A ray of light is incident at an angle of 60^ | vedclass

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(D) For a light ray to retrace its path after reflection from the silvered second face, it must strike the second face normally (at an angle of $90^{\

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(D) For a light ray to retrace its path after reflection from the silvered second face, it must strike the second face normally (at an angle of $90^{\circ}$ to the surface).Let $i = 60^{\circ}$ be the angle of incidence on the first face, $r_1$ be the angle of refraction at the first face, and $A = 30^{\circ}$ be the prism angle.Since the ray strikes the second face normally, the angle of refraction at the second face is $r_2 = 0^{\circ}$.From the geometry of the prism, we know that $A = r_1 + r_2$.Substituting the values, $30^{\circ} = r_1 + 0^{\circ}$, so $r_1 = 30^{\circ}$.Using Snell's Law at the first face: $\mu = \frac{\sin(i)}{\sin(r_1)}$.Substituting the values: $\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$.$\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.Thus, the refractive index of the material of the prism is $\sqrt{3}$.

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