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(D) At point $A$,by Snell's law:$1 \cdot \sin 45^{\circ} = \mu \cdot \sin r$$\sin r = \frac{1}{\mu \sqrt{2}}$  ..... $(i)$At point $B$,for total inter

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$\sqrt{\frac{3}{2}}$

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(D) At point $A$,by Snell's law:$1 \cdot \sin 45^{\circ} = \mu \cdot \sin r$$\sin r = \frac{1}{\mu \sqrt{2}}$  ..... $(i)$At point $B$,for total internal reflection to occur at the vertical face,the angle of incidence $i_1$ must be greater than or equal to the critical angle $C$.$\sin i_1 = \frac{1}{\mu}$From the geometry of the triangle,$i_1 = 90^{\circ} - r$.Therefore,$\sin(90^{\circ} - r) = \frac{1}{\mu} \Rightarrow \cos r = \frac{1}{\mu}$  ..... $(ii)$Using the identity $\sin^2 r + \cos^2 r = 1$:$\left(\frac{1}{\mu \sqrt{2}}\right)^2 + \left(\frac{1}{\mu}\right)^2 = 1$$\frac{1}{2\mu^2} + \frac{1}{\mu^2} = 1$$\frac{1 + 2}{2\mu^2} = 1$$\frac{3}{2\mu^2} = 1 \Rightarrow \mu^2 = \frac{3}{2}$$\mu = \sqrt{\frac{3}{2}}$

$A$ 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

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