JEE Main 2023 — Ray Optics Question with Solution

Let $C$ be the critical angle.

Then, $\mathrm{sinC}=\frac{1}{\mu }$, where, $\mu$ is the refractive index of glass.

So, $\mathrm{sinC}=\frac{1}{\sqrt{2}}\Rightarrow$$\mathrm{C}=45°$

As the angle of incidence is equal to the critical angle, so using Snell's law,

 $\mathrm{sinC}=\mu \sin \theta$

$\Rightarrow \frac{1}{\sqrt{2}}=\sqrt{2 }\sin \theta$

$\Rightarrow \theta =30°$

Now, lateral displacement of the ray is 

$x=t \sin (i-r) \sec r$

$\Rightarrow x=\sqrt{3}\sin \left(45°-30°\right)\sec 30°$

$\Rightarrow x=\sqrt{3}(0.26)\left(\frac{2}{\sqrt{3}}\right)$

$\Rightarrow x=0.52 \mathrm{cm}$

$\Rightarrow x=52\times {10}^{-2} \mathrm{cm}$