In the diagram,the ray passing through the pr | vedclass

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(D) Given that the ray inside the prism is parallel to the base,the angle of refraction $r$ at the first surface must be equal to the base angle of th

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2}}$

Vedclass · Answer

(D) Given that the ray inside the prism is parallel to the base,the angle of refraction $r$ at the first surface must be equal to the base angle of the prism.From the geometry of the triangle,the base angle is $45^{\circ}$,so $r = 45^{\circ}$.The angle of incidence $i$ is given as $60^{\circ}$.According to Snell's Law,$\mu = \frac{\sin i}{\sin r}$.Substituting the values,$\mu = \frac{\sin 60^{\circ}}{\sin 45^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{3}}{2} 	imes \sqrt{2} = \sqrt{\frac{3}{2}}$.Thus,the refractive index is $\sqrt{\frac{3}{2}}$.

In the diagram,the ray passing through the prism is parallel to the base. The refractive index of the material of the prism is:

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