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(B) Let the angle of the prism be $A$. Since the triangle is isosceles with $AB=AC$,the base angles are $\angle B = \angle C = (180^{\circ}-A)/2 = 90^

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$36$

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(B) Let the angle of the prism be $A$. Since the triangle is isosceles with $AB=AC$,the base angles are $\angle B = \angle C = (180^{\circ}-A)/2 = 90^{\circ} - A/2$.$1$. The ray is incident normally on face $AB$,so it enters the prism undeviated.$2$. At face $AC$ (which is silvered),the angle of incidence $i_1$ is equal to the angle $A$ of the prism.$3$. After reflection at $AC$,the angle of reflection is also $i_1 = A$. The angle the reflected ray makes with the face $AC$ is $90^{\circ}-A$.$4$. In the triangle formed by the ray and the sides of the prism,the angle at the face $AB$ is $90^{\circ}-2A$.$5$. At the second reflection on face $AB$,the angle of incidence $i_2$ is $90^{\circ}-(90^{\circ}-2A) = 2A$.$6$. The ray then strikes the base $BC$ normally. In the triangle formed by the ray and the base $BC$,the angle at $B$ is $90^{\circ}-2A$. Since the ray is perpendicular to $BC$,we have $90^{\circ}-2A + (90^{\circ}-A/2) = 90^{\circ}$.$7$. Solving for $A$: $90^{\circ} - 2.5A = 0$,which gives $A = 36^{\circ}$.

The principal section of a glass prism is an isosceles triangle $ABC$ with $AB=AC$. The face $AC$ is silvered. $A$ ray of light is incident normally on the face $AB$ and after two reflections,it emerges from the base $BC$ perpendicular to the base. Angle $BAC$ of the prism is (in $^{\circ}$)

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