A narrow pencil of parallel light falls normally on a solid...

For refraction at a single spherical surface, the formula is $\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$.

Here, the incident light is a parallel pencil of rays, so $u = -\infty$. The light enters from air ($\mu_1 = 1$) into the transparent sphere ($\mu_2 = \mu$). The radius of curvature of the first surface is $R = +r$.

Substituting these values into the formula:

$\dfrac{\mu}{v} - \dfrac{1}{-\infty} = \dfrac{\mu - 1}{r}$

$\dfrac{\mu}{v} = \dfrac{\mu - 1}{r}$

$v = \dfrac{\mu r}{\mu - 1}$

For the rays to be focused at the opposite surface of the sphere, the image distance $v$ from the first surface must be equal to the diameter of the sphere, which is $2r$.

Equating the two expressions for $v$:

$\dfrac{\mu r}{\mu - 1} = 2r$

$\mu = 2(\mu - 1)$

$\mu = 2\mu - 2$

$\mu = 2$