Images formed by flat mirrors

An image formed by a flat mirrorappears to come from behind themirror. The light rays do not actuallypass through the image. Such animage is called avirtual image.

The position and size of an imageformed by a flat mirror can be easilyfoundby constructinga ray diagramasfoundby constructinga ray diagramasshown on the right. The simple raydiagram reveals that the distance of theimage from the front face of the mirroris equal to that of the object from themirror:

qp =

Sometimes, we write: pq −=

where a negative sign is introduced in the equation to indicate that the object and the image are on opposite sides of the mirror.

Lateral and longitudinal magnifications

Lateral magnificationis defined as the ratio of the image height to theobject height, which is obviously equal to 1 for a flat mirror:

1' ==

hh

M

Longitudinal magnification is defined as theratio of the image length to the object lengthalongthedirectionof thenormalto themirror: Object Image

Mirror

alongthedirectionof thenormalto themirror:

1−=∆∆=

pq

Mn

It is also obvious thatMn has a magnitude of 1.The minus sign is introduced to indicate thefront-back reversal of the image.

p q∆p ∆q

Object Image

h h’

Images formed by concave spherical mirrors

If an object is placed on the principal axisat a point beyond the C from the mirror, its

The line passing through the center ofcurvature C and the vertex V of the mirroris call theprincipal axis or, sometimes, theoptic axis.

image is formed by the reflected raysbetween C and V However, the reflectedrays do not intercept the principal axis atthe same point in general, resulting in theformation of a blurred image. Such aneffect is called spherical aberration.Since the reflected light rays pass throughthe image and diverge from there, we callsuch an imagereal image.

C

O I

The mirror equation for paraxial rays

P QC

R h

δ

θ

φφ

u u’

φ+=θ u φ+θ='u& ⇒ θ=+ 2'uu

δ−=p

hutan δ−=θ

Rh

tanδ−=q

hu'tan

For small u, u’ and θ will also be small. Also δ << p, q, R. Under this condition, we have:

p

qph

u ≈qh

u ≈' Rh≈θ

Hence, Rqp211 =+

Rays from a point on the principal axis making sufficiently small angleswith the principal axis are calledparaxial rays. These rays, afterreflection, intersect at the same point Q on the principal axis, whosedistance from the vertex of the concave mirror can be calculated from theequation just derived.

Lateral & longitudinal magnifications of concave mirror

The lateral magnificationM ofa concave mirror can beobtained from a ray diagram asshown on the right:

pq

hh

M −== '

The minus sign is introduced to indicate that the image is inverted.The minus sign is introduced to indicate that the image is inverted.

The longitudinal magnification Mn is defined as before (i.e. Mn = ∆q/ ∆p = dq/dp). Using the mirror equation for paraxial rays, we have

022 =−−qdq

pdp ⇒ 2

2

2

Mpq

dpdq

Mn

−=−==

Focal point & focal length of concave mirror

When p → ∞, all rays from the object areparallel to one another, and the mirrorequation becomes:

Rq211 =+∞ ⇒

2R

q =

The image distance is hence equal to one-half of the radius ofcurvatureR. We call this image point thefocal point F, and the image distancethefocal length f, where

2R

f =

The mirror equation can therefore be written as:fqp111 =+

Convex spherical mirrors

The ray diagram below shows the reflection of rays bya convex mirror. The image is virtual because thereflected rays appear to originate at the image point,which is behind the mirror. Notice also that the imageis upright and smaller than the object.

Sign conventions for mirrors

The equation derived for concave mirrors can be used forconvex mirrors, with the application of the following signconventions:

Both R andf are positive if center of curvature is in front of mirror(concave mirror).

Both R and f are negativeif center of curvature is in back of mirrorBoth R and f are negativeif center of curvature is in back of mirror(convex mirror).

q is positive if image is in front of mirror (real image).q is negativeif image is in back of mirror (virtual image).

Image is upright if M is positive.Image is inverted if M is negative

p > f q is positive⇒ image is realM is negative⇒ image is inverted

pfq

111 −=

p = f q & M arebothinfinitive

pq

hh

M −== '

For concave mirror(f > 0)

p = f q & M arebothinfinitive

p < f q is negative⇒ image is virtualM is positive⇒ image is upright

q is always negative⇒ image is virtual

M is always positive⇒ image is upright

For convex mirror(f < 0)

Ray diagrams for mirrors

1. Ray 1 is drawn from thetop of the object // to theprincipal axis and isreflected through the focalpointF.

2. Ray 2 is drawn from thetop of the object throughthe focal point and isreflected// to the principalreflected// to the principalaxis.

3. Ray 3 is drawn from thetop of the object throughthe center of curvature andis reflected back on itself.

Two rays are sufficient to locate the image of the object.

Example : What type of mirror is required to form an image, ona wall of 3 m from the mirror, of the filament of a headlightlamp 10 cm in front of the mirror and what is the height of theimage if the height of the object is 5 mm?

p = 10 cm, q = 300 cm, hence

R2

3001

101 =+ ⇒ R= 19.4 cm

⇒A concave mirror is required.

3010

300 −=−=−=p

qm

Therefore h’ = - 5 x 30 = -150 mm.

Example : A wire frame in the form of a small cube 3 cm on aside is placed with its center on the axis of a concave mirror ofradius of curvature 30 cm. The sides of the cube are parallel orperpendicular to the axis. The face toward the mirror is 60 cmto the left of the vertex. Find (a) the position of the image oftheright face of the cube, (b) the lateral magnification, and (c) thelongitudinal magnification.

(a) p = 60 cm, R = 30 cm

3021

601 =+

q⇒ q = 20 cm

60 cm

⇒ h’ = - 3 x 1/3 = - 1 cm

(b)31−=−=

pq

M

The negative value means that the right face is inverted.

(c)9

12 −=−= MM n

The negative value means that the image of the leftface is nearer to the mirror than that of the right face.

⇒ ∆q = ∆p x Mn = - 3 x 1/9 = - 1/3 cm

ImageWire frame object

ac b’

c’b a’

Example : A 2-m tall man stands at 3 m from the vertex of aspherical convex mirror. The focal length of the mirror is 30cm. Find the location of the image and the magnification.

p = 300 cm, f = 30 cm, hence

3011

3001 −=+

q⇒ q = - 27.3 cm

⇒ The image is virtual and behind the mirror.

091.0300

3.27 ==−=pq

M

Therefore the image is upright and has a height of 0.182 m.

Images formed by refraction at spherical surfaces

2211sinsin θ=θ nn (Snell’s law)

For paraxial rays, α is small and so are θ1 and θ2, we have

2211θ⋅=θ⋅ nn

From the ray diagram, we observe that

β+α=θ1γ+θ=β 2&

From the equations above, we obtain

β−=γ+α )(1221

nnnn

Butpd=α≈αtan

⇒ nnnn −=+

Rd=β≈βtan

qd=γ≈γtan

⇒

Rnn

qn

pn 1221 −

=+

Lateral and longitudinal magnifications

ph=θ≈θ 11 tansin

qh'

tansin 22 −=θ≈θ

Hence qhn

phn '

21 −=

h

h’θ1

θ2

C

n1

n2

Hence qp−=

p q

Therefore the lateral magnification is pnqn

hh

M2

1' −==

The longitudinal magnification is 2

1

2 Mn

nM n −=

Sign conventions for refractions at spherical surfaces

The equation derived for the convex spherical surface can beused for the concave spherical surface. One must, however,carefully follow the sign conventions given below:

p is positive if object is in front of surface (real object).p is negativeif object is in back of surface (virtual object).

q is positive if image is in back of surface (real image).q is negativeif image is in front of surface (virtual image).

R is positive if center of curvature is in back of convex surface.R is negativeif center of curvature is in front of concave surface.

Example : One end of a cylindrical glass rod (n = 1.5) is groundto a hemispherical surface of R = 20 mm. Find the imagedistance of a point object on the axis of the rod, 80 mm to the leftof the vertex. If the height of the object is 1 mm, calculate theheight of its image. The rod is in air.

R= 20 mm, p = 80 mm, n1 = 1, n2 = 1.5

2015.15.1

801 −=+

q⇒ q = 120 mm

2080=+

q⇒ q = 120 mm

1805.1

1201

2

1 −=×

×−=−=pn

qnM 11' −=×= Mh mm⇒

The image has the same size as the object but is inverted.

Images formed by flat refracting surfaces

The equation derived for spherical refracting surfaces will become one for the flat refracting surface when R is set to infinity, i.e.

Rnn

qn

pn 1221 −

=+ 021 =+qn

pn

R= ∞

Since q is negative, the image is virtual.

The equation can be written as:

pnn

q1

2−=

Lateral and longitudinal magnifications of flat refractin g surfaces

n1 n2

P Q

h1 θ1

θ2

h2

qp

2211sinsin θ=θ nn Snell’s law

For paraxial rays, the Snell’s law can be written as

2211tantan θ≈θ nn

p2211

qh

nph

n 22

11

−=or 12

1

1

2 =−==pnqn

hh

M⇒ Lateral magnification

Longitudinal magnification1

2

nn

dpdq

Mn

−==

Thin lensesConsider a biconvex lens asshown in the figure.I1 isthe virtual image of objectO formed by surface 1 andI2 is the image ofI1 formedby the surface 2.

111

11R

nqn

p−=+111 Rqp

222

11R

nqp

n −=+But

12 qtp −= (note that q1 is –ve)

From these equations, we get (for negligible t )

−−=+2121

11)1(

11RR

nqp

−−=+21

11)1(

11RR

nqp

or

Focal length of thin lens

The focal lengthf of a thin lens may be defined as the imagedistance of a point object on the lens axis at infinite distancefrom the lens.

f ff f

−−=+21

11)1(

11RR

nqp

Substituting p =∞ in the thin-lens equation

we get

−−=21

11)1(

1RR

nf fqp

111 =+&

Converging lenses & diverging lenses

Although the derivation of the lens equation was based on aconverging lens model, the equation is also applicable for diverginglenses.

The focal length of a converging lens is positive, while thatof adiverging lens is negative.

One can also tell whether a lens is converging or diverging from itsshape. A converginglens is thicker at the centerthan at the edge,shape. A converginglens is thicker at the centerthan at the edge,whereas a diverging lens is thinner at the center than at the edge.

Converging lenses Diverging lenses

Sign conventions for thin lenses

p is positive if object is in front of lens (real object).p is negativeif object is in back of lens (virtual object).

q is positive if image is in back of lens (real image).q is negativeif image is in front of lens (virtual image).q is negativeif image is in front of lens (virtual image).

R1 & R2 are positive if center of curvature is in back of lens.R1 & R2 are negativeif center of curvature is in front of lens.

Ray diagrams of thin converging lenses

o Ray 1 is drawn parallelto the principal axis.After being refracted bythe lens, this ray passesthrough the focal pointon the back side of thelens.

o Ray 2 is drawn throughthe center of the lensand continues in astraight line.

o Ray 3 is drawn through that focal point on the front side of the lens (oras if coming from the focal point if p < f) and emerges from the lensparallel to the principal axis

Ray diagrams of thin diverging lenses

o Ray 1 is drawn parallel to the principal axis. After being refracted bythe lens, this ray emerges such that it appears to have passedthroughthe focal point on the front side of the lens (This apparent directionis indicated by the dashed line in the figure).

o Ray 2 is drawn through the center of the lens and continues inastraight line.

o Ray 3 is drawn toward the focal point on the back side of the lensand emerges from the lens parallel to the principal axis.

Lateral & longitudinal magnifications of thin lenses

From the ray diagrams, it is easy to see that the lateral magnification issimply

pq

hh

M −== '

For real images(q is positive)M is negative,which implies that theFor real images(q is positive)M is negative,which implies that theimage is inverted. For virtual images (q is negative)M is positive,which implies that the image is upright.

Taking the derivative of the lens equation, we have 022 =−−qdq

pdp

Hence the longitudinal magnification is 22

2

Mpq

dpdq

Mn

−=−==

Example : A converging lens of diameter 5 cm and focal length3 cm is placed at a spot under the Sun where the intensity ofsunlight is 2 kW/m2. (a) What is the power produced by theSun at its image? (b) What is the light intensity at the image?(c)What would be the required lens specifications for the lightintensity to be at least double of value obtained in (b)? Assumethat the disk of the Sun subtends an angle of 0.533o at theEarth.

a) The power produced by the Sun at it image is simply theproduct of the intensity of the Sun at the spot and the area ofthe lens:

P = 2 x ππππ (0.05/2)2 = 0.003927 kW

b) The intensity at the image is equal to the power divided by thearea of the image.

I = P

A

Since the Sun is at infinity, theimage is at the focal point and itsarea is:

0.533o

f

000612.02180

533.02

=

××

π×π= fA cm2

0.003927P64167

0001.0000612.00.003927 =

×==

A

PI kW/m2

c. To double the intensity, we may choose a lens withd = 5 x 1.414cm and f = 3 cm, or one withd = 5 cm and f = 3/1.414 cm.Generally, we need a lens whosed / f = (1. 414 x 5/3) .

Systems of thin lenses

If a system of lenses consists of several thin lenses, the imageformed by the system can be obtained by the following procedure.First, the image formed by the 1st lens is located as if the otherlenses were not present. The image formed by the 1st lens is thentaken to be the object for finding the image formed by the 2nd lensas if the rest of the lenses were not there. The same procedureisrepeatedfor all thelenses,andtheimageformedby thelast lensisrepeatedfor all thelenses,andtheimageformedby thelast lensisthe image of the system.

O

I’

I

A system of two thin lenses

O

I´I

d

pq

q’

f1 f2

p q’

1

1

'

11

fqp=+

2

11

'

1

fqqd=+

−

2

1

1

111

fqfp

pfd

=+

−−⇒

For d = 0, we have 21

1111

ffqp+=+

Lens aberrations

• The simple lens equation we have derived relating the objectdistance, the image distance and the focal length is based ontheparaxial approximation and negligible lens thickness. Whentheseconditions are satisfied, all rays from a point source focusat a singlepoint, producing a sharp image. If they do not hold, blur images areformed.

• The focal length of a lens depends upon its index of refraction,which varies with wavelength. Therefore if the light from an objectwhich varies with wavelength. Therefore if the light from an objectis not monochromatic (i.e. not having one color or a singlewavelength), a lens forms a number of colored images, which lie indifferent positions and are of different sizes even for paraxial rays.

• The departures of an actual image from the prediction of the simpleequation derived earlier are calledaberrations. Those caused by thevariation of the refraction index with wavelength are referred to aschromatic aberrations. The others, which would arise even if thelight were monochromatic, are referred to as monochromaticaberrations.

Chromatic aberrations

You may recall this so calledlens makers’ formula

−−=21

11)1(

1RR

nf

and the fact that the index ofrefraction generally decreasesrefraction generally decreaseswith increasing wavelength.

It is clearly that the focal length is generally decreases with decreasing wavelength.

Monochromatic aberrations

There are five different monochromatic aberrationsand they are all related to the failure of the paraxialapproximation for lenses of finite aperture(diameter).

o Spherical aberrationo Spherical aberrationo Comao Astigmatismo Curvature of fieldo distortion

Spherical aberration

This is caused by the failure of rays from a pointobject on the principal axis to converge to a pointimage. Rays passing through the lens close to itscenter are focused farther away than rays passingthrough a circular zone near its rim.

Circle of least confusion

Coma

This is the blurring of an image of a point objectoff the principal axis into a comet shape. Raysfrom the off-axis point object passing throughdifferent circular zones of the lens form circularimages of different sizes. These circular imagesarealsodisplacedwith respectto eachother.arealsodisplacedwith respectto eachother.

Astigmatism

Astigmatism results from the failure of a singlecircular zone of a lens to focus the rays from anoff-axis point object at a single point.

Circle of least confusion

Curvature of field

Curvature of field refers to the effect where thefocal surface is curved.

Distortion

Distortion refers to deformation of an image.There are two kinds of distortion:barrel distortion,in which magnification decreases with distancefrom the axis, andpincushion distortion, in whichmagnification increases with distance from theaxis.

Barrel distortion Pincushion distortion

Example : An object 2 cm high is placed at 40 cm to the left of aconverging lens with a focal length of 30 cm. A diverging lenswith a focal length of –20 cm is placed 110 cm to the right of theconverging lens. Determine the final position and lateralmagnification of the final image. Is the final image real orvirtual? Is the final image upright or inverted?

Lens 1: p1 = 40 cm, f1 = 30 cm

3011

401 =+

q⇒ 120

1+=q cm 3

40120

1

11

−=−=−=pq

M3040

1q 120

1+=q 40

11 p

Lens 2: p2 = 110 – 120 = -10 cm, f2 = -20 cm

2011

101

2−=+− q ⇒ 20

2+=q cm 2

1020

2

22 +=−−=−=

pq

M

621

−== MMMfinal

Hence, the final image is real and inverted.

film

Aperture

Lens

Camera

LENS – converging; itsdistance from the filmcan be adjusted by meansof a mechanism forfocusing purpose.

APERTURE – a diaphragm of adjustable

Shutter

APERTURE – a diaphragm of adjustablediameter; the intensity of light reaching the film isproportional to the square of its diameter; when asmall aperture diameter is used, only light fromthe central portion of the lens reaches the film andspherical aberration is reduced.

SHUTTER – a mechanical device that is opened for selected timeintervals called exposure time. Typical shutter speeds are1/30 s. 1/60s, 1/125 s and 1/250 s.

The f-number

The intensityI of the light reaching the film is proportionalto the square of the diameterD of the aperture and inverselyproportional to the area of the image. The area of the imageis proportional toq2, andq2 ≈ f 2 becausep >> f. Hence

2 1DI =∝ ( )22 /

1

DffD

I =∝

f /D is called the f-number. Typical f-numbers marked ona camera for aperture settings aref/2.8, f/4, f/5.6, f/8, f/11,and f/16. Note that successivef-numbers are different by afactor of√2. Hence, changing thef-number from a value tothe next larger one will decreaseI by a factor or 2.

The eye

Vitreoushumor

PupilIris

Crystallinelens

Ciliarymuscle

Crystalline lens – a capsule containing afibrous jelly, hard at the center andprogressively softer at the outer portions; itis held in placed by ligaments that attach itto theciliary muscle.

Aqueoushumor

Aqueous humor – a clear liquid betweenthe cornea and the crystalline lens.

Cornea – a tough, transparent membranecovering the front part of the eye.

Cornea

Retina

Opticalnerve

Vitreous humor – a thin jelly, consisting of largely water, that fill the eyeball behindthe crystalline lens.

Iris – a muscular diaphragm front of the crystalline lens; it regulates the amount oflight entering the eye by dilating the pupil (an opening at the center of the iris) in low-light conditions and contracting it in high-light conditions.

to theciliary muscle. nerve

Retina – a film of nerve fibers; it consists of millions of sensitive receptors called rodsand cones; when stimulated by light, these receptors send impulses via the opticalnerve to the brain.

Accommodation

This is the process by which the eye focuses on an objectby varying the shape of the crystalline lens. The normaleye can focus sharply on an object at any distance frominfinity up to about 25 cm in front of the eye. This ismade possible by the action of the crystalline lens andtheciliary muscleto which it is attached. Whenrelaxed,theciliary muscleto which it is attached. Whenrelaxed,the normal eye is focused on objects at infinity, i.e., thesecond focal point is at the retina. When it is desired toview an object nearer than infinity, the ciliary muscletenses and this action makes the crystalline lens to bulgea bit, and its focal length decreases, resulting in theimage being focused on the retina.

The extremes of the range over which distinct vision ispossible are known as thefar point and thenear point ofthe eye.

Far point - It is infinity for normal eye.

Near point - The position of the near point evidentlydepends on the extent to which the curvature of thecrystalline lens may be increasedin accommodation.crystalline lens may be increasedin accommodation.This gradually diminishes with age as the crystallinelens loses its flexibility. For this reason the near pointgradually recedes as one grows older. This recession ofnear point with age is calledpresbyopia, and is notconsidered a defect of vision, since it proceeds at aboutthe same rate in all normal eyes. It can be correctedwith converging lenses.

Myopia

Myopia is a common defect of vision and is commonly known asnearsightedness. In the myopic eye, the eyeball is too long from front toback in comparison with the focal length of the lens-cornea system, andrays from an object at infinity are focused in front of the retina. Hencethe far point of a myopic eye is near than infinity. When it looks at anobject that lie beyond the eye’s far point, the image is formed in front ofthe retina, resulting in blurred vision. Myopia can be corrected with adiverging lens.

HyperopiaHyperopia is a common defect of vision and is commonly known asfarsightedness. In the hyperopic eye, the eyeball is too short from frontto back in comparison with the focal length of the lens-cornea system,and rays from an object at infinity would be focused behind the retina.The hyperopic eye may focus by accommodation. This works fordistant objects, but the light from nearby objects cannot bebrought to asharp focus at the retina because the near point is much farther away.The refracting power of the lens-cornea system is insufficient to focusthe light from all but distantobjectssatisfactorily. Hyperopiacan bethe light from all but distantobjectssatisfactorily. Hyperopiacan becorrected with a converging lens.

Astigmatism

Astigmatism refers to the defect, in which the surface of thecornea is not spherical, but is more sharply curved in oneplane than another. It makes it impossible, for example, tofocus clearly on the horizontal and vertical bars of awindowat thesametime.windowat thesametime.

Astigmatism can be corrected with lenses that have differentcurvatures in two mutually perpendicular directions.

Diopter

Optometrists and ophthalmologists usually prescribe lensesmeasured indiopters.

The powerP of a lens in diopters equals the inverse of thefocal length in meters.

fP

1=

For example, a converging lens of focal length +20 cmhas a power of +5.0 diopters, and a diverging lens of focallength – 40 cm has a power of –2.5 diopters.

Example : The near point of a certain eye is 100 cm in front ofthe eye. What lens should be used to see clearly an object 25 cmin front of the eye?

The function of this lens is to “move” the object from 25 cm to 100 cm so that the light from the object can be focused on the retina.

p = +25 cm and q = -100cmHence, we have

f

1

100

1

25

1 =−

+ ⇒ f = 33.3 cm

That is, a converging lens of focal length 33.3 cm is required

Example : The far point of a certain eye is 1 m in front of the eye.What lens should be used to see clearly an object a infinity?

The purpose of this lens is to “move” the object from infinity to 1 m so that the light from the object can be focused on the retina.

p = +∞cm and q = -100 cmHence, we have

f

1

100

11 =−

+∞ ⇒ f = -100 cm

That is, a diverging lens of focal length -100 cm is required

The simple magnifier

The apparent size of an object is determined by the size of itsretinal image, which in turn depends upon the angleθ0

subtended by the object at the eye. However, an eye cannotfocus on an object closer than its near point (the near point fora normal eye is ~ 25 cm). Thereforeθ0 is maximum at thenear point.

To further increase theTo further increase theapparent angular size of anobject, a converging lens canbe used with the objectplaced at the focal point orcloser to form a virtual imageat a distance between the nearpoint (~25 cm) and the farpoint (infinity).

h’

θF

h

hh θ0' θ0

Angular magnification of the simple magnifierThe angular magnificationm of a magnifier is defined as the ratio of theangle subtended by an object with a lens in used to the angle subtended y theobject placed at the near point with no lens in used.

ph

phm

25

25/

/tantan

00

==θθ≈

θθ=

If the object is placed at the focal point of the magnifier, the image isformed at infinity, and we have the minimum magnification:

The object position that gives a sharp virtual image that subtends the largest angle (maximum angular magnification) can b found by the equation:

fp

1

25

11 =−

+f

fp

+=

25

25⇒

Hence, the maximum angular magnification isf

m25

1max +=

fm

25min = ( f in units of cm)

The compound microscope

From the equation derived for the angularmagnificationm of the simple magnifier,it appears thatm may be made as large asdesired by decreasing the focal length.However, the aberrations of a simpledouble convex lens set a limit tom ofabout 2x to 4x. Greater magnificationcanbeachievedby combiningtwo lenses

Eyepiece

I1canbeachievedby combiningtwo lensesin a device calledcompound microscopeof merelymicroscopy.

FE

Objectivelens

Object

I2

F0

A microscopy consists of anobjective lens(which is used to form a real and enlargedimage of the object) and aneyepiece lens(the function of which is to produce avirtual and magnified final image fromreal image created by the objective lens).

Angular magnification of the compound microscope

The overall angular magnification of a compound microscopeis theproduct of the lateral magnification of the objective lens and the angularmagnification of the eyepiece lens:

eomMM =

Since the image formed by the objective lens is real and inverted, Mo isnegative and has a magnitude equal to the ratio of the image distance to theobjectivedistance,(i.e. - q /p ). Theobjectis usuallyplacedvery closed

The eyepiece lens is a simple magnifier and its angular magnification is 25/fe.

Therefore, the overall angular magnification M iseo

o

ffq

M25

−=

objectivedistance,(i.e. - qo/po). Theobjectis usuallyplacedvery closedto the first focal point of the objective lens. Thereforepo ≈ fo, wherefo isthe focal length of the objective lens. Hence

o

oo f

qM −=

The refracting telescope

The optical system of a refractingtelescope is essentially the same asthat of a compound microscope,namely it consists of an objectivelens and an eyepiece lens. The twolenses are arranged so that the imageof the objective forms a real andinverted image very close to theinverted image very close to thefocal point of the eyepiece. Theeyepiece then forms an enlargedvirtual image from the this realimage formed by the objective.

Since the object is essentially at infinity, its image formed by the objective is atthe objective’s focal point. Hence the distance between thetwo lenses is equalto sum of the focal lengthfo of the objective and that of the eyepiecefe.

Angular magnification of the refracting telescope

The angular magnification of the refracting telescope is given by:

0θθ=m

Whereθ0 is the angle subtended by the object at the objective andθ is theangle subtended by the final image at the viewer’s eye. From the diagramshownin theearlierslide,it canbenotedthatshownin theearlierslide,it canbenotedthat

ofh'

tan00

=θ≈θ

Therefore, the angular magnification of the refracting telescope is

e

o

ff

m −=

efh'

tan =θ≈θand

The reflecting telescope

The figure to the right showsa design of the reflectingtelescope. The light raysfrom a distant object arereflected by a parabolicmirror and focused, by meansof a small flat mirror, at apoint near the focal point ofthe eyepiece.

Since no objective lens isused in such a design,problems associated withchromatic aberration arevirtually eliminated.