Physics · Geometrical Optics

JEE Advanced 2019 — Paper 1 — Question 16

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

Question figure

Answer: 50

Numerical answer — enter this value.

Step-by-step solution

$1.5 \sin \theta_{\mathrm{C}}=1.44 \sin 90^{\circ}$

$\sin \theta_{\mathrm{C}}=\frac{24}{25}$

$\ell=\frac{\mathrm{x}}{\sin \theta_{\mathrm{C}}}=\frac{25}{4} \mathrm{x}$

total length for light to travel

$\ell^{\prime}=\frac{25}{4} \times 9.6=10 \mathrm{~m}$

$\therefore$ time $=\frac{\ell^{\prime}}{\mathrm{c} / 1.5}=5 \times 10^{-8} \mathrm{~s} \Rightarrow 50 \times 10^{-9} \mathrm{~s}$

$\mathrm{t}=50.00$

Solution figure

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Physics
Chapter: Geometrical Optics
Topic: Reflection of Light and Plane Mirrors

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