Question 37: Ray Optics

Question

A monochromatic light is incident from air on a refracting surface of a prism of angle 75$$^\circ $$ and refractive index $${n_0} = \sqrt 3 $$. The other refracting surface of the prism is coated by a thin film of material of refractive index n as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of $$	heta  \le 60^\circ $$. The value of n2 is .............

Accepted Answer

For TIR at coating,
Applying Snell's law at first surface $$\sin 	heta  = \sqrt 3 \sin (75^\circ  - {	heta _c})$$
For limiting condition, at $$	heta $$ = 60$$^\circ $$
$$\sin 60^\circ  = \sqrt 3 \sin (75^\circ  - {	heta _c})$$
$${{\sqrt 3 } \over 2} = \sqrt 3 \sin (75^\circ  - {	heta _c})$$
$${1 \over 2} = \sin (75^\circ  - {	heta _c})$$
$$ \Rightarrow $$ $$sin30^\circ  = \sin (75^\circ  - {	heta _c})$$
$$30^\circ $$ = $$75^\circ  - {	heta _c}$$ $$ \Rightarrow $$ $${	heta _c}$$ = 45$$^\circ $$
Now since ÃÂ¸
c
< 60
o
, therefore total internal reflection takes place. For total internal reflection at the other refracting surface, using SnellÃ¢ÂÂs law, we have
$${n \over {\sqrt 3 }} = {1 \over {\sqrt 2 }} \Rightarrow {n^2} = {3 \over 2} = 1.50$$

Ray Optics

Choose the correct answer: