Question

There is a small black dot at the centre C of a solid glass sphere of refractive index $\mu$. When seen from outside comment on the location of the dot.
A. Away from C for all values of $\mu$
B. At C for all values of $\mu$
C. At C for $\mu = 1.5$, but away from C for $\mu \ne 1.5$
D. At C only for $\sqrt 2 < \mu < 1.5$

Answer 1

The ray of light from the centre of the sphere strikes the surface of a sphere at ${90^ \circ }$, it is the same as the radius. We know that the radius of a sphere falls normally on the surface of the sphere. In refraction at curved surfaces we take the radius at the point of incidence as the normal. So, in this case the incident ray and normal coincides. Which means the angle between the normal and the incident ray is zero. Snell's law states that the refractive index of a medium is the ratio of the sine of angle of incidence to the sine of angle of refraction. Refractive index can also be calculated as the ratio of speed of light in one medium to the speed of light in another.
Using all this information, we can find the answer to the question.

Complete answer:
Given a black dot is at the centre C of a solid glass sphere.
The refractive index of the sphere is $\mu$.
Let us create the diagrammatic representation of the situation.

The rays originating from the black dot will strike the surface of the sphere at ${90^ \circ }$. This is because any line drawn from the centre of a sphere towards the surface will be the radius and it will always strike the surface at an angle ${90^ \circ }$ .
We know that in refraction at curved surfaces we take the normal at the point of incidence as the line joining the point to the centre of curvature of the sphere of which the curved surface is a part. Hence, here the incident ray and the normal are the same.They coincide with each other which means the angle between the normal and the incident ray is zero. The angle of incidence $i$ is equal to zero.
Snell's law states that the refractive index of a medium is the ratio of the sine of angle of incidence to the sine of angle of refraction.
$n = \dfrac{{\sin i}}{{\sin r}}$
Where $i$ is angle of incidence and $r$ is the angle of refraction.
Refractive index can also be calculated as the ratio of speed of light in one medium to the speed of light in another.
$n = \dfrac{{{v_1}}}{{{v_2}}}$ where ${v_1}$ is the speed of light in the first medium and ${v_2}$ is the speed of light in the second medium.
Therefore,
$\Rightarrow \dfrac{{\sin i}}{{\sin r}} = \dfrac{{{v_1}}}{{{v_2}}}$
$\Rightarrow {v_2}\sin i = {v_{_1}}\sin r$
Here since angle of incidence is equal to ${90^ \circ }$, $\sin i = 0$
$\Rightarrow {v_2} \times 0 = {v_1}\sin r$
$\Rightarrow \sin r = 0$
Since speed in both medium cannot be zero $\sin r$ must be equal to zero.$\therefore r = 0$.
That is, the angle of refraction is zero. That is the angle between the refracted ray and the normal should be zero. So, the ray will travel without any deviation. This is the case when the ray from origin strikes the surface in any direction. Therefore, the position of the dot will be the same from any direction. Hence, the dot will appear at the centre itself.

So, the answer is option B.

Note:
We know that any line drawn from the centre of curvature to the surface of the sphere will be normal. Since the dot is at the centre all the rays originating from it at any direction will fall on the surface of the sphere at ${90^ \circ }$ . But if the dot is at any other point inside the sphere the position of the dot as seen from outside will be different.