in the directionABCD and leaves via ring Y and brush Q.

(i) In side AB of the coil, the direction of current is fromA to B and the direction of magnetic field is from N to Spole. So, by applying Fleming�s left hand rule to the

side AB of the coil we find that it will experience a forcein the upward direction.(ii) In side DC of the coil, the direction of current is fromC to D towards but the direction of magnetic fieldremains the same from N to S pole as shown in figure.So, by applying Fleming�s left hand rule to the side DC

of the coil, we find that. It will experience a force in thedownward direction.

(iii) We find that the force acting on the side AB of thecoil is in the upward direction whereas the force actingon the side DC of the coil is in the downward direction.These two equal, opposite and parallel forces actingon the two sides of the coil form a couple (torque) androtate the coil in the anticlockwise direction.

(iv) While rotating, when the coil reaches the verticalposition, then the brushes P and Q will touch the gapbetween the two commutator rings and current to thecoil is cut off. Though the current to the coil is cut offwhen it is in the exact vertical position, the coil doesn�tstop rotating because it has already gained momentumdue to which it goes beyond the vertical position.(v) When the coil goes beyond the vertical position, thetwo commutator�s half rings automatically change

contact from one brush to the other. This reverses thedirection of current through the coil which, in turn,reverses the direction of forces acting on the two sidesof the coil. The side AB of the coil will now be on the left

hand side with a downward force on it, whereas sideDC of the coil will come on the right hand side with anupward force on it. In this position also a couple actson the coil which rotates it in the same direction(anticlockwise direction). This process is repeatedagain and again and the coil continues to rotate aslong as the current is passing. This is how an electricmotor works.


CONSERVATION OF ELECTRIC CHARGE

Whenever two bodies are charged by rubbing, one getspositively charged and the other gets negativelycharged. The net charge on the two bodies, however,remains zero�the same as that before rubbing. In other

words, charge is conserved. It can neither be creatednor be destroyed. The only thing that happens onrubbing is that charged particles (electrons) gettransferred from one body to the other.

In some phenomena, charged particles are created.But even then the conservation of charge holds. Forexample, a free neutron converts itself into an electronand the proton taken together is also zero. So, there isno change in the conversion of a neutron to an electronand a proton.

COULOMB�S LAW

Charles Augustine de Coulomb studied the interactionforces of charged particles in detail in 1784. He used atorsion balance. On the basis of his experiments heestablished Coulomb�s law. According to this law the

magnitude of the electric force between two pointcharges is directly proportional to the product of themagnitude of the two charges and inverselyproportional to the square of the distance betweenthem and acts along the straight line joining the twocharges.

In mathematical terms, the force that each of the twopoint charges q1 and q2 at a distance r apart exerts onthe other can be expressed as�

F = 221

r

qqk

This force is repulsive for like charges and attractivefor unlike charges.

Where k is a constant of proportionality. k = 04

1

,

here 0 is absolute permittivity of free space.The force is directed along the line joining the centresof the two charged particles.For any other medium except air, free space or vacuumcoulomb�s law reduces to

F = 1 22

q q14 r

= Permittivity of the mediumand = 0r

r = relative Permittivity or dielectric constant of the me-dium.Coulomb�s law is based on physical observation and

it is not logically derived from any other concept.

ILLUSTRATIONS

1. Find out the electrostatics force between two pointcharges placed in air (each of +1 C) if they areseparated by 1m .

Sol. Fe = 2

21

r

qkq = 2

9

1

11109 = 9×109 N

Note : From the above result we can say that 1 Ccharge is too large to realize. In nature, charge isusually of the order of C

2. A particle of mass m carrying charge q1 is revolving

around a fixed charge �q2 in a circular path of radius

r. Calculate the period of revolution and its speedalso.

Sol.04

1 2

21

r

qq = mr2 = 2

2

T

mr4'

T2 = 21

220

qq

)mr4(r)4(

or T = 4r 21

0

qq

mr

and also we can say that

20

21

r4

qq

=

rmv2

V = mr4

qq

0

21

PROPERTIES OF ELECTRIC FIELD INTENSITY

(i) It is a vector quantity. Its direction is the sameas the force experienced by positive charge.

(ii) Electric field due to positive charge is alwaysaway from it while due to negative charge alwaystowards it.

(iii) Its S.. unit is Newton/Coulomb.

(iv) Electric force on a charge q placed in a regionof electric field at a point where the electric field

intensity is E

is given by EqF

.

Electric force on point charge is in the samedirection of electric field on positive chargeand in opposite direction on a negative charge.

(v) It obeys the superposition principle, that is, thefield intensity at a point due to a system of charges isvector sum of the field intensities due to individual

ELECTRICITY



point charges.

321 EEEE

+ .....

(vi) It is produced by source charges. The electricfield will be a fixed value at a point unless we changethe distribution of source charges.

3. Five point charges, each of value q are placed on fivevertices of a regular hexagon of side L. What is themagnitude of the force on a point charge of value �

q

coulomb placed at the centre of the hexagon?

qq C

qqDE

O

-q

q

BA

F

L

Sol. If there had been a sixth charge +q at the remainingvertex of hexagon force due to all the six charges on �q at O would be zero (as the forces due to individualcharges will balance each other), i.e.,

0FR

Now if f

is the force due to sixth charge and F

due toremaining five charges.

F

+ f

= 0 i.e. F

= � f

or, |F| = |f| = 04

1 2L

qq = 2

2

0 L

q4

1

NetF

= COF

= 2

2

L

q4

1

along CO

4. Calculate the electric field intensity which would bejust sufficient to balance the weight of a particle ofcharge �10 c and mass 10 mg.

Sol. As force on a charge q in an electric field E

is

F

q = q E

So according to given problem

q E

W

A

Fe

|W| |F| q

i.e., |q|E = mg

i.e., E = |q|mg

= 10 N/C., in downward direction.

ELECTROSTATIC EQUILIBRIUM

The position where the resultant force on a chargedparticle becomes zero is called equilibrium position.

(a) Stable Equilibrium :

A charge is initially in equilibrium position and isdisplaced by a small distance. If the charge tries toreturn back to the same equilibrium position then thisequilibrium is called position of stable equilibrium.

(b) Unstable Equilibrium :

If charge is displaced by a small distance from itsequilibrium position and the charge has no tendencyto return to the same equilibrium position. Instead itgoes away from the equilibrium position.

(c) Neutral Equilibrium :

If charge is displaced by a small distance and it is stillin equilibrium condition then it is called neutralequilibrium.

5. Two equal positive point charges 'Q' are fixed at pointsB(a, 0) and A(�a, 0). Another test charge q

0 is also

placed at O(0, 0). Show that the equilibrium at 'O' is(i) stable for displacement along X-axis.(ii) unstable for displacement along Y-axis.

Sol. (i)

Initially AOF

+ BOF

= 0 |F| AO

= |F| BO

= 2

0

a

KQq

When charge is slightly shifted towards + x axis by asmall distance x, then.

|F| AO

< |F| BO


Therefore the particle will move towards origin (itsoriginal position) hence the equilibrium is stable.(ii) When charge is shifted along y axis

After resolving components net force will be along yaxis so the particle will not return to its originalposition so it is unstable equilibrium. Finally thecharge will move to infinity.

ELECTRIC LINES OF FORCE (ELOF)

The line of force in an electric field is an imaginaryline, the tangent to which at any point on itrepresents the direction of electric field at the givenpoint.

(a) Properties :

(i) Line of force originates out from a positivecharge and terminates on a negative charge. Ifthere is only one positive charge then lines startfrom positive charge and terminate at . If thereis only one negative charge then lines start from and terminates at negative charge.

(ii) The electric intensity at a point is the numberof lines of force streaming through per unit areanormal to the direction of the intensity at that point.The intensity will be more where the density oflines is more.

(iii) Number of lines originating (terminating) from(on) is directly proportional to the magnitude ofthe charge.

(iv) ELOF of resultant electric field can never intersectwith each other.

(v) Electric lines of force produced by staticcharges do not form close loop.

(vi ) E lec tr ic l ines of force end or startperpendicularly on the surface of a conductor.

(vii) Electric l ines of force never enter intoconductors.

6. If number of electric lines of force from charge qare 10 then find out number of electric lines offorce from 2q charge.

Sol. No. of ELOF charge10 q 20 2qSo number of ELOF will be 20.

7. A charge + Q is fixed at a distance of d in front ofan infinite metal plate. Draw the lines of forceindicating the directions clearly.


Sol. There will be induced charge on two surfaces ofconducting plate, so ELOF will start from +Q chargeand terminate at conductor and then will again startfrom other surface of conductor.

ELECTRIC FLUX

Consider some surface in an electric field E

. Let us

select a small area element dS on this surface.

The electric flux of the field over the

area element is given by dE =

ds.E

Direction of dS is normal to the surface. It is along

n�or d

E = EdS cos

or dE = (E cos ) dS

dS E

or dE = E

n dS

where En is the component of electric field in the

direction of dS .

If the electric field is uniform over that area then

E = SE

(a) Physical Meaning :

The electric flux through a surface inside an electricfield represents the total number of electric lines offorce crossing the surface in a direction normal thesurface. It is a property of electric field

(b) Unit :

(i) The SI unit of electric flux is Nm2 C�1 (gauss) or Jm2 C�1.

(ii) Electric flux is a scalar quantity. (It can be positive,negative or zero)

8. The electric field in a region is given by,

jE54

iE53

E 00

with E0 = 2.0 × 103 N/C. Find the

flux of this field through a rectangular surface of area0.2m2 parallel to the Y�Z plane.

Sol. E = SE

=

jE

5

4iE

5

300

. i�2.0 = CmN

2402

ELECTRICAL RESISTANCE

The property of a substance by virtue of which itopposes the flow of electric current through it istermed as electrical resistance. Electrical resistancedepends on the size, geometry, temperature andinternal structure of the conductor.

We known that, vd =

meE

=

meV

I = Anevd

= Anem

eV

I = mVAne2

2Ane

mIV

R =

2Ane

mIV

R = 2Ane

m

R = A

=

RA

= 2ne

m

is called resistivity (it is also called specific

resistance), and =

2ne

m =

1, is cal led

conductivity. Therefore current in conductors isproportional to potential difference applied across

its ends. This is Ohm's Law. Units: 11m

also called siemens m�1.

9. If a copper wire is stretched to make its radiusdecrease by 0.15%, Find the percentage increasein resistance (approximately).

Sol. Due to stretching resistance changes are in the ratio4

2

1

1

2

r

r

R

R

or 4rR

orrr

4RR

%15.04 = 0.60%


EFFECT OF STRETCHING OF A WIRE ON RESISTANCE

In stretching, the density of wire usually does notchange. Therefore

Volume before stretching = Volume after stretching

2211 AA

and2

1

1

2

1

2

AA

RR

If information of lengths before and after stretching

is given, then use 1

2

2

1

AA

2

1

2

1

2

R

R

If information of radius r1 and r

2 is given then use

2

1

1

2

AA

2

2

1

1

2

A

A

R

R

4

2

1

r

r

CONDUCTIVITY :

(a) Reciprocal of resistivity of a conductor is calledits conductivity. It is generally represented by .

(b)

1

(c) Unit : 11 metre.ohm

EFFECT OF TEMPERATURE ON RESISTANCE

AND RESISTIVITY

The resistance of a conductor depends upon thetemperature. As the temperature increases, therandom motion of free electrons also increases. Ifthe number density of charge carrier electronsremains constant as in the case of a conductor, thenthe increase of random motion increases theresistivity. The variation of resistance with temperatureis given by the following relation

t1RR 0t

where Rt and R

0 are the resistance at t0C and 00C

respectively and is called as temperature coefficientof resistance of the substance.

or tR

RR

0

0t

This constant

If R0 = 1 ohm, t = 10C, then

0t RR

Thus, the temperature coefficient of resistance isequal to the increase in resistance of a conductorhaving a resistance of one ohm on raising itstemperature by 10C. The temperature coefficient ofresistance may be positive or negative.

From calculations it is found that for most of the

metals the value of is nearly C/273

1 0. Hence

substituting in the above equation

273t

1RR 0t

273T

R273

t273R 00

where T is the absolute temperature of the conductor.

TR t

Thus, the resistance of a pure metallic wire is directlyproportional to its absolute temperature.

The graph drawn between the resistance Rt and

temperature t is found to be a straight line

Rt

tºC

R0

The resistivity or specific resistance varies withtemperature. This variation is due to change inresistance of a conductor with temperature. Thedependence of the resistivity with temperature isrepresented by the following equation.

t10t

With the rise of temperature the specific resistanceor resistivity of pure metals increases and that ofsemi-conductors and insulators decrease.

The resistivity of alloys increases with the rise oftemperature but less than that of metals.

On applying pressure on pure metals, its resistivitydecreases but on applying tension, the resistivityincreases.

The resistance of alloys such as eureka, manganinetc., increases in smaller amount with the rise intemperature. Their temperature coefficient ofresistance is negligible. On account of their highresistivity and negligible temperature coefficient ofresistance these alloys are used to make wires forresistance boxes, potentiometer, meter bridge etc.,

The resistance of semiconductors, insulators,


electrolytes etc., decreases with the rise intemperature. Their temperature coefficients ofresistance are negative.

On increasing the temperature of semi conductors alarge number of electrons get free after breaking theirbonds. These electrons reach the conduction bandfrom valence band. Thus conductivity increases orresistivity decreases with the increase of free electrondensity.

10. A wire has a resistance of 2 ohm at 273 K and aresistance of 2.5 ohm at 373 K. What isthe temperature coefficient of resistance of thematerial?

Sol. 273373225.2

TTR

RR

00

0

K/105.2200

5.0 03

SUPER CONDUCTOR AND ITS APPLICATIONS

Prof. K. Onnes in 1911 discovered that certain metalsand alloys at very low temperature lose theirresistance considerably. This phenomenon is knownas super-conductivity. As the temperature decreases,the resistance of the material also decreases, but whenthe temperature reaches a certain critical value (calledcritical temperature or transition temperature), theresistance of the material completelydisappears i.e. it becomes zero. Then the materialbehaves as if it is a super-conductor and there will beflow of electrons without any resistance whatsoever.The critical temperature is different for different material.It has been found that mercury at critical temperature4.2 K, lead at 7.25 K and niobium at critical temperature9.2 K become super-conductor.

Applications of super conductors :

(i) Super conductors are used for making very strongelectromagnets.

(ii) Super conductivity is playing an important role inmaterial science research and high energy particlephysics.(iii) Super conductivity is used to produce very highspeed computers.

(iv) Super conductors are used for the transmission ofelectric power.

WHEATSTONE BRIDGE

Wheatstone bridge is an arrangement of four resistorsin the shape of a quadrilateral which can be used tomeasure unknown resistance in terms of theremaining three resistances.

The arrangement of Wheatstone bridge is shown infigure below. Out of four resistors, two resistances R1,R2 and R3, R4 are connected in series and are joined inparallel across two points a and c. A battery of emf E isconnected across junctions a and c and a galvanometer(G) between junction b and d. The keys K

1 and K

2 are

used for the flow of current in the various branches ofbridge.

Principle of Wheatstone Bridge :

When key K1 is closed, current i from the battery is

divided at junction a in two parts. A part i1 goes throughR1 and the rest i2 goes through R3. When key K

2 is

closed, galvanometer shows a deflection.

The direction of deflection depends on the value ofpotential difference between b and d. When the valueof potential at b and d is same, then no current will flowthrough galvanometer. This condition is known as thecondition of balanced bridge or null deflectioncondition. This situation can be obtained by choosingsuitable values of the resistances. Thus, in nulldeflection state, we have :

Va � V

b = V

a � V

d

or i1 R

1 = i

2 R

3...(i)

Similarly :V

b � V

c = V

d � V

c

or i1 R

2 = i

2 R

4...(ii)

On dividing equation (i) by (ii), we get

21

11

RiRi

= 42

32

Ri

Rior

4

3

2

1

R

R

RR

...(iii)

Equation (iii) states the condition of balanced bridge.Thus, in null deflection condition the ratio ofresistances of adjacent arms of the bridge are same.The resistor of unknown resistance is usuallyconnected in one of the arm of the bridge. Theresistance of one of the remaining three arms isadjusted such that the galvanometer shows zerodeflection. If resistance of unknown resistor is R4. Then

R4 = (R3)

1

2

RR

For better accuracy of the bridge one should choose

resistances R1, R2, R3 and R4 of same order.


KIRCHHOFF'S LAWS

(a) Kirchhoff�s Current Law (Junction law) :

This law is based on law of conservation of charge.It states that "The algebraic sum of the currentsmeeting at a point of the circuit is zero" or totalcurrent entering a junction is equal total currentleaving the junction.

in = out.

It is also known as KCL (Kirchhoff's current law).

(b) Kirchhof f �s Voltage Law (Loop law) :

�The algebraic sum of all the potential differences

along a closed loop is zero. IR + EMF =0�. The

closed loop can be traversed in any direction. Whiletraversing a loop if potential increases, put a positivesign in expression and if potential decreases puta negative sign.

V1

V2 + V

3 V

4 = 0. Boxes may contain resistor

or battery or any other element (linear or nonlinear).It is also known as KVL

11. Figure shows, current in a part of electrical circuit, whatwill be the value of current (i) ?

2 A 1 A

1.3 A

i 3 A

2 A

P Q RS

Sol.

2 A 1 A

1.3 A

i 3 A

2 A

P Q RS

i 1i 2

i 3

From KCL, current at junction P, i1 = 2 + 3 = 5 A

From KCL, current at junction Q, i2 = i

1 + 1 = 5 + 1 = 6 A

From KCL, current at junction R, i3 = i

2 � 2 = 6 � 2 = 4 A

From KCL, current at junction S, i = i3 � 1.3 = 4 � 1.3

= 2.7 A

12. In the circuit shown, calculate the value of R in ohmthat will result in no current through the 30 V battery.

Sol. Applying KVL in loop CEFDC50 � iR � 20 i = 0

i = R20

50

20R

50V

10

i

i E

FD iB

A C

Potential drop across R = Potential drop across ABiR = 30

R20

50

.R = 30

R = 30

GROUPING OF CELLS

(a) Cells in Series :

BE ,r1 1 E ,r2 2 E ,r3 3

E rn n

A BE ,req eq

Equivalent EMFE

eq = E

1 + E

2 + ......... + E

n (write EMF�s with polarity)

Equivalent internal resistance req

= r1 + r

2 + r

3 + r

4 + ........ r

n

If n cells each of emf E, are arranged in series and if r isinternal resistance of each cell, then total emf = nE

R

E,r E,r E,r E,r

IUpto n

A B

So current in the circuit, I = nrR

nE

There may by two cases :

(i) If nr << R, then I =RnE

= n × current due to one cell.

So, series combination is advantageous.


(ii) If nr >> R, then I =rE

= current due to one cell.

So, Series combination is not advantageous.

Note : If polarity of m cells is reversed, then equivalente.m.f. = (n�2m) E while the equivalent resistance is

still nr + R, so current in R will be

i = Rnr

E)m2n(

(b) Cells in Parallel :

If m cells each of emf E and internal resistance r beconnected in parallel and if this combination isconnected to an external resistance then the emf ofthe circuit is E.

Internal resistance of the circuit = mr

.

and I = rmRmE

mr

R

E

There may by two cases :

(i) If mR << r, then I =r

mE = m × current due to one cell.

So, Parallel combination is advantageous.

(ii) If mR >> r, then I = RE

= current due to one cell.

So, parallel combination is not advantageous.

If emf and internal resistances of each cell are different,then,

Eeq =n21

nn2211

r1r1r1rErErE

/.....///.....//

for two cells E = 21

1221

rrrErE

(Use emf with polarity)

E1

E2

E3

En

r2

r3

rn

r1

(c) Cells in Multiple Arc :

n = number of rowsm = number of cells in each row.

mn = N (total number of identical cells) :The combination of cells is equivalent to single cell of

emf = mE and internal resistance = n

mr

Current I =

nmr

R

mE

For maximum current, nR = mr

or R = n

mr = internal resistance of the equivalent

battery.

Imax

= R2

mEr2

nE .

using mn = N in above equation we get number of

rows n = RNr

9. 9 cells, each having the same emf and 3 ohm internalresistance, are used to draw maximum current throughan external resistance of 3 ohm. find the combinationof cells.

Sol. For the condition of maximum current number of rows

n = RNr

so n = 3

39 = 3

so combination will be like 3 rows and 3 cells in eachrow.

BATTERY

Battery is an arrangement that creates a constantpotential difference between its terminals. It is acombination of a number of cells in series.

The impact of battery :

With the discovery of voltaic cell, it was soon realisedthat if one constructs a number of cells and joins thenegative terminal of one with the positive terminal ofthe other and so on, then the emf (which is the potentialdifference between the electrodes in an open circuit)of the combination of cells will be the sum of the emf�sof the individual cells. This observation led to a burst ofscientific activity in 1802. Humphrey Davy, an Englishchemist, made a battery of 60 pairs of zinc and copper


plates. The large emf thus produced, was used to gethigh current, which could melt iron and platinum wires.By 1807, he had a battery of almost 300 plates withwhich he was able to decompose chemical salts. Thisled to the discovery of new elements.By 1808, Davy had assembled 2,000 pairs of plates.With this battery, he created electric arcs andsucceeded in extracting the elements like barium,calcium and magnesium from their compounds. Thus,electricity took a front seat in exploring the nature ofmatter.

GALVANOMETER

Galvanometer is a simple device, used to detect thecurrent, to find direction of current and also to comparethe currents.With the help of galvanometer we make twoimportant devises known as Ammeter and voltmeteras discussed below.

(a) Ammeter :

Ammeter is an electrical instrument which measuresthe strength of current in �ampere� in a circuit. Ammeter

is a pivoted coil galvanometer which is alwaysconnected in series in circuit so that total current (to bemeasured) may pass through it. For an ammeter ofgood quality, the resistance of its coil should be verylow so that it may measure the strength of currentaccurately (without affecting the current passingthrough the circuit). The resistance of an ideal ammeteris zero (practically it should be minimum). So, tominimize the effective resistance of an ammeter, a lowvalue resistance (shunt) as per requirement isconnected in parallel to the galvanometer to convert itto ammeter of desired range.In electric circuit, the positive terminal of an ammeteris connected to positive plate and negative terminal isconnected to negative plate of battery.

Desired value of shunt depends on the range(measurable maximum current) of ammeter convertedfrom galvanometer.If pivoted galvanometer of resistance G is to measurecurrent i (as an ammeter) then from figure.

Gi ig ig i

is S

ig G = (i � i

g) S or S = )ii(

Gi

g

g

Where ig is an amount of current required for fulldeflection in galvanometer. By using a low value ofresistance S (shunt) in parallel to the galvanometer(resistance G), the effective resistance of

converted ammeter RA = S)(GGS

becomes very low..

NOTE :

Shunt : If anyhow, the flowing current throughgalvanometer becomes more than its capacity, the coilhas possibility of burning due to heat produced byflowing current. Secondly, its pointer may break up dueto impact with �stop pin� as its proportional deflection

as per amount of flowing current.

In order to minimize these possibilities a low resistancewire (or strip) is connected in parallel with galvanometer,which is known as shunt.

(b) Voltmeter :

It is an electrical instrument which measures thepotential difference in �volt� between two points of

electric circuit. It�s construction is similar as that of

ammeter. The only difference between ammeter andvoltmeter is that ammeter has its negligible(approximately zero) resistance so that it may measurecurrent of circuit passing through it more accuratelygiving the deflection accordingly, while the voltmeterpasses negligible current through itself so that potentialdifference developed due to maximum current passingthrough circuit may be measured. Therefore, anappropriate value of high resistance is required to beconnected in series of galvanometer to convert it into avoltmeter of desired range.Voltmeter is connected in parallel to the electric circuit.

If a galvanometer of resistance G is to be convertedinto a voltmeter of range V, then required value of highresistance RH will be

V = ig (R

H + G)

or RH =

g

VI � G

G

i

ig

RH

V

Connecting this value of high resistance in the seriesof galvanometer, it will be converted to a voltmeter ofrange V. After connecting high resistance RH in seriesof galvanometer of resistance G, the effective resistanceof voltmeter becomes RV = (RH + G) very high (high incomparison to G).

Ideal voltmeter has infinite resistance of its own. Whenideal voltmeter is connected parallel to a part of anelectric circuit, it passes zero amount of current throughitself from the circuit so that measurement of potentialdifference across the points of connection may beperfectly accurate.


LIGHT

PROPERTIES OF LIGHT

Human eye can see a wavelength of about 400 nm to

760 nm of the electromagnetic spectrum.

Light is an electromagnetic wave. It consist of varying

electric field and magnetic field.

Light carries energy and momentum.

Speed of light in vacuum is 3 × 108 m/s.

The formula v = f is applicable to light.

1m10 m�3

3600A° 7800A°10 m�9

X-ray U.V. visiblelight

Infrared

Microwave

Radiowave

10 m3

10 m�14

cosmic ray

10 m�12

10 m�11

When light gets reflected in some medium, it suffers

no change in frequency, speed and wavelength.

Frequency of light remains unchanged when it gets

reflected or refracted.

PLANE MIRROR

(a) Image Formed by a Plane Mirror :

The image of an object A is formed at Awith the help of

plane mirror (MM)

M M'

A'

A N C

D

B

i r1

2

3

4

AD = D'A

So the image of an object formed by the plane mirror is

at same distance behind the plane mirror as the object

is in front of it.

(b) Characteristics of Image Formed by a

Plane Mirror :

(i) It is of the same size as that of the object.

(ii) It is at same distance behind the mirror as the object

is in front of it.

(iii) It is laterally inverted.

(iv) It is virtual and erect.

(c) Points to Remember :

(i) Focal length of a plane mirror is infinity.

(ii) Power of a plane mirrors is zero.

(iii) If keeping the incident ray fixed, the mirror is ro-

tated through an angle , about an axis in the plane of

mirror, the reflected ray is rotated through an angle 2

Reflected ray

Incident ray

M

M'

M'

M

I rayncident

Reflected ray

(iv) As every part of a mirror forms a complete image

of an extended object and due to super-position of

images brightness will depend on its light reflecting

area, a large mirror gives more bright image than a

small one. This in turn also implies that if a portion of

a mirror is obstructed, complete image will be formed

but of reduced brightness.

(v) Though every part of a mirror forms a completeimage of an object, we usually see only that part of itfrom which light after reflection from the mirror reachesour eye. That is why :

(A) To see his full image in a plane mirror a personrequires a mirror of at least half of his height.

(B) To see a complete wall behind himself a personrequires a mirror of at least (1/3rd) the height of walland he must be in the middle of wall and mirror.

(vi) Deviation is defined as the angle between direc-tions of incident ray and emergent ray.

= 180 � (i + r) = 180 � 2i

ir

Plane mirror



(vii) If an object moves towards (or away from) a plane

mirror at speed v, the image will also approach (or

recede) at same speed v i.e., the speed of image rela-

tive to object will be 2v.

(viii)In reflection the speed, wavelength and frequency

of light does not change.

But the amplitude or intensity of the reflected ray is

less than that of the incident ray.

(ix) Plane mirrors are used in sextant, Kaleidoscope,

Periscope

(x) If angle between two mirrors is then after two

consecutive reflection

total deviation = 1 + 2 = 2 � 2

(xi) A thick plane mirror forms number of images, due

to multiple reflection of light. Out of these images, sec-

ond image is the brightest and the intensity of other

images goes on decreasing.

ILLUSTRATIONS

1. Light incident normally on a plane mirror retraces back-

ward. The mirror is deflected through an angle 30°.

What is the displacement of the reflected spot of light

on a screen placed 2 m away?

Sol. When mirror turns by = 30° the reflected ray turns by

2 = 60°.

From figure, displacement of the reflected spot,

d = 2.0 tan 2

= 2.0 tan 60° = 2.0 × 3 m

= 2 3 m

d

s 2.0 m M

2. A boy is 1.6 m tall and can see his image in a plane

mirror fixed on a wall. His eyes are 1.4 m from the floor

level. Find the minimum length of the mirror. Will an-

swer depend on the eye level?

Sol. A ray from feet enters the eye after striking the mirror atM2 so height of M2 from ground level is half of the eye

level from ground i.e. 1.42

= 0.7 m

Similarly a ray from the head enters the eye afterstriking the mirror at M1 so position of M1 from the headlevel is half of the distance between head and eyelevel.

Wall

M1

Mirror

PH (Head)

F (Feet)GroundO

M2

Eye level

i.e.1.6 1.4

2

= 0.22

= 0.1 m

So, minimum length of mirror is(1.6 � (0.7 + 0.1)) = 0.8 m. Ans.

Minimum length of mirror is independent of the eyelevel because it is equal to the half the top height of theboy.Note : Positioning of the mirror of minimum length(equal to the half of the total height) will depend uponthe eye level.

(d) Lateral Inversion :

Letter L appears to be inverted or reversed, i.e. there isan interchange of left and right sides of the image andthe object.

Eg. : If a man stands in front of a plane mirror his righthand appears to be the left hand of the image.

(e) Number of Images formed when the object

is placed between Two Plane Mirrors :

When two plane mirrors are placed facing each otherat an angle and an object is placed between them,multiple images are formed as a result of multiplereflections.

If

360º is even then the number of image formed,

n =

360º � 1.


If

360º is odd then :

Case I : If the object lies symmetrically, then

n =

360º � 1.

Case II : If the object lies asymmetrically, then

n =

360º.

Case III : If

360º is equal to fraction then number of

images=[n] i.e. only integer part.

MIRROR FORMULA

v1

u1

f1

RELATION BETWEEN RADIUS OF CURVATURE

AND FOCAL LENGTH OF SPHERICAL MIRRORS

(a) Concave Mirror :

2R

f

(b) Convex Mirror :

2R

f

3. The sun (diameter d) subtends an angle radians atthe pole of a concave mirror of focal length f. What isthe diameter of the image of the sun formed by themirror?

Sol. Since the sun is very distant, u is very large and so(1/u) is practically zero.

So1

0v =

1f

i.e. v = � f

i.e., the image of sun will be formed at the focus andwill be real, inverted and diminished.Now as the rays from the sun subtend an angle radians at the pole, then in accordance with figure.

= A 'B'FP

= df

[where d is the diameter of the image of the sun].i.e., d = f.

� Newton�s formula

If x and y are the distances (along the principal axis) ofthe object and image respectively from the principalfocus, thenxy = f 2, where f is the focal length.

� Paraxial rays :The rays of light which fall near the pole making small

angles with the principal axis of the mirror

� Marginal rays :The rays of light which are parallel and travel far away

from the principal axis of the mirror.

� Spherical Aberration : It is the inability of a spherical

mirror to bring all the rays of a beam of light to its focus

is called spherical aberration. It can be eliminated by

using a paraboloid mirror.

PRINCIPLE OF REVERSIBILITY OF LIGHT

�If the path of a ray of light is reversed after suffering a

number of refraction and reflection, then it retraces its

path� This is known as principle of reversibility of light.

Figure : Reversibility of light

A ray of light (AB) travelling in air medium (1) strikes the

surface of water medium (2) at B and bends towards

the normal NN�. The refracted ray BC strikes a plane

mirror M normally as shown above. After reflection it

will retrace the path i.e. along CBA path.

COMPOUND SLAB

A compound slab is made of two or more media (say

water and glass) bounded by parallel faces and is

placed in air. A compound slab can be made by placing

a glass tray completely filled with water on a glass

slab.


When an incident ray AB travelling in air (medium 1)strikes the water surface (medium 2) at B, it is refracted

along BC. In figure ABN = i (incident angle) and

BC'N = r1 (angle of refraction).

A

C

Figure : Lateral shifting of light in compound slab

Now the ray BC acts as an incident ray for the surfaceseparating glass slab and water. So the incident rayBC after striking this surface at C is refracted along CD

in glass (medium 3). 1BCN = r1, which is equal to

angle of refraction, now acts as angle of incidence.

DCN1 = r

2 = angle of refraction

The ray CD acts as an incident ray for the surfaceseparating glass slab and air. So the incident ray CDafter striking this surface at D is refracted along DE in

air. The rays DE and AB are parallel, so DE'N2

= ABN = i. In this case, 2CDN = r2.

NOTE : Incident ray AB and emergent ray DE will beparallel.

Factors on which lateral shift depends :

(i) Lateral shift varies directly with the thickness of glassslab.

(ii) Lateral shift varies directly with the incident angle.

(iii) Lateral shift varies directly with the refractive indexof glass slab.

(A) Emergent ray is parallel to the incident ray if thereis same medium on both sides light is laterally stifled

by, d = t sin(i r)

cosr

= 1

)ri(t = t×i

ir

1 = t×i

11

(B) Emergent ray is not parallel to incident ray if themedium on both the sides of slab are different.

TOTAL INTERNAL REFLECTION

The phenomenon of reflection when a ray of lighttravelling from a denser to rarer medium is sent backto the same denser medium, provided when it strikesthe interface of the denser and the rarer media at anangle greater than the critical angle, is called totalinternal reflection.

When a ray of light falls on the interface separatingdenser and rarer medium, it is refracted as shown infigure. As the angle of incidence increases, the refractedray bends towards the interface. At a particular angle ofincidence, the, refracted light travels along the interfaceand the angle of refraction becomes 90º. The angle of

incidence for which angle of refraction becomes 900 iscalled critical angle iC.

w

a =

º90sinisin c sin i

c =

wa

1

Figure : Ray diagram showing total internal reflection

When the angle of incidence becomes greater thanthe critical angle, there is no refracted light and all thelight is reflected in the denser medium. Thisphenomenon is known as total internal reflection.

(a) Conditions for total Internal Reflection :

(i) The light should travel from denser to rarer medium.

(ii) The angle of incidence must be greater than thecritical angle for the given pair of media.

IMPORTANT NOTE :

During total internal reflection of light, the whole incidentlight energy is reflected back to the parent opticallydenser medium.

(i) Critical angle of a medium depends upon thewavelength of light.

Critical angle wavelength :

Greater the wavelength, greater will be the criticalangle. Thus, critical angle of a medium will bemaximum for red colour and minimum for violet colour.

(ii) Critical angle depends upon the nature of the pair


of media. Greater the refractive index, lesser will bethe critical angle.

(iii) Image formed due to total internal reflection is muchbrighter because total light is reflected back into thesame medium and there is no loss in intensity of light.

(b) Some Phenomena due to total Internal

Reflection :

(i) Working of Porro Prism :

A right angled isosceles prism called Porro-Prism canbe used in periscope or binocular.

The refractive index of glass is 1.5 and the criticalangle is equal to 41.8º. When the ray of light fallson the face of a right angled prism at angle greaterthan 41.8º, it will suffer total internal reflection.Right angled prisms used to bend the light through90º and 180º are shown in figure (a) and (b) respectively.A right angled prism used to invert the image of anobject without changing its size as shown in figure.

C

P

Q

A

45º

45ºB

P'Q'

(a)

Figure : Working of porro prism

Additional Information :

Mirrors can also be used for bending the rays of light.But the intensity of the beam reflected by mirrors is lowbecause even a highly polished mirror does not reflectthe whole light. On the other hand, in Porro-prism thewhole light is reflected. Therefore, there is no loss inintensity of light and hence image is bright.

(ii) Sparking or brilliance of a diamond

The refractive index of diamond is 2.5 which gives, thecritical angle as 24º. The faces of the diamond are cutin such a way that whenever light falls on any of thefaces, the angle of incidence is greater than the criticalangle i.e. 24º. So when light falls on the diamond, it

suffers repeated total internal reflections. The lightwhich finally emerges out from few places in certaindirections makes the diamond sparkling.(iii) Shining of air bubble in water

The critical angle for water-air interface is 48º 45. Whenlight propagating from water (denser medium) isincident on the surface of air bubble (rarer medium) atan angle greater than 480 45�, the total internal reflectiontakes place. Hence the air bubble in water shinesbrilliantly.

I > Ic

Figure : Shining of air bubble in water

(iv) Mirage :

Mirage is an optical illusion of water observed generally in

deserts when the inverted image of an object

(e.g. a tree) is observed along with the object itself on a hot

day.

Figure : A mirage formation in deserts

Due to the heating of the surface of earth on a hot day,

the density and hence the refractive index of the layers

of air close to the surface of earth becomes less. The

temperature of the atmosphere decreases with height

from the surface of earth, so the value of density and

hence the refractive index of the layers of air at higher

altitude is more. The rays of light from distant objects

(say a tree) reaches the surface of earth with an angle

of incidence greater than the critical angle. Hence the

incident light suffers total internal reflections as shown

in the figure. When an observer sees the object as

well as the image he gets the impression of water

pool near the object.

(A) The mirage formed in hot regions is called inferiormirage.

(B) Superior mirage is formed in cold regions. This

type of mirage is called looming.

(v) Optic pipe and optical fibres

Optical fibre is extremely thin (radius of few microns)

and long strand of very fine quality glass or quartz

coated with a thin layer of material of refractive index

less than the refractive index of the strand.


(If refractive index of the core is say 1.7 then refractive

index of the coating is 1.5). The coating or

surrounding of optical strands is known as cladding.

The sleeve containing a bundle of optical fibres is called

a light pipe.

When light falls at one end of the optical fibre, it gets

total internally refracted into the fibre. The refracted ray

of light falls on the interface separating fibre and coating

at an angle which is greater than the critical angle. The

total internal reflection takes place again and again as

shown in figure below. The light travels the entire length

of the fibre and arrives at the other end of the fibre

without any loss in its intensity even if the fibre is

rounded or curved.

Figure : Structure of optical fibre

Uses of Optical Pipe :

(i) Optical fibres are used to transmit light without any

loss in its intensity over distances of several

kilometer.

(ii) Optical fibres are used in the manufacture of medical

instruments called endoscopies. Light pipe is inserted

into the stomach of the human being. Light is sent

through few optical fibres of the light pipe. The reflected

light from the stomach is taken back through the

remaining optical fibres of the same light pipe. This

helps the doctors to see deeply into the human body.

Hence the doctor can visually examine the stomach

and intestines etc. of a patient.

(iii) They are used in telecommunication for transmitting

signals. A single fibre is able to transmit multiple

signals (say3000) simultaneously without interference,

whereas the electric wire can preferably transmit one

signal at a time.

(iv) Optical fibres are used to transmit the images of

the objects.

(v) Optical fibres are used to transmit electrical signals

from one place to another. The electrical signals are

converted into light by special devices called

transducers, then these light signals are transmitted

through optical fibres to distant places.

REFRACTION AT SPHERICAL SURFACES

Spherical refracting surface is a refracting mediumwhose curved surface is a part of sphere.For paraxial rays incident on a spherical surface sepa-rating two media.

Ruv1212

P C IOu v

where light moves from the medium of R.I. 1 to themedium of R.I 2.

4. There is a small air bubble inside a glass sphere(=1.5) of radius 10 cm. The bubble is 4.0cm belowthe surface and is viewed normally from the outside.Find the apparent depth of the bubble.

c

O

I

A P

Sol. The observer sees the image formed due to refractionat the spherical surface when the light from the bubblegoes from the glass to the air.Here u = �4.0 cm, R= �10 cm, 1 = 1.5 and 2 = 1We have,

Ruv1212

or, cm105.11

cm0.45.1

v1

or, cm0.45.1

cm105.1

v1

or , v = �3.0 cm

Thus, the bubble will apear 3 cm below the surface.


PROOF OF LENS FORMULA

Relation between object distance u, image distance v

and focal length f is : f1

u1

v1

.

NOTE : Lens maker formula :

f1

=

21medium

lens

21 R1

R1

1R1

R1

)1(

(where is absolute refractive index of lens material)

Spherical Aberration in Lenses :

The inability of a lens to focus all the rays of light fallingon it at a single point is known as spherical aberrationin the lens.

NOTE :

1. Intensity or brightness of the image is proportional tothe square of the aperture of the lens, i.e, I A2. that iswhy, the brightness of the image produced by a lenswhich is half painted black reduces to half. However,size of image remains the same because every part ofa lens forms a complete image of an object.

2. If a lens is cut horizontally into two equal halves asshown, then intensity of transmitted light becomes

half and aperture of lens becomes 1

2 of its initial

value.

3. If a lens is cut vertically into two equal halves as shown,then intensity of transmitted light and aperture of thelens remains same.

4. Minimum distance between a real object and realimage formed by a convex lens of focal length f is of.

5. A convex lens of refractive index 2 behaves as aconvex lens in a medium of refractive index 1 (<2)and diverging lens in a medium of refractive index1(>2).

6. A concave lens of refractive index 2 behaves as aconcave lens in a medium of refractive index 1 (<2)and converging lens in a medium of refractive index1(>2).

HUMAN EYE

The human eye is one of the most sensitive senseorgan of sight which enables us to see the wonderfulworld of light and colour around us. It is like a camerahaving a lens system and forming an inverted, realimage on a light sensitive screen inside the eye. Thestructure and working of the eye is as follows :

(a) Structure and Working of Human Eye :

The human eye has the following parts :

(i) Sclera : It is the outer part of eye which protects theeye. It is hard, opaque and white in colour.(ii) Cornea : It is a transparent spherical membranecovering the front of the eye.

(iii) Iris : It is a coloured diaphragm between the corneaand lens.

(iv) Pupil : It is a small hole in the iris.

(v) Eye lens : It is a transparent lens made of jelly likematerial. ( = 1.396)

(vi) Ciliary muscles : These muscles hold the lens inposition.

(vii) Retina : It is a back surface of the eye.

(viii) Blind spot : It is a point at which the optic nerveleaves the eye. An image formed at this point is not sent to the brain.

(ix) Aqueous humor : It is a clear liquid region betweenthe cornea and the lens. ( = 1.336)

(x) Vitreous humor : The space between eye lens andretina is filled with another liquid called vitreoushumor. ( = 1.336)

In the eye, the image is formed on the retina bysuccessive refractions at the cornera, the aqueoushumor, the lens and the vitreous humor. Electricalsignals then travel along the optic nerve to the brain tobe interpreted. In good light, the yellow spot is mostsensitive to detail and the image is automaticallyformed there.

(b) Power of Accommodation :

The images of the objects at different distances fromthe eye are brought to focus on the retina by changingthe focal length of the eye-lens, which is composed offibrous jelly-like material, can be modified to someextent by the ciliary muscles.

(c) Near Point and Far Point :

The nearest point at which a small object can be seendistinctly by the eye is called the near point. For a normaleye, it is about 25 cm and is denoted by the symbol D.


With advancing age, the power of accommodation ofthe eye decreases as the eye lens gradually loses itsflexibility. For most of the old persons aged nearly 60years, the near point is about 200 cm and correctiveglasses are needed to see the nearby objects clearly.

The farthest point upto which our eye can see objectsclearly, without any strain on the eye is called the farpoint. For a person with normal vision, the far point isat infinity.

(d) Least Distance of Distinct Vision :

The minimum distance of an object from the eye atwhich it can be seen most clearly and distinctly withoutany strain on the eye, is called the least distance ofdistinct vision. For a person with normal vision, it isabout 25 cm and is represented by the symbol D, i.e.

Least distance of distinct vision = D = 25 cm.(e) Persistence of Vision :

The image formed on the retina of the eye does notfade away instantaneously, when the object is removedfrom the sight. The impression (or sensation) of theobject remains on the retina for about (1/16)th of asecond, even after the object is removed from the sight.This continuance of the sensation of eye is called thepersistence of vision.

Let a sequence of still pictures is taken by a moviecamera. If the sequence of these still pictures isprojected on a screen at a rate of 24 images or moreper second then the successive impressions of theimages on the screen appear to blend or mergesmoothly into one another. This is because an image(or a scene) on the screen appears just before theimpression of previous image on the retina is lost.Hence, the sequence of images blend into one anothergiving the impression of a moving picture. This principleis used in motion picture projection or in cinematography.

(f) Colour-Blindness :

The retina of our eye has large number of light sensitivecells having shapes of rods and cones. Therod-shaped cells respond to the intensity of light withdifferent degrees of brightness and darkness whereasthe cone shaped cells respond to colours. In dim lightrods are sensitive, but cones are sensitive only in brightlight. The cones are sensitive to red, green and bluecolours of light to different extents.

Due to genetic disorder, some persons do not possesssome cone-shaped cells that respond to certainspecific colours only. Such persons cannot distinguishbetween certain colours but can see well otherwise.Such persons are said to have colour-blindness.Driving licenses are generally not issued to personshaving colour-blindness.

(g) Colour Perception of Animals :

Different animals have different colour perception dueto different structure of rod shaped cells and coneshaped cells. For example, bees have some cone-

shaped cells that are sensitive to ultraviolet. Thereforebees can see objects in ultraviolet light and canperceive colours which we cannot do.

Human beings cannot see in ultraviolet light as theirretina do not have cone-shaped cells that are sensitiveto ultraviolet light.

The retina of chicks have mostly cone shaped cellsand only a few rod shaped cells. As rod shaped cellsare sensitive to bright light only, therefore, chicks wakeup with sunrise and sleep in their resting place by thesunset.

(h) Cataract :

Sometimes due to the formation of a membrane overthe crystalline lens of some people in the old age, theeye lens becomes hazy or even opaque. This is calledcataract. It results in decrease or loss in vision of theeye. Cataract can be corrected by surgery leading tonormal vision.

DEFECTS OF VISION AND THEIR CORRECTION

People with normal vision can focus clearly on verydistant objects. We say their far point is at infinity.

People with normal vision can also focus clearly onnear objects upto a distance of 25 cm. We say theirnear point is at a distance of 25 cm from the eye.

But there are some defects due to eye irregularitieswhich are as follows :

(a) Short Sightedness (or Myopia) :

A person with myopia can see nearby object clearly butcan see distant objects distinctly. A myopic personhas the far point nearer than infinity. In a myopic eye,the image of distant object is form in front of the retinathis defect is arise due to :(i) Excessive curvature of eye lens.(ii) Elongation of eye ball.This defect can be corrected by using a concave lensof suitbale power.


Ray diagram in case of short sightedness

The rays of light from distant object are diverged by the

concave lens so that final image is formed at the retina.

If the object is very far off (i.e. u ~ ), then focal length

of the concave lens is so chosen that virtual image of

the distant object is formed at the far point F of the

short-sighted eye. Therefore rays of light appear to

come from the image at the far point F of the short-

sighted eye and not from the more distant object.

Correction of short sightedness by concave lens

Note that focal length of the lens for a short-sighted

person is equal to the negative value of the person�s

far point.

(b) Far Sightedness (or Hyperopia or

Hypermetropia) :

A person with hypermetropia can see distant obects

clearly but can not see near by objects distinctly.

A hypermetropic person has the near point farther away

from minimum distance of distinct vision (i.e. 25 cm).

In a hypermetropic eye, the image of a nearby object is

formed bhind the retina. This defects arises because:

(i) Increase in focal length of eye lens.

(ii) The eye ball has become to small.

This defect can be corrected by using by convex lens of

suitable power.

The converging lens of correct focal length will causethe virtual image to be formed at the actual near pointof the farsighted person�s eye as shown in figure (c).

25 cm

Short eyeball

(a)

Normalnear point

25 cm

NN'

Near point ofdefective eye

(b)

N

25 cm

NN'

Near point ofdefective eye

(c)Correction of far sightedness by convex lens

(c) Presbyopia :

This defect arises with aging. A person suffering fromthis defect can see neither nearby objects nor distantobjects clearly/distinctly. This is because the power ofaccommodation of the eye decreases due to thegradual weakening of the ciliary muscles anddiminishing flexibility of the eye lens.

This defect can be corrected by using bi-focal lenses.Its lower part consists of a convex lens and is used forreading purposes whereas the upper part consists ofa concave lens and is used for seeing distant objects.(d) Astigmatism :

A person suffering from this defect cannot simultaneouslyfocus on both horizontal and vertical lines of a wire gauze.

Normal Wire Gauge

Normal Wire Gauge Wire Gauge withDistorted Vertical Lines


Normal Wire GaugeDistorted Vertical Lines

Wire Gauge withDistorted Horizontal Lines

This defect arises due to the fact that the cornea is notperfectly spherical and has different curvatures forhorizontally and vertically lying objects. Hence, objectsin one direction are well focused whereas objects inthe perpendicular direction are not well focused. Thisdefect can be corrected by using cylindrical lenses.The cylindrical lenses are designed in such a way soas to compensate for the irregularities in the curvatureof cornea.

REFRACTION THROUGH A PRISM

A homogeneous solid transparent and refractingmedium bounded by two plane surfaces inclined at anangle is called a prism.

3-D view

Refraction through a prism:View from one side

ir1 r2

e

incident rayemergent ray

A

P

Q RBase

N'

N N" NN' & N'N" are normalsX

Y

Z

(a) Characteristics of Prism

(i) PQ and PR are refracting surfaces.

(ii) QPR = A is called refracting angle or the angleof prism (also called Apex angle).

(iii) = angle of deviation

(iv) For refraction of a monochromatic ( single wavelength) ray of light through a prism:

r1 + r

2 = A

= i + e A .......(iv)

(v) Variation of versus i (shown in diagram).

For one (except min) there are two values ofangle of incidence. If i and e are interchanged thenwe get the same value of because of reversibilityprinciple of light

e = imax

(vi) There is one and only one angle of incidencefor which the angle of deviation is minimum.

(vii) When = min

, the angle of minimum deviation,then i = e and r

1 = r

2, the ray passes symmetrically

w.r.t. the refracting surfaces. We can show bysimple calculation that

min

= 2i � A

where imin

= angle of incidence for minimumdeviation, and r = A/2.

rel = 2A2

A

sin

sin m

.......(v)

where rel

= gssurroundin

prism

Also min

= ( 1) A(for small values of A)

(viii) For a thin prism ( A 10o) and for small valueof i, all values of

= ( 1 ) A .......(vi) is different for different coloursso for red colour,

R = (

R 1 ) A

For violet colour, V

= ( V 1 ) A

and for yellow colour Y

= ( Y 1 ) A

ILLUSTRATION

From the figure, find the deviation caused by a prismhaving

refracting angle 4º and refractive index 23

.

Sol. = (23

� 1) × 40= 20

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