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(B) Applying Snell's Law on the $1^{	ext{st}}$ surface: $\sin(60^{\circ}) = n \sin(r_1) \Rightarrow \sin(r_1) = \frac{\sqrt{3}}{2n}$.For an equilater

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$2$

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(B) Applying Snell's Law on the $1^{	ext{st}}$ surface: $\sin(60^{\circ}) = n \sin(r_1) \Rightarrow \sin(r_1) = \frac{\sqrt{3}}{2n}$.For an equilateral prism,the prism angle $A = 60^{\circ}$,so $r_1 + r_2 = 60^{\circ}$,which implies $r_2 = 60^{\circ} - r_1$.Applying Snell's Law on the $2^{	ext{nd}}$ surface: $n \sin(r_2) = \sin(	heta)$.Substituting $r_2$: $\sin(	heta) = n \sin(60^{\circ} - r_1) = n [\sin(60^{\circ}) \cos(r_1) - \cos(60^{\circ}) \sin(r_1)]$.Since $\sin(r_1) = \frac{\sqrt{3}}{2n}$,then $\cos(r_1) = \sqrt{1 - \frac{3}{4n^2}} = \frac{\sqrt{4n^2 - 3}}{2n}$.Substituting these into the equation for $\sin(	heta)$:$\sin(	heta) = n [\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{4n^2 - 3}}{2n} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2n}] = \frac{\sqrt{3}}{4} (\sqrt{4n^2 - 3} - 1)$.Differentiating with respect to $n$: $\cos(	heta) \frac{d	heta}{dn} = \frac{\sqrt{3}}{4} \cdot \frac{1}{2\sqrt{4n^2 - 3}} \cdot 8n = \frac{\sqrt{3} n}{\sqrt{4n^2 - 3}}$.At $n = \sqrt{3}$,$	heta = 60^{\circ}$,so $\cos(60^{\circ}) \frac{d	heta}{dn} = \frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{4(3) - 3}} = \frac{3}{\sqrt{9}} = 1$.$\frac{1}{2} \cdot \frac{d	heta}{dn} = 1 \Rightarrow \frac{d	heta}{dn} = 2$. Thus,$m = 2$.

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