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(A) From the geometry of the prism,the light ray enters the prism normally to the hypotenuse face. Thus,the angle of incidence at the first surface is

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) From the geometry of the prism,the light ray enters the prism normally to the hypotenuse face. Thus,the angle of incidence at the first surface is $i_1 = 0^{\circ}$,which implies the angle of refraction $r_1 = 0^{\circ}$.Given the prism angle $A = 30^{\circ}$,we use the relation $r_1 + r_2 = A$. Substituting $r_1 = 0^{\circ}$,we get $r_2 = 30^{\circ}$.Applying Snell's law at the second surface (the vertical face): $\mu \sin r_2 = 1 	imes \sin e$,where $\mu = \sqrt{3}$ and $e$ is the angle of emergence.$\sqrt{3} \sin 30^{\circ} = \sin e$$\sqrt{3} 	imes \frac{1}{2} = \sin e$$\sin e = \frac{\sqrt{3}}{2}$Therefore,$e = 60^{\circ}$.

Find the value of the angle of emergence from the prism. The refractive index of the glass is $\sqrt{3}$. (in $^{\circ}$)

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