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Question

(B) Let the angle of the prism be $A$. Since the ray is incident normally on face $AB$, it enters the prism without deviation.At the silvered face $AC

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) Let the angle of the prism be $A$. Since the ray is incident normally on face $AB$, it enters the prism without deviation.At the silvered face $AC$, the angle of incidence $i_1$ is equal to the prism angle $A$ (as the normal to $AC$ makes an angle $A$ with the normal to $AB$).By the law of reflection, the angle of reflection is also $i_1 = A$.In the triangle formed by the ray and the faces, the angle at the base $AB$ is $90^{\circ} - A$. Thus, the angle of incidence at the second reflection on face $AB$ is $i_2 = 90^{\circ} - (90^{\circ} - 2A) = 2A$.Finally, the ray emerges from the base $BC$ normally. In the triangle near vertex $B$, the angles are $B$, $90^{\circ} - i_2$, and $90^{\circ}$. Since $AB = AC$, the base angles are $B = C = (180^{\circ} - A) / 2 = 90^{\circ} - A/2$.From the geometry of the path, the sum of angles in the triangle formed by the ray inside the prism is $A + 2A + 2A = 180^{\circ}$.$5A = 180^{\circ} \implies A = 36^{\circ}$.

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. The angle $BAC$ of the prism is: (in $^{\circ}$)

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