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(D) Given: Apex angle $A = 45^o$. Grazing incidence means the angle of incidence $i = 90^o$. The angle of emergence $e = 45^o$.From Snell's law at the

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$(5)^{1/2}$

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(D) Given: Apex angle $A = 45^o$. Grazing incidence means the angle of incidence $i = 90^o$. The angle of emergence $e = 45^o$.From Snell's law at the first surface: $\sin i = \mu \sin r_1 \Rightarrow \sin 90^o = \mu \sin r_1 \Rightarrow \sin r_1 = \frac{1}{\mu}$.From Snell's law at the second surface: $\sin e = \mu \sin r_2 \Rightarrow \sin 45^o = \mu \sin r_2 \Rightarrow \frac{1}{\sqrt{2}} = \mu \sin r_2 \Rightarrow \sin r_2 = \frac{1}{\mu \sqrt{2}}$.We know that $r_1 + r_2 = A = 45^o$. Therefore,$r_1 = 45^o - r_2$.Taking sine on both sides: $\sin r_1 = \sin(45^o - r_2) = \sin 45^o \cos r_2 - \cos 45^o \sin r_2$.Substituting the values: $\frac{1}{\mu} = \frac{1}{\sqrt{2}} \cos r_2 - \frac{1}{\sqrt{2}} \sin r_2$.Since $\sin r_2 = \frac{1}{\mu \sqrt{2}}$,then $\cos r_2 = \sqrt{1 - \sin^2 r_2} = \sqrt{1 - \frac{1}{2\mu^2}} = \frac{\sqrt{2\mu^2 - 1}}{\sqrt{2}\mu}$.Substituting these into the equation: $\frac{1}{\mu} = \frac{1}{\sqrt{2}} \left( \frac{\sqrt{2\mu^2 - 1}}{\sqrt{2}\mu} \right) - \frac{1}{\sqrt{2}} \left( \frac{1}{\sqrt{2}\mu} \right) = \frac{\sqrt{2\mu^2 - 1}}{2\mu} - \frac{1}{2\mu}$.Multiplying by $2\mu$: $2 = \sqrt{2\mu^2 - 1} - 1 \Rightarrow 3 = \sqrt{2\mu^2 - 1}$.Squaring both sides: $9 = 2\mu^2 - 1 \Rightarrow 2\mu^2 = 10 \Rightarrow \mu^2 = 5 \Rightarrow \mu = \sqrt{5}$.

For a prism of apex angle $45^o$,it is found that the angle of emergence is $45^o$ for grazing incidence. Calculate the refractive index of the prism.

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