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(B) For refraction at the first spherical surface,the formula is $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,$\mu_1 = 1$ (air),

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(B) For refraction at the first spherical surface,the formula is $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.Here,$\mu_1 = 1$ (air),$\mu_2 = \mu$ (sphere),$u = -x$,and $v$ is the image distance formed by the first surface.If $x = \infty$,then $u = -\infty$,so $\frac{\mu}{v} - \frac{1}{-\infty} = \frac{\mu - 1}{R}$,which gives $v = \frac{\mu R}{\mu - 1}$.This image acts as a virtual object for the second surface. The distance of this image from the second surface is $v' = 2R - v = 2R - \frac{\mu R}{\mu - 1} = \frac{2\mu R - 2R - \mu R}{\mu - 1} = \frac{(\mu - 2)R}{\mu - 1}$.For the second surface,refraction occurs from $\mu$ to $1$. The pole is at the right side,so the object distance $u' = +v' = \frac{(\mu - 2)R}{\mu - 1}$ and the final image is at the pole,so $v_{final} = 0$.The formula is $\frac{1}{v_{final}} - \frac{\mu}{u'} = \frac{1 - \mu}{-R}$.Substituting the values: $\frac{1}{0} - \frac{\mu}{\frac{(\mu - 2)R}{\mu - 1}} = \frac{1 - \mu}{-R}$.For the image to be at the pole,the rays must be incident normally on the second surface,which happens if the rays are parallel inside the sphere. This implies the first surface must focus the rays at infinity,or simply,for a sphere,if $x = \infty$,the rays enter parallel to the axis and focus at the pole if $\mu = 2$.

The figure shows a transparent sphere of radius $R$ and refractive index $\mu$. An object $O$ is placed at a distance $x$ from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface. If $x = \infty$,then the value of $\mu$ is

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