In the figure shown,for an angle of incidence | vedclass

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

In the figure shown,for an angle of incidence $45^{\circ}$ at the top surface,what is the minimum refractive index $\mu$ needed for total internal reflection at the vertical face?

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