Question 69 of 295

A light ray falls on a square glass slab as shown in the diagram. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is equal to :

A$\frac{(\sqrt{2}+1)}{2}$B$\sqrt{\frac{5}{2}}$C$\frac{3}{2}$D$\sqrt{\frac{3}{2}}$

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Answer: (D) $\sqrt{\frac{3}{2}}$

At point A by Snell's law
$$\mu=\frac{\sin 45^{\circ}}{\sin r} \Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}$$
At point $\mathrm{B}$, for total internal reflection,
$$\sin i_1=\frac{1}{\mu}$$


From figure, $i_1=90^{\circ}-r$
$$\begin{aligned} & \therefore\left(\sin 90^{\circ}-r\right)=\frac{1}{\mu} \\ & \Rightarrow \cos r=\frac{1}{\mu} \end{aligned}$$
Now $\cos r=\sqrt{1-\sin ^2 r}=\sqrt{1-\frac{1}{2 \mu^2}}$
$$=\sqrt{\frac{2 \mu^2-1}{2 \mu^2}}$$
From eqs (ii) and (iii)
$$\frac{1}{\mu}=\sqrt{\frac{2 \mu^2-1}{2 \mu^2}}$$
Squaring both sides and then solving, we get
$$\mu=\sqrt{\frac{3}{2}}$$

JEE Main Physics practice question on Ray Optics (Critical angle and total internal reflection), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.