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

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

Here, the light is incident from air ($\mu_1 = 1$) to the transparent sphere ($\mu_2 = \mu$).

Since the incident light is a parallel pencil, the object distance is $u = -\infty$.

The radius of curvature of the surface is $R = +r$.

The light is to be focused at the centre of the sphere, which is at a distance $r$ from the pole of the first surface. So, the image distance is $v = +r$.

Substituting these values into the refraction formula:

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

$\dfrac{\mu}{r} - 0 = \dfrac{\mu - 1}{r}$

$\mu = \mu - 1$

$0 = -1$

This leads to a mathematical contradiction, which implies that no such refractive index exists.

Conceptually, any ray passing through the centre of a sphere must travel along the normal to the surface. For a parallel beam, only the central axial ray is normal to the surface. All other parallel rays will have a non-zero angle of incidence and will be refracted at a non-zero angle of refraction, meaning they will deviate from the normal and miss the centre. Thus, it is not possible to focus a parallel pencil of light at the centre of the sphere.