Question 388: Ray Optics

Question

A planar structure of length L and width W is made of two different optical media of refractive indices n
1
= 1.5 and n
2
= 1.44 as shown in figure. If L >> W, a ray entering from end AB will emerge from end CD. CD only if the total internal reflection condition is met inside the structure. For L = 9.6 m, if the incident angle $$	heta $$ is varied, the maximum time taken by a ray to exit the plane CD is t $$ 	imes $$ 10
-9
s, where, t is ................
[Speed of light, c = 3 $$ 	imes $$ 10
8
m/s]

Accepted Answer

According to total internal reflection (TIR),
$$1.5\sin {	heta _c} = 1.44\sin 90^\circ $$
$$\sin {	heta _c} = {{1.44} \over {1.50}} = {{24} \over {25}}$$
$$ 	herefore $$ $$\sin {	heta _c} = {x \over d} = {{24} \over {25}} \Rightarrow d = {{25x} \over {24}}$$
$$ 	herefore $$ Total length travelled by light,
$$ 	herefore $$ $$t = {S \over {\left( {{c \over {{n_1}}}} ight)}} = {{10} \over {{{3 	imes {{10}^8}} \over {1.5}}}} = {1 \over 2} 	imes {10^{ - 7}} = 5 	imes {10^{ - 8}}$$
t = 50 ns $$ \Rightarrow $$ t = 50 $$ 	imes $$ 10
-9

Ray Optics

Integer / Numeric Answer Type