Diffraction -Single Slit

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8/7/2019 Diffraction -Single Slit

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b

S

S’

A

B

According to geometrical optics, region AB of the screen SS’ should be illuminated and

remaining portion should be dark

Diffraction

However illumination in the region of geometrical shadow is observed showing bendingof light round the edges as shown above schematically.

S’

S

D K Rai, JIITU, Noida



Diffraction is the apparent `bending' of light waves

around obstacles in its path which can be explained by

Huygen's principle.

It is defined as the phenomenon of bending of light

round the corner of an obstacle and their spreading into

the region of geometrical shadow. The resulting

distribution of light intensity resulting in dark and

bright fringes is called diffraction pattern.




The diffraction phenomenon is usually divided into two

categories:1) Fresnel diffraction

2) Fraunhofer diffraction

1. The source of light and the screen are at finite

distance from the diffracting aperture.

S

S’

Point source




2. The source and the screen are at infinite distance from the

aparture

Point source

f f

f



http://slidepdf.com/reader/full/diffraction-single-slit 6/25D K Rai, JIITU, Noida



ΔA

A1

A2

A1’b

B

B1

B2

θP

f

θ




s

L2

A

B

A1 A1’

L1

P

CO

b

θ

θ

B1

D

Path difference, Δ, between waves from A and B reaching P = BDFrom ΔABD, Δ = bsinθ and thus phase difference will be δ = (2π/λ)bsinθ




Let AB be imagined to be divided into n (very large) number of equal parts, each part being source of secondary wavelets.

The phase difference between any two successive parts of slit AB

would be

φ = (1/n)(2π/λ)bsinθ (a)

Thus interference at P would be due to superposition of n number of waves having same amplitude a (say) and constant phase

difference between successive waves given by equation (a)




Resultant of n simple harmonic waves of equal amplitude

And periods and phases increasing in arithmetic progressions

φ

2φ

3φ

(n-1)φ

O A

B

C

D

E

P

δa

R




Resolving the amplitude parallel and perpendicular to OA

)2....(..................................................

)1sin(...3sin2sinsinsin)1...(..................................................

)1cos(...3cos2coscoscos

ϕ ϕ ϕ ϕ ϕ

ϕ ϕ ϕ ϕ δ

−++++=

−+++++=

naaaaR

naaaaaR

Multiply eqution (1) with 2sinφ/2

2

sin)1cos(2...

2

sin3cos2

2sincos2

2sin2

2sincos2

ϕ ϕ

ϕ ϕ

ϕ ϕ

ϕ ϕ δ

−++

++=

naa

aaR

( ) ( )[ ] BABABA sincos2sinsin =−−+

We know that




⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

−−−

+−++−−

++−−++

=

)2

)1sin[(

)2

)1sin[(...)2

2sin(

)2

2sin()2

sin()2

sin(2

sin2

2sincos2

ϕ ϕ

ϕ ϕ

ϕ ϕ

ϕ ϕ ϕ ϕ ϕ ϕ ϕ

ϕ δ

n

na

R

ϕ ϕ ϕ

δ ⎟⎠

⎞⎜⎝

⎛ −+=

2

1sin

2sin

2sincos2 naaR

( ) ( )

2

12cos

2

12sin2

2sincos2

ϕ ϕ

ϕ ϕ

ϕ δ

−−−+=

nn

aR

( )

( ))3........(2

1

cos

2sin

2sin

cos

2

1cos

2sin2

2sincos2

ϕ

ϕ

ϕ

δ

ϕ ϕ ϕ δ

−=⇒

−=

n

na

R

nnaR




( ) )4....(..........2

1sin

2sin

2sinsin

ϕ ϕ

ϕ

δ −= n

n

aR

Squaring and adding equation (3) and (4)

)5....(..........

2sin

2sin

2

22

2

ϕ

ϕ na

R =

( ))3........(2

1

cos

2sin

2sin

cos

ϕ

ϕ

ϕ

δ

−

=

n

na

R

Similarly



Dividing (4) by (3) we have ( )

( )2

1

2

1tantan

ϕ δ

ϕ δ

−=⇒

−=

n

n

Thus if we have large number of vibration as

( )( )( )⎥

⎦

⎤⎢⎣

⎡−−−

ϕ

ϕ

1cos,

,cos,cos

nωt ...

ωt ωt

Then resultant vibration will be

( )

( ) ]1

2

1cos[

2sin

2sin

]

2

1cos[

2sin

2sin

ϕ ω

ϕ

ϕ

ϕ ω

ϕ

ϕ

−−=

−−=

nt

na

nt

na




Where amplitude R of the resultant wave will be

2sin

2sin

ϕ

ϕ ⎟⎠⎞⎜⎝ ⎛ =

n

aR

θ λ π θ

λ π ϕ sinsin21

22bb

n

nn →=

Therefore,

)sinsin(

sinsin

θ λ

π

θ λ

π

bn

b

aR

⎟⎠

⎞⎜⎝

⎛

=

θ λ

π

β sinb=Let

)sin(

sin

n

aR

β

=




For large value of n sinβ/n ≈ β/n.

na][ sinsin ===⇒ AAnaRβ

β β

β

Thus resultant intensity at P which is proportional to the squareof the resultant amplitude R is given by,

2

2

02

222 sinsin

β

β

β

β I ARI ===




Position of maxima and minima

PRINCIPLE MAXIMAThe resultant amplitude is given by

⎥⎦

⎤⎢⎣

⎡+−+−=

+−+−==

......!7!5!3

1

.....]!7!5!3

[sin

642

753

β β β

β β β β

β β

β

A

AAR

R will be maximum if the negative terms vanish.This is possible when

0

0sin

=⇒

==

θ

λ

θ π β

b

So for maximum value of resultant amplitude (R) = A

And thus Imax = I0 = A2

This is called principal maxima.D K Rai, JIITU, Noida



MINIMA

From equation again, intensity is minimum

when sinβ=0 and β ≠ 0

λ θ

π λ

θ π β

π β

π π π π π β

nb

nb

n

n

±=⇒

±==⇒

=±=⇒

±±±±±=

sin

sin

..1,2,3,....n where,

...,4,3,2, i.e.

sin

β

β AR =

So the condition for minima isλ θ nb ±=sin

Where n=1,2,3…. gives the condition of 1st, 2nd, 3rd, minima. Here n ≠0, because n = 0 for θ = 0 corresponds principle maxima.

so first minima will occur at

and 2nd minima at

⎟⎠

⎞⎜⎝

⎛ ±= −

b

λ θ 1sin

⎟⎠

⎞⎜⎝

⎛ ±=

−

b

λ

θ

2

sin1


Secondary maxima



Secondary maxima

2

222 sin

β

β ARI ==

Apply the method of finding maxima and minima

In order to determine the position of secondary maxima differentiate the

above equation with respect to β

0sincossin2

sin2cossin2

2sincossin2sin

2

2

3

22

4

222

2

22

=⎥

⎦

⎤⎢

⎣

⎡ −=

⎥⎦⎤⎢

⎣⎡ −=

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ =∴

β

β β β

β

β

β β β β β

β

β β β β β

β

β

β β

A

A

AA

d

d

d

dI

0sin

=β

β So either 0

sincos2

=−

β

β β β or

β β

β β β β

tan

0sin-cosor 0sin

===⇒




But sinβ=0 is the condition for principal maxima (corresponding to β = 0)

and for minima (corresponding to β = ± nπ) therefore, the position of

secondary maxima can be obtained by the equation:β = tan β

One of the roots of this equation is β = 0 but it corresponds to the

principal maxima. The other roots can be found by determining the

points of intersections of the curves y = β and y = tan β by graphical

method as shown below:




From the graph, other roots which satisfy the equation are

β = ±3π/2, ± 5π/2, ± 7π/2,……=

(the more exact values are = 1.430π, 2.46π, 3.47π, 4.471π…)

So the direction of secondary maxima are approximately:

( )

( )

( )2

12sin

2

12sin

1,2,3.... nwhere,2

12

λ θ

π

λ

θ π

π β

+±=⇒

+±=⇒

=+±=

nb

nb

n

2)12(π

+± n

( )

( )⎟

⎠

⎞⎜

⎝

⎛ +=⇒

+±=⇒

−

b

n

nb

2

12sin

212sin

1 λ θ

λ θ




INTENSITY CALCULATIONS



INTENSITY CALCULATIONS

2

222 sin

β

β ARI ==

Intensity of principle maxima:

for β = 0 Imax = I0 = A2

Intensity of diffraction pattern is given as

The intensity of secondary maxima:

Substituting β = ±3π/2 (first secondary maxima),

100

1

02

2

2

2

22

2

22

1

22

22

1

2222

9

4

2

3

2

3sin

sin

I I

I

I

I A

AAAI

=⇒=⇒

==

=

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ==

π π

π

β

β

Thus I1 is 4.5% of I0




Similarly, substituting β = ±5π/2 (2nd

secondary maxima)

621

626225

4

)2

5(

2

5sin

sin

0

2

0

2

2

2

2

22

2

22

2

=⇒

=====

I

I

I AAA

AI

π π

π

β

β

Thus I2 is 1.61% of I0.

Intensity of other secondary maxima can also be similarly calculated.

From the above values of intensities it is obvious that intensity of

secondary maxima decreases as the order increases.


Diffraction -Single Slit

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