JEE Advance - Physics (2019 - Paper 1 Offline - No. 14)
A planar structure of length L and width W is made of two different optical media of refractive indices n1 = 1.5 and n2 = 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 $$\theta $$ is varied, the maximum time taken by a ray to exit the plane CD is t $$ \times $$ 10-9 s, where, t is ................
[Speed of light, c = 3 $$ \times $$ 108 m/s]
[Speed of light, c = 3 $$ \times $$ 108 m/s]
Answer
50
Explanation
According to total internal reflection (TIR),
$$1.5\sin {\theta _c} = 1.44\sin 90^\circ $$
$$\sin {\theta _c} = {{1.44} \over {1.50}} = {{24} \over {25}}$$
$$ \therefore $$ $$\sin {\theta _c} = {x \over d} = {{24} \over {25}} \Rightarrow d = {{25x} \over {24}}$$
$$ \therefore $$ Total length travelled by light,
$$ \therefore $$ $$t = {S \over {\left( {{c \over {{n_1}}}} \right)}} = {{10} \over {{{3 \times {{10}^8}} \over {1.5}}}} = {1 \over 2} \times {10^{ - 7}} = 5 \times {10^{ - 8}}$$
t = 50 ns $$ \Rightarrow $$ t = 50 $$ \times $$ 10-9
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