2 Solution

20131_221hw2S

2.

3. An achromatic doublet consists of two thin lenses having focal lengths f1 and f2. The doublet has an overall focal length of 2f1.

The refractive indices of the lens materials are given in the table below.

a) Calculate the refractive index of the material of the second lens for

= 656.3 nm (i.e. nC).

1 2

1 2

0f f

is the condition for achromatic doublet. Eq.1

e q 1

1 2 e q

1 1 1 s in c e f 2 f

f f f we should have

2 1

1 2 1

1 1 1 th e re fo re f 2 f

f f 2 f

use this condition in Eq.1 1 2

1 2

1 1

0 h e n c e 2f 2 f

F c F c

D D1s t m a te r ia l 2 n d m a te r ia l

n n n n2

n 1 n 1

c

c

1 .8 2 n1 .8 1 1 .7 92 o n e o b ta in s n 1 .7 8 fo r th e 2n d m a te ria l.

1 .8 0 1 1 .8 0 1

b) Now, consider two right-angled isosceles prisms cemented as shown below. These prisms are manufactured from the same

materials described above. A beam of light containing three wavelengths C, D and F only is sent parallel to the base. Carry out

the necessary calculations and sketch (on the given figure) how the rays corresponding to different wavelengths exit from the

second prism. (Note: if you are not sure about your answer to part (a) take nC=1.78 for the second prism material).

The incident ray enters the 1st prism with a zero angle of incidence, it meets

the boundary at an angle of 45˚.

For D : Since nD is the same for both prisms ray passes from 1st

prism to 2nd with no deviation and meets the final surface perpendicularly

hence leaves the prism with no deviation at all.

For F : Since (nF)1<(nF)2 at the interface ray bends upwards(closer

to the normal) and meets the final surface with a small angle of incidence.

At this point critical angle for total internal refection is:

0

c c

1s in 3 3

1 .8 2 is much greater than the angle of incidence

so no total reflection takes place hence ray exits the prism bending

downwards.

For C: Since (nC )1>(nC )2 at the interface ray bends downwards(away from the normal) and meets the final surface with

a small angle of incidence. At this point critical angle for total internal refection is: 0

c c

1s in 3 4

1 .7 8 is much

greater than the angle of incidence so no total reflection takes place hence ray exits the prism bending downwards.

Note that refractive indices of the prisms are very close for all wavelengths the critical angle total internal refection is

around 80˚>angle of incidences at the interface between two prisms, we expect no total internal reflection at the junction.

nC nD nF

Lens 1(f1), Prism1(see part b) 1.79 1.80 1.81

Lens 1(f2), Prism2(see part b) ? 1.80 1.82

45˚

45˚

Prism 1

Prism 2

D

F

6 . The objective of a microscope is a plano-convex lens with n=1.5 and R=12.5mm.The ocular is a symmetrical convex

lens with n=1.5 and R=25mm. If a small object placed at 27mm from the objective is viewed with a relaxed eye; find

the tube length and the magnification.

0

0

e

e

1 1 1(1 .5 1)( ) f = 2 5 m m

f 1 2 .5

1 1 1(1 .5 1)( ) f = 2 5 m m

f 2 5 2 5

Objective produces an image:

1 1 1 2 7 2 5

s '= = 3 3 7 .5 m m2 5 2 7 s ' 2

Relaxed eye means final image is at infinity: therefore image given by the objective must be at the focal poinr of the ocular. Hence

the tube length L= e

3 3 7 .5 m m + f 3 3 7 .5 2 5 3 6 2 .5m m

e

s ' 2 5 (c m ) 3 3 7 .5 m m 2 5 0 m mM ( ) = -1 2 4 .8

s f 2 7 m m 2 5 m m

4. Design an achromatic doublet of crown glass 520/636 and flint glass 805/255 that has an overall focal

length of 30cm;

a) when the crown glass is equi-convex,

b) when the flint glass has one flat surface.