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(C) Let the radius of the sphere be $R$. The light source is at the surface,so the object distance $u = -2R$ from the second surface (the diameter of

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$2$

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(C) Let the radius of the sphere be $R$. The light source is at the surface,so the object distance $u = -2R$ from the second surface (the diameter of the sphere).Using the refraction formula at a spherical surface: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R'}$Here,$\mu_1 = \mu$ (refractive index of the sphere),$\mu_2 = 1$ (refractive index of air),$v = \infty$ (since the emergent beam is parallel),$u = -2R$,and the radius of curvature $R' = -R$ (as the center of curvature is to the left of the second surface).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$$\mu = 2\mu - 2$$\mu = 2$

$A$ point source of light at the surface of a sphere causes a parallel beam of light to emerge from the opposite surface of the sphere. The refractive index of the material of the sphere is

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