A parallel beam of light emerges from the opp | vedclass

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(C) Let the radius of the sphere be $R$. The point source is at $O$ on the surface of the sphere. The light rays travel through the sphere and emerge

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$2$

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(C) Let the radius of the sphere be $R$. The point source is at $O$ on the surface of the sphere. The light rays travel through the sphere and emerge from the opposite surface at $P$ as a parallel beam.Applying the refraction formula at the spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,the light travels from the sphere (refractive index $\mu$) to the air (refractive index $1$).Thus,$\mu_1 = \mu$,$\mu_2 = 1$.The object distance $u$ from the surface $P$ is the diameter of the sphere,so $u = -2R$.The image is formed at infinity,so $v = \infty$.The radius of curvature of the surface at $P$ is $R$,but since the center of curvature is to the left of $P$,the radius is taken as $-R$.Substituting these values: $\frac{1}{\infty} - \frac{\mu}{-2R} = \frac{1 - \mu}{-R}$.$0 + \frac{\mu}{2R} = \frac{\mu - 1}{R}$.$\frac{\mu}{2} = \mu - 1$.$1 = \mu - \frac{\mu}{2} = \frac{\mu}{2}$.$\mu = 2$.

$A$ parallel beam of light emerges from the opposite surface of the sphere when a point source of light lies at the surface of the sphere. The refractive index of the sphere is

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