One face of a glass prism is silver-polished. | vedclass

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(B) Given: Angle of incidence $i = 45^{\circ}$, Prism angle $A = 30^{\circ}$.Since the ray retraces its path after reflection from the silvered face,

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$\sqrt{2}$

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(B) Given: Angle of incidence $i = 45^{\circ}$, Prism angle $A = 30^{\circ}$.Since the ray retraces its path after reflection from the silvered face, it must strike the silvered face normally (at $90^{\circ}$).In the triangle formed by the ray inside the prism, the angles are $A = 30^{\circ}$, the angle at the silvered face is $90^{\circ}$, and the angle of refraction $r$ at the first face.The sum of angles in a triangle is $180^{\circ}$, so $r + 90^{\circ} + 30^{\circ} = 180^{\circ}$.$r = 180^{\circ} - 120^{\circ} = 60^{\circ}$.Wait, looking at the geometry: the angle of refraction $r$ is the angle with the normal. The angle inside the prism at the first face is $90^{\circ} - r$.Thus, $(90^{\circ} - r) + 90^{\circ} + 30^{\circ} = 180^{\circ}$.$210^{\circ} - r = 180^{\circ} \implies r = 30^{\circ}$.Using Snell's Law: $\mu = \frac{\sin i}{\sin r} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}}$.$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

One face of a glass prism is silver-polished. $A$ light ray falls at an angle of $45^{\circ}$ on the other face. After refraction, it is subsequently reflected from the silvered face and then retraces its path. The refracting angle of the prism is $30^{\circ}$. The refractive index of the prism is:

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$A$ prism has a refracting angle of $30^\circ$ and a refractive index of $\sqrt{2}$. One of its faces is silvered. At what angle of incidence $i$ should a ray be incident on the other face so that it retraces its path after reflection from the silvered face?

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