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(C) Let the angle of refraction at the top surface be $r$. By Snell's Law at the top surface:$1 \cdot \sin(i) = \mu \cdot \sin(r) \implies \sin(r) = \

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$\sqrt{1 + \sin^2(i)}$

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(C) Let the angle of refraction at the top surface be $r$. By Snell's Law at the top surface:$1 \cdot \sin(i) = \mu \cdot \sin(r) \implies \sin(r) = \frac{\sin(i)}{\mu}$For total internal reflection at the vertical surface,the angle of incidence at that surface must be at least the critical angle $C$. The angle of incidence at the vertical surface is $(90^\circ - r)$.Thus,$90^\circ - r \ge C$,where $\sin(C) = \frac{1}{\mu}$.For the minimum refractive index,we take the limiting case: $\sin(90^\circ - r) = \sin(C) = \frac{1}{\mu}$.$\cos(r) = \frac{1}{\mu} \implies \cos^2(r) = \frac{1}{\mu^2}$.Using $\sin^2(r) + \cos^2(r) = 1$:$\left(\frac{\sin(i)}{\mu}\right)^2 + \frac{1}{\mu^2} = 1$$\sin^2(i) + 1 = \mu^2$$\mu = \sqrt{1 + \sin^2(i)}$

In the figure shown,for an angle of incidence $i$ at the top surface,the minimum refractive index $\mu$ needed for total internal reflection at the vertical surface will be

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