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JEE MAINS 2023
25.01.23
Question
A ray of light is incident from air on a glass plate having thickness $\sqrt{3} \mathrm{~cm}$ and refractive index $\sqrt{2}$. The angle of incidence of a ray is equal to the critical angle for glass-air interface The lateral displacement of the ray when it passes through the plate is $\_\_\_\_$ $\times 10^{-2} \mathrm{~cm}$. (given $\sin 15^{\circ}=0.26$ )
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Solution
Sol. $$ \begin{aligned} & \sin \mathrm{c}=\frac{1}{\sqrt{2}} \\ & \mathrm{c}=45^{\circ} \\ & \sin \mathrm{c}=\mu \sin \theta \\ & \frac{1}{\sqrt{2}}=\sqrt{2} \sin \theta \\ & \theta=30^{\circ} \end{aligned} $$ Lateral displacement: $$ \begin{aligned} & x=t \sin (1-r) \sec r \\ & x=\sqrt{3} \sin \left(45^{\circ}-30^{\circ}\right) \sec 30^{\circ} \\ & x=\sqrt{3}(0.26)\left(\frac{2}{\sqrt{3}}\right) \\ & x=0.52 \mathrm{~cm} \\ & x=52 \times 10^{-2} \mathrm{~cm} \end{aligned} $$
Question Tags
JEE Main
Physics
Hard
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