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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 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 $$c=3\times 10^8$$ m/s]

A
50.00
Solution
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Correct option is A. 50.00
$$1.5\sin \theta_c=1.44\sin 90^o$$
$$\sin \theta_c=\dfrac{1.44}{1.50}=\dfrac{24}{25}$$
$$\therefore \sin\theta_c=\dfrac{x}{d}=\dfrac{24}{25}$$
$$d=\dfrac{25x}{24}$$
$$\therefore$$ Total length travel by light $$=\dfrac{25}{24}\times 9.6=10m$$
$$\therefore t=\dfrac{S}{\left(\dfrac{C}{n_1}\right)}=\dfrac{10}{\dfrac{3\times 10^8}{1.5}}=\dfrac{1}{2}\times 10^{-7}=5\times 10^{-8}$$
$$t=50ns$$
$$t=50\times 10^{-9}$$
$$\therefore$$ Answer$$=50$$.

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