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Question

(A) Refraction occurs at two surfaces of the glass sphere.For the first surface,the refraction formula is $\frac{\mu}{v} - \frac{1}{u} = \frac{\mu-1}{

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$\frac{R(2-\mu)}{2(\mu-1)}$

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(A) Refraction occurs at two surfaces of the glass sphere.For the first surface,the refraction formula is $\frac{\mu}{v} - \frac{1}{u} = \frac{\mu-1}{R}$.Given $u = -\infty$,we have $\frac{\mu}{v_1} = \frac{\mu-1}{R}$,which gives $v_1 = \frac{\mu R}{\mu-1}$. This image $I_1$ is formed at a distance $v_1$ from the first surface $P_1$.For the second surface,the object distance $u_2$ is $v_1 - 2R = \frac{\mu R}{\mu-1} - 2R = \frac{\mu R - 2\mu R + 2R}{\mu-1} = \frac{R(2-\mu)}{\mu-1}$.Using the refraction formula at the second surface: $\frac{1}{v} - \frac{\mu}{u_2} = \frac{1-\mu}{-R} = \frac{\mu-1}{R}$.Substituting $u_2 = \frac{R(2-\mu)}{\mu-1}$:$\frac{1}{v} = \frac{\mu-1}{R} + \frac{\mu(\mu-1)}{R(2-\mu)} = \frac{\mu-1}{R} \left( 1 + \frac{\mu}{2-\mu} \right) = \frac{\mu-1}{R} \left( \frac{2-\mu+\mu}{2-\mu} \right) = \frac{\mu-1}{R} \left( \frac{2}{2-\mu} \right)$.Therefore,$v = \frac{R(2-\mu)}{2(\mu-1)}$.Thus,the distance of the image from the outer edge $P_2$ is $\frac{R(2-\mu)}{2(\mu-1)}$.

$A$ narrow parallel beam of light falls on a glass sphere of radius $R$ and refractive index $\mu$ at normal incidence. The distance of the image from the outer edge is given by

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