JEE Advanced 2019 · Paper 1 · Physics Q18

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PhysicsNumerical+3Question 18 of 54

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 $L \gg W$ , 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 $t\times10^{-9}$ s, where $t$ is ______.

[Speed of light $c = 3\times10^8$ m/s]

$A slab of length L: a core of index n₁ between two cladding layers of index n₂, total width W. A ray from air hits the end face AB at angle θ, refracts into the core and zigzags along it by total internal reflection towards the far end CD.$

Your answer

Answer and worked solution

Official answer: 49

Slowest ray. A ray takes longest when it zigzags most steeply, i.e. when it meets the core-cladding boundary exactly at the critical angle:
$\sin\theta_c=\frac{n_2}{n_1}=0.96.$

Path and time. The ray then makes angle $\theta_c$ with the boundary normal, so its path length is $L/\sin\theta_c$ :
$t=\frac{n_1L}{c\sin\theta_c}=\frac{1.5\times9.6}{3\times10^8\times0.96}=5.0\times10^{-8}\text{ s}.$

Check that this ray can enter. It needs $\sin\theta=n_1\cos\theta_c=1.5\times0.28=0.42<1$ , which is fine.

Answer: 50 (key: 49 to 51)

Topics:Total internal reflection Refraction at plane surfaces

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JEE Advanced 2019 · Paper 1, Physics question 18