A prism having a refractive index 1.414 and a | vedclass

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

Vedclass · Accepted Answer

$45$

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(D) For a light ray to retrace its path after reflection from a silvered surface,it must strike the silvered surface normally (at an angle of $90^{\circ}$ to the surface).Since the light ray strikes the silvered surface normally,the angle of refraction at the second surface is $r_{2} = 0^{\circ}$.In a prism,the sum of the angles of refraction is equal to the prism angle: $r_{1} + r_{2} = A$.Given $A = 30^{\circ}$ and $r_{2} = 0^{\circ}$,we get $r_{1} = 30^{\circ}$.Applying Snell's Law at the first surface: $n = \frac{\sin i}{\sin r_{1}}$.Given $n = 1.414 = \sqrt{2}$,we have $\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.$\sin i = \sqrt{2} 	imes \sin 30^{\circ} = \sqrt{2} 	imes \frac{1}{2} = \frac{1}{\sqrt{2}}$.Therefore,$i = 45^{\circ}$.

$A$ prism having a refractive index $1.414$ and a refracting angle $30^{\circ}$ has one of its refracting surfaces silvered. $A$ beam of light incident on the other refracting surface will retrace its path if the angle of incidence is: (in $^{\circ}$)

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