EP225 Note No. 1Geometrical Optics

1.1 IntroductionIn geometrical optics, the wave nature of light waves is largely ignored. The merit of

such simpli�cation is that light re�ection by mirrors and light refraction by lenses can beanalyzed by following light beams (rays). At a boundary between two dielectric media, suchas air/glass boundary, light is re�ected and refracted, but both re�ected and refracted wavesare assumed to travel along straight lines. In practice, di¤raction makes any light beamsspatially dispersed. However, for su¢ ciently short propagation distances, di¤raction e¤ectcan be ignored.

1.2 Light VelocityAs we will study later, light is an electromagnetic wave that can propagate even in

vacuum. Electromagnetic waves can be accommodated in a medium capable of storing bothelectric and magnetic energy. Its velocity in vacuum is given by

c =1

p"0�0

' 3:0� 108 m s�1 (1)

where "0 = 8:85�10�12 F m�1 is the vacuum electric permittivity and �0 = 4��10�7 H m�1

is the vacuum magnetic permeability. In air, the relative permittivity is very close to unityand the light velocity above may be used without too much of error. The electric energydensity associated with an electric �eld E (V m�1) in vacuum is

1

2"0E

2 (J m�3)

and magnetic energy associated with the magnetic �eld H (A m�1) is

1

2�0H

2 (J m�3)

It is not surprising that the light velocity is determined by the basic electromagnetic constants"0 and �0: In a material medium with a permittivity "; the velocity is accordingly modi�edas

c0 =1

p"�0

(2)

In normal insulating matter, c0 < c holds and the ratio

n = c=c0 (> 1) (3)

is de�ned as the index of refraction. Water has n ' 1:3 and glass has n = 1:3�1:6 dependingon materials contained in glass and the wavelength of light.The velocity of light is large but �nite and attempts to measure it were made as early as

in the 17th century. Huygens, using the astronomical observation made earlier by Roemer(Danish astronomer), deduced c ' 2:3� 108 m s�1: Roemer observed and recorded apparentchange in the rotation frequency of a moon Io about Jupiter. (This is a Doppler e¤ect.)Fairly accurate value of c ' 3� 108 m s�1 was found by Bradley using the aberration e¤ect,

1

namely, tilting of the angular location of distant stars due to the motion of the Earth in thedirection perpendicular to the line of sight. Laboratory measurements of light velocity was�rst achieved by Fizeau using a rotating toothed wheel.

Quiz 1: Bradley observed tilting of light from a distant star at �� = 20" (1" = 2�=(360 �60 � 60) = 4:85 � 10�6 rad) when the Earth motion was perpendicular to the direction oflight propagation. The Earth velocity is approximately v = 3�104 m s�1: (This follows fromv = R! where R is the sun-earth distance R = 1:49� 108 km and ! is the angular velocity! = 2�=(365� 24� 3600) rad s�1:) Find the light velocity c using the aberration formula

�� ' v

c� 1:

Quiz 2: What is the relative permittivity of diamond if its index of refraction is 2.4?

Quiz 3: What is the apparent depth of a pool 2 m deep? The index of refraction of water isn = 1:3:

1.3 Re�ection and Refraction of Light Beam

Fig. 1.1. Re�ection and refraction at a dielectric boundary.

Light beam incident on a dielectric surface is subject to both re�ection and refraction.The re�ection angle is unchanged from the incident angle �1: (This follows from momentumconservation of light waves.) The refraction angle �2 obeys the Snell�s law,

sin �2sin �1

=c2c1=n1n2

(4)

where c1 is the speed of light in medium 1 and c2 is that in medium 2. This can be seenfrom the change in the wavelengths,

from �1 =c1�to �2 =

c2�;

and continuity of the component �= sin � (the horizontal thick line) in each medium asillustrated in Fig. 1.2.

2

Fig. 1.2. Derivation of Snell�s law. Note that �1= sin �1 = �2= sin �2:

The index of refraction is not constant but depends on the wavelength. In glass andplastics, it is larger at blue color (shorter wavelength) and decreases by a few % at red colorover the visible spectrum region � = 400 nm (violet) to 700 nm (red). It may sound ratherweak dependence, but is large enough to cause the so-called chromatic aberration of lenses(dependence of the focal length on colors which causes blur in images). Color splitting ina prism, shown in Fig. 1.3, is caused by the same dispersive nature of light waves in glass.An achromatic lens consists of several lenses made of di¤erent glass materials to correctchromatic aberration as we will study later.

Fig. 1.3. Splitting of white light (containing all wavelengths) into color spectrum by aprism. Rainbow is caused by the same mechanism. Note that violet light is refracted morethan red light. (Color spectrum on CD or DVD is caused by a di¤erent mechanism:

di¤raction.)

Both re�ection and refraction are consistent with a more general physics principle ofminimum action, namely, the light ray follows a path of minimum transit time. Natureknows how to be most economical.

3

1.4 Total Re�ection

Fig. 1.4. Total re�ection occurs when n1 > n2 and incident angle �1 exceeds the criticalangle.

Total re�ection of light occurs when the index of refraction n1 is larger than n2; e.g.,when light emerges from water to air. The maximum refraction angle in this case is foundfrom

sin �2 =n1n2sin �1 = 1 or �2 = 90�

or the minimum incident angle issin �1 �

n2n1

For the incident angle �1 larger than the critical angle,

�c = arcsin

�n2n1

�; n1 > n2 (5)

the refraction angle becomes complex (unphysical) and light is totally re�ected. (Light wavesin the medium 2 becomes evanescent.) Propagation of light wave in optical �bers dependson this principle. Re�ection is literally total without approximation in contrast to re�ectionat a mirror surface which depends on the conductivity of coated metal (aluminum, silver)but can never be total.Example 1: In the geometry shown, �nd the range of the incident angle � for which totalre�ection of light occurs at the vertical surface.

4

Solution: Let the refracted angle at the horizontal surface be �2: From Snell�s law,

sin �2 =1

1:3sin �1

The incident angle at the vertical surface is

�3 =�

2� �2

The condition for total re�ection is

sin �3 >1

1:3; �3 > 50:3

�

Therefore,�2 < 39:7

�

andsin �1 < 1:3� sin(39:7�) = 0:83

or�1 < 56:1

�

Example 2: Find the region to be covered on the surface of a glass cube of side 5 cm sothat a spot at the center cannot be seen. Assume n = 1:5 for the glass cube.Solution: The critical angle for total re�ection is

sin �c =1

1:5

Therefore, a circular patch of radius

r = 2:5 cm � sin �c= 1:67 cm

centered on each surface will make the spot invisible.

5

Example 3: A point light source is at a depth h from the surface of water. Show that thefraction of light power emerging out of water is independent of the depth h and given by

1

2

1�

r1� 1

n2

!where n = 1:3 is the index of refraction of water.

Solution: Light within the critical angle � < �c will escape into air. The solid angle in theregion � < �c is

= 2�

Z �c

0

sin �d�

= 2�(1� cos �c)= 2�

�1�

p1� sin2 �c

�= 2�

1�

r1� 1

n2

!The total solid angle about the light source is 4�: Therefore, the fraction of light powerescaping into air is

1

4�2�

1�

r1� 1

n2

!

=1

2

1�

r1� 1

n2

!= 0:18

Quiz 4: On hot summer day, the highway surface becomes a mirror with total re�ection forlight travelling nearly parallel to the surface. Explain in terms of the index of refraction ofair which increases with the air density.

PrismFor a given apex angle of a prism, the deviation angle between the incident ray and �nal

refracted ray can be uniquely determined. Consider a triangular prism with apex angle �: Aray is incident at an angle i as shown. The refraction angle r can be found from the Snell�slaw,

sin r =1

nsin i

or

r = sin�1�1

nsin i

�The ray is incident on the second surface at angle r0: Since

r + r0 = �

we �nd

r0 = �� r = �� sin�1�1

nsin i

�6

Applying Snell�s law again for the second refraction

sin i0 = n sin r0 = n sin

��� sin�1

�1

nsin i

��we �nally �nd

� = i+ i0 � �

= i� �+ sin�1�n sin

��� sin�1

�1

nsin i

���(6)

The minimum deviation angle �min occurs when d�=di = 0 which yields

sin i = n sin��2

�= n sin r

In this case, the following relationship holds

i = i0; r = r0 = �=2

and

n sin��2

�= sin

��+ �min

2

�(7)

The index of refraction of the prism can be determined by measuring the minimum deviationangle �min:

ir

i'r'

α

α

δ

Fig. 1.4. Double refraction in a triangular prism.

α

δmin

7

Fig. 1.5. Minimum deviation angle �min occurs when the refraction geometry is symmetricabout the plane �=2: In this case,

n sin��2

�= sin

��+ �min

2

�holds.

1.6 Spherical Mirrors

Fig. 1.6. Re�ection at a spherical surface.

Mirrors exploit re�ection of light at spherical surfaces. Incident and re�ection angles arewith respect to the curvature radius R as shown in Fig. 1.5. In the �gure, the angles �; �and are related through

�+ 2� = �; �+ � =

Eliminating �; we obtain�+ � = 2

If the aperture of the mirror h is much smaller than the curvature radius R; all angles aresmall and we may approximate them by

� ' tan� = h

o; � ' tan � = h

i; ' tan = h

R

where o is the object distance from the mirror and i is the location of the image. Cancellingthe common factor h;

1

o+1

i=2

R(8)

For a given object location o; the location of its image i can be found from this formula.Note that in the course of derivation, we have assumed a small mirror aperture, h� R:For an object far away from the mirror, o!1; the image is formed at

i =R

2

Light rays parallel to the axis are focused at a distance R=2 from the mirror. This point iscalled the focal point and

f =R

2(9)

is de�ned as the focal length. The sign convention for o; i; and f is as follows:

8

o; i > 0 real object and image in front of a mirroro; i < 0 virtual object and image behind a mirrorR; f > 0 concave mirrorR; f < 0 convex mirror

A negative object distance (virtual object) occurs when the light ray hits the axis behindthe mirror as shown.

Fig. 1.7. Virtual object distance o < 0 occurs when the incidient ray hits the axis behindthe mirror.

The lateral (or angular) magni�cation m is de�ned by

m = � io: (10)

A positive magni�cation means that the image is erect relative to the object, and negativem means that the image is inverted. The axial or longitudinal magni�cation can be foundby di¤erentiating the mirror formula,

�doo2� dii2= 0;

di

do= �

�i

o

�2= �m2

Example 4: Determine the image distance and magni�cation when (a) o = 15 cm, f = 10cm; (b) o = �15 cm, f = 10 cm, (c) o = 20 cm, f = �15 cm. Draw rays for each case.Solution:

(a) From1

15+1

i=1

10

we �ndi = 30 cm

The magni�cation is

m = � io= �2

The image is twice the object size and inverted.

9

(b) From

� 115+1

i=1

10

i = 6 cm

The magni�cation is

m = � 6

�15 = +0:4

(c)1

20+1

i= � 1

15

yieldsi = �8:57 cmm = +0:429

Quiz 5: Do ray tracing for paraxial beams falling on a hemispherical mirror. Observe thatthe rays near the edge are not focused.1.6 Spherical Aberration, Parabolic MirrorsIn spherical mirrors, light rays near the mirror edge are not focused as shown in Fig.

1.8 (top). This causes blurring of image due to spherical aberration. Parabolic mirror(lower �gure) is not subject to this defect and in astronomical re�ective telescopes, parabolicmirrors are used. In astronomical telescopes and microwave antennas, the image distance isat in�nity and the concept of �xed focal point works well.

10

R

Fig. 1.8. Re�ection of paraxial rays at a spherical (top) and parabolic (bottom) mirrorsurfaces.

Spherical aberration in a spherical mirror can be corrected to some extent by placing ane¤ective diverging lens. (It is not a simple diverging lens, however.) This invention wasmade by Schmidt. Note that in spherical aberration, light rays near the edge hit the axis attoo short distances form the mirror surface exactly as in myopia (shortsightedness in humaneyes) and use of e¤ective diverging lens to correct spherical aberration is understandable.The correction lens requires sophisticated calculations and manufacturing procedure.

11

1. 7 Refraction at Spherical Surface, Lens Formula

Fig. 1.9. Refraction at a spherical dielectric boundary.

As in re�ection at a spherical surface, incident and refraction angles are with respect tothe curvature radius. Since

�+ = �i; � + �r =

if the height h is small (small aperture h and small angles), Snell�s law gives

sin �isin �r

' �i�r= n

or�+ n� = (n� 1)

Approximating � ' tan� = h=o; � ' tan � = h=i; ' tan = h=R; we obtain

1

o+n

i= (n� 1) 1

R(11)

The sign convention is

o > 0 object in front of the surfaceo < 0 object behind the surfacei > 0 image behind the surfacei < 0 image in front of the surfaceR > 0 convex surface toward rayR < 0 concave surface toward ray

and the magni�cation is de�ned as

m = � io

(12)

:

12

Fig. 1.10. Refraction by a lens.

For a lens, refraction at both surfaces should be considered. Refraction at the �rst surfacewith curvature radius R1 obeys

1

o+n

i0= (n� 1) 1

R1

where for the case shown in Fig. 1.10, the image is virtual with negative image distancei0 < 0. Refraction at the second surface with curvature radius R2 can be found by invertingthe direction of light beam. i in Fig. 1.9 can be regarded as object distance outside the glassand i0 + l > 0 as the image distance in the glass,

1

i+

n

ji0j+ l = (n� 1)1

R2

where l is the thickness of the lens. For a thin lens with l � R1;2; l can be ignored, andaddition of the two equations reduces to

1

o+1

i= (n� 1)

�1

R1+1

R2

�=1

f(13)

where f is the focal length of the lens. f > 0 is for converging lenses and f < 0 is fordiverging lenses. If the sign convention for the curvature radius is adopted,

1

o+1

i= (n� 1)

�1

R1� 1

R2

�=1

f(14)

In Fig. 1.10, R1 > 0 and R2 < 0; and thus f > 0 (converging lens).Example 5: Find the location and nature of the image of a �sh at the center of a sphericalbowl �lled with water. The index of refraction of water is 4/3 and the bowl has a radius R.Solution: In the refraction formula, the object (�sh) is in water. Then

n

R+1

i= (n� 1) 1

R

If n = 4=3; we havei = �R

which means that the image is formed at a distance �R from the bowl surface toward air, ordistance R toward water. Thus the image is formed at the bowl center, the location of the

13

object itself. The image is virtual because the image distance is negative. The magni�cationin this case is

m = � i

o=n= +

4

3

The image is erect and is magni�ed by 4/3. Homework: Con�rm the result by ray tracing.Example 6: Physical meaning of focusing. Show that the transit time for light fromthe object to its image is independent of the path length OHI provided the height h is smallcompared with the curvature radius R:

Solution: The transit time is given by

� =OHc+ n

HIc

whereOH =

p(o+ �)2 + h2

HI =p(i� �)2 + h2

However,R2 = (R� �)2 + h2

If � and h are small,

� =h2

2R;

and thus

OH ' o+ h2

2

�1

o+1

R

�HI ' i+ h

2

2

�1

i� 1

R

�(Use is made of the binomial expansion,

p1 + x ' 1 + 1

2x

for jxj � 1:) Therefore, the transit time reduces to

� =o

c+ni

c+h2

2c

�1

o+n

i� (n� 1) 1

R

�=

o

c+ni

c

14

where the refraction formula1

o+n

i= (n� 1) 1

R

is recalled. � is independent of the height h to order h2:

Example 7: Design converging and diverging lenses having focal lengths of �25 cm givenglass material of n = 1:5:Solution: For the converging lens,

1

R1� 1

R2=

1

n� 11

f

=1

12:5

There are many combinations of R1 and R2 to satisfy this. A few cases are shown.

1. R1 = 12:5 cm, R2 =1; (one surface is �at),

2. R1 = �R2 = 25 cm, (with equal curvature radii),

3. R1 = 15 cm, R2 = �75 cm, etc.

For the diverging lens,1

R1� 1

R2= � 1

12:5

1. R1 = �12:5 cm, R2 =1:

2. R1 = 25 cm, R2 = +8:33 cm, etc.

Example 8: Find the location of the �nal image and magni�cation.

30 cm ­10 cm

50 cm 40 cm

O

Solution: The image distance i due to the �rst lens can be found from

1

50+1

i=1

30; i = 75:0 cm.

The object distance for the second lens is �35 cm (virtual), and the location of its imagecan be found from

� 135+1

i0= � 1

10; i0 = �14:0 cm

15

The location of the �nal image is 14.0 cm to the left of the diverging lens. The magni�cationis

m = m1m2

=

��7550

�����14:0�35

�= +0:60

The image is virtual and erect.

Example 9: Find the location of the �nal image and magni�cation.

Solution: The image distance i due to the lens is

1

30+1

i=1

20; i = 60 cm

The object distance for the mirror is at �20 cm (virtual), and the mirror formula gives

� 120+1

i0=1

10; i0 = 6:67 cm

Since the image due to the mirror is real, there is further refraction by the lens,

1

33:33+1

i00=1

20; i00 = 50:0 cm

The �nal image is at 50 cm to the left of the lens and the magni�cation is

m = (�2)� 6:6720

��� 50

33:33

�= +1:0

The image is real and erect.

1.8 Achromatic LensesThe index of refraction n of glass, plastics and water depends on the wavelength. Color

splitting in a prism and also in rainbow is due to nonuniform refraction angles of di¤erentcolors. In general, blue light is refracted more than red light. (See p. 246 for typical glass andquartz.) Prism action in lenses is unavoidable and strictly speaking, images of di¤erent colorsare formed at di¤erent locations which makes the image blurred. This is called chromaticaberration.

16

Fig. 1.11. E¤ective focal length of a compond lens (doublet) is f = f1f2=(f1 + f2): If thetwo lenses are separated by t; the focal length is

1

f=1

f1+1

f2� t

f1f2

Chromatic aberration can be corrected by combining more than one lens with di¤erentglass materials. A simple compound achromatic lens consists of one converging and onediverging lens. The e¤ective focal length f of a compound lens is given by

1

f=1

f1+1

f2(15)

This can be seen from the �nal image position of a ray parallel to the axis. The image due tothe �rst lens is at i0 = f1 which becomes a virtual object distance for the lens f2: Therefore,

� 1f1+1

i=1

f2

and the image distance, which is the e¤ective focal length of the compound lens, is

1

i=1

f=1

f1+1

f2

Let us consider a compound lens as shown in Fig. 1.11. The converging lens is made of amaterial with indices of refraction at red nR and at blue nB: Similarly, those of diverging lensare n0R and n

0B: The curvature radii are R1; R2 and R3 and R4; respectively. The e¤ective

focal length of the compound lens for red color is

1

fR+1

f 0R= (nR � 1)

�1

R1� 1

R2

�+ (n0R � 1)

�1

R3� 1

R4

�(16)

and for blue color, it is

1

fB+1

f 0B= (nB � 1)

�1

R1� 1

R2

�+ (n0B � 1)

�1

R3� 1

R4

�(17)

By requirement, these focal lengths must be equal to correct chromatic aberration. We thushave

(nB � nR)�1

R1� 1

R2

�= (n0R � n0B)

�1

R3� 1

R4

�(18)

Since nB > nR and n0B > n0R; we see that�

1

R1� 1

R2

��1

R3� 1

R4

�< 0

which indicates that one lens should be converging and the other diverging.

17

Fig. 1.12. Achromatic compound lens.

Example 10: Design an achromatic lens of focal length 50 cm using lenses made of crownglass and �int glass. Assume a surface of common curvature radius and one �at surface asshown. Crown glass has nB = 1:510; nR = 1:505 and Flint glass nB = 1:630; nR = 1:615:

Fig. 1.13. Compound achromatic lens.

Solution: From Eq. (18), we have

0:005

�1

R1� 1

R2

�= �0:015

�1

R2� 1

1

�;

1

R1� 1

R2= �3 1

R2;

R2 = �2R1:From the formula for a compound lens, Eq. (16),

1

50= 0:51

�1

R1� 1

R2

�+ 0:63

1

R2;

1

50=0:51

R1+0:12

R2:

Solving the simultaneous equations, we �nd

R1 = 22:5 cm, and R2 = �45 cm

1.9 Matrix Method

18

In a sophisticated compound lens, there are several lenses to correct for chromatic aber-ration and calculation based on ray tracing becomes too cumbersome. The matrix methodgreatly facilitates lens and mirror calculations as shown below. Consider refraction at oneboundary with a curvature radius R: A light beam propagating at an angle � in mediumwith index of refraction n is refracted and in medium with n0 it propagates at angle �0: Theangles are related through the Snell�s law,

�+ =n0

n( + �0)

or noting = h=R;

�0 =� nn0� 1� hR+n0

n�

At the surface, the beams intersect the surface at the same height h = h0 in both media.Then �

h0

�0

�=

1 0� n

n0� 1� 1R

n

n0

!�h�

�(19)

The matrix for a �at glass slab of thickness t only causes change in the height, and itsmatrix is simply

T =

�1 t0 1

�(20)

R

φ

φ'

n n'

h γ

Fig. 1.14. Refraction at a spherical surface: de�nition of �; �0; h = h0:

Consider a thick lens with thickness t, index of refraction ng; and curvature radii R1 andR2: The refraction matrix at the �rst (entrance) surface is

R1 =

0@ 1 0�1

ng� 1�1

R1

1

ng

1A (21)

the transmission matrix over the thickness t is

T =

�1 t0 1

�(22)

and refraction matrix at the second (exit) surface is

R2 =

0@ 1 0

(ng � 1)1

R2ng

1A (23)

19

The total matrix (system matrix) of the lens is given by the product of three matrices,

R2TR1 =

0@ 1 0

(ng � 1)1

R2ng

1A� 1 t0 1

�0@ 1 0�1

ng� 1�1

R1

1

ng

1A=

0B@ 1� t

R1

�1� 1

ng

� t

ng

� (ng � 1)�1R1� 1

R2

�� (ng�1)2t

R1R2ng1 +

t

R2ng(ng � 1)

1CA�

�A BC D

�The focal length can be found by demanding � = 0 in the matrix�

h0

�0

�=

�A BC D

��h0

�(24)

which yieldsh0 = Ah; �0 = Ch:

Then the focal length is

f 0 = �h0

�0= �A

C

In the present case,

f 0 = �AC= �

1 +t

R1

�1

ng� 1�

1

R2(ng � 1) +

1

R1

�ng +

t

R2(ng � 1)

��1

ng� 1� (25)

The e¤ective focal length of thick lens is normally de�ned by the negative inverse of thecomponent C

1

f= � 1

C= (ng � 1)

�1

R1� 1

R2

�+(ng � 1)2 tR1R2ng

=1

f1+1

f2� t

ngf1f2(26)

f 0 is measured from the vertex of the second spherical glass surface while f is measured fromthe so-called principal plane which is at a distance

h2 =f

ngf1t (27)

to the left of the surface. For the �rst spherical surface, the principal plane is at

h1 =f

ngf2t (28)

to the right of the �rst vertex. When t is negligible (thin lens), we recover

1

f=1

f 0= (ng � 1)

�1

R1� 1

R2

�=1

f1+1

f2

20

R R

t

1 2

gnf'f

H2

Fig. 1.15. The focal position of a thick lens. f 0 in Eq. (??) is measured from the vertex ofthe exit surface, while f in Eq. (24) is from the principal plane of the second surface.

Example 11: A thick converging lens has R1 = 10 cm, R2 = �5 cm, t (thickness) = 1 cm,and ng = 1:5:

4. (a) Determine the lens matrix.

(b) Find the e¤ective focal length f and the focal position f 0 of paraxial ray comingfrom the left. Determine the location of the principal plane for the second surface.

(c) When an object is placed at 20 cm in front of the spherical surface with curvatureradius R1 = 10 cm, where is its image formed?

(d) Find the magni�cation and nature of the image.

Solution: The lens matrix is

L =

0@ 1 0

(ng � 1)1

R2ng

1A� 1 t0 1

�0@ 1 0�1

ng� 1�1

R1

1

ng

1A=

1 0

(1:5� 1) 1�5 1:5

!�1 10 1

�0@ 1 0�1

1:5� 1�1

10

1

1:5

1A=

�0:966 67 0:666 67�0:146 67 0:933 33

�The e¤ective focal length is

f = � 1C= � 1

�0:146 67 = 6:82 cm

The focal position measure from the second vertex is

f 0 = �AC= 6:59 cm

The principal plane is at6:82� 6:59 = 0:23 cm

to the left of the second vertex. This agrees with

h2 =ft

ngf1=6:82 (cm)� 10 (mm)

1:5� 20 (cm) = 2:3 mm

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Note that the focal length of the �rst surface is

f1 =R1

ng � 1= 20 cm

If an object is placed at o = 20 cm in front of the �rst surface (R1 = 10 cm), the object-lenssystem matrix becomes �

0:966 67 0:666 67�0:146 67 0:933 33

��1 200 1

�With an image at i after the second surface, the entire matrix reduces to�

1 i0 1

��0:966 67 0:666 67�0:146 67 0:933 33

��1 200 1

�=

�0:966 67� 0:146 67i 20:0� 2: 0i

�0:146 67 �2: 0

�The image location i can be found from the condition B = 20:0 � 2: 0i = 0; or i = 10 cm.The magni�cation is

A = 0:966 67� 0:146 67i = �0:5The image is real and inverted. Note that the condition B = 0 is required for forming afocused image independent of the incident angle �:

The matrix for a spherical mirror can be worked out in a similar manner. The mirrormatrix is

M =

�1 0� 2R1

�where the sign convention for mirrors should be adopted, R > 0 for concave mirror andR < 0 for convex mirror. Consider a simple example. If an object is placed at 30 cm infront of a convex mirror with focal length of �20 cm, its image is formed at 12 cm behindthe mirror. The magni�cation is m = � (�12) =30 = 0:4: This can be analyzed using thepertinent matrices as follows.

M =

�1 i0 1

��1 0� 2R1

��1 o0 1

�=

�1 i0 1

��1 0+ 120

1

��1 300 1

�=

�1 + 1

20i 30 + 5

2i

120

52

�The image location can be found by demanding

B = 30 +5

2i = 0

which yieldsi = �12 cm

The magni�cation is

A = 1 +1

20i = 1� 12

20= 0:4

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The system matrix for Example 9 (lens-mirror combination) is given by (homework)

M =

�1 i0 1

��1 0� 120

1

��1 400 1

��1 0�151

��1 400 1

��1 0� 120

1

��1 300 1

�=

�1 �50 + i0 1

�The image location can be readily found to be 50 cm to the left of the lens. In this example,the magni�cation is unity irrespective of the object position.

Example 12: Show that if the separation distance d of two lenses with focal lengths f1and f2 is chosen as

d =f1 + f22

(29)

chromatic aberration can be corrected. In this case, two lenses can be made of the sameglass.

Solution: The e¤ective focal length of two lenses f1 and f2 separated by d is given by

1

feff=

1

f1+1

f2� d

f1f2

= (ng � 1)�1

R1� 1

R2+1

R3� 1

R4

�� d (ng � 1)2

�1

R1� 1

R2

��1

R3� 1

R4

�This follows from Eq. (??) where ng should be replaced by n = 1 (in air). Chromaticaberration originates from the dependence of the index of refraction on the wavelengthng(�): Therefore the condition to avoid chromatic aberration can be found from

dfeffd�

= 0

ord

dng

�1

feff

�= 0

This yields

1

R1� 1

R2+1

R3� 1

R4� 2d (ng � 1)

�1

R1� 1

R2

��1

R3� 1

R4

�= 0

and thus

d =1

2 (ng � 1)

1

R1� 1

R2+1

R3� 1

R4�1

R1� 1

R2

��1

R3� 1

R4

� = f1 + f22

This works only for converging lenses. The trick was known to Huygens in the 17th century(Huygens�achromatic eyepiece for telescopes).1.10 Optical Instruments

23

1. Myopia correction Myopia (nearsightedness) can be corrected by placing a diverginglens in front of the eye. The focal length should be about negative of the far distanceof the eye. For example, if a person is unable to focus on objects further than 50 cm,the focal length of the diverging lens should be �50 cm or �2:0 D (Diopter = 1=f withf in meters).

5. Hyperopia correction Hyperopia (farsightedness) can be corrected by a converginglens. If the near point of hyperopia eye is dnear; the focal length of the lens can befound from

1

f (cm)+

1

dnear (cm)=

1

25 cm

where 25 cm is the average natural image distance of human eye.

6. Telescope A telescope consists of an object mirror or lens and eyepiece lens. For anobject at a large distance, its image is formed at the focal distance of the objectivemirror (or lens). The function of the eyepiece lens is to collimate the rays emergingfrom the image. The magni�cation is given by the ratio of the angles,

m = ���= �fo

fe; (30)

where fo is the focal distance of the objective mirror (or lens) and fe is that of eyepiece.Note that fe can be either converging (fe > 0 for inverted image) or diverging (fe < 0for erect image) lens. However, the optical depth of a telescope having a divergingeyepiece lens is small and is seldom used except for toys. Binoculars have convergingeyepiece lenses but produce erect images. Image reversal for erect image is achievedby combination of two prisms as shown in Fig. 17.

Fig. 1.16. Telescope. The eyepiece can be a diverging lens for erect image.

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Fig. 1.17. Image inversion in binoculars (with converging eyepieces) using two 90� prisms.

2. Microscope A microscope consists of objective and eyepiece lenses both converging.An object is to be placed slightly beyond the focal point of the objective lens so thata large real image is formed. The eyepiece simply acts as a magnifying glass. For thispurpose, the image should be formed at a point close to the focal point of the eyepieceso that an erect, virtual image is formed. The magni�cation is approximately given by

m = � Lfo

25

fe; (31)

where L is the length of the microscope and 25 cm is the natural object distance ofnormal human eye.

Magni�cations of telescopes and microscopes are limited by the inevitable di¤ractione¤ect. For telescopes, the resolving power (the minimum angular resolution of two objects)is approximately given by

�� ' �

D;

where D is the diameter of the objective mirror. A larger mirror collects more light andimages will be brighter but this is a secondary merit. A more important merit is that alarger diameter increases resolving power for observing stars at further distances.

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