Question

A transparent sphere of radius R and refractive index $\mu $ is kept in air. At what distance from the surface of the sphere should a point object be placed on the principal axis so as to form a real image at the same distance from the second surface of the sphere?
A) $\dfrac{R}{\mu }$
B) $\mu R$
C) $\dfrac{R}{{\mu - 1}}$
D) $\dfrac{R}{{\mu + 1}}$

Answer 1

Hint: We can draw a ray diagram by keeping the point object at a certain distance from the first surface of the sphere to obtain its image at infinity (as the image is real) from the second surface.
As there is change in the medium, refraction will take place, so we can use the equation for refraction through a spherical surface to find the required distance of the object. Refractive index of air is always taken as 1.

Complete Step by step answer: Sphere is a denser medium than air and thus light will suffer refraction while moving from air to the sphere. The the ray diagram for an object and its image through the spherical surface can be given as:

The object is kept at O at distance u from the first surface of the sphere and its image is formed at I at distance v from the second surface of the sphere. For the image to be real, it should be at infinite distance from the sphere. By sign conventions, the distances on the left are taken as negative an hence the distance of object (u) is taken as negative in the ray diagram.
Equation for the refraction through a spherical can be used to calculate the distance of object from the sphere:
$\dfrac{{{\mu _2}}}{v} + \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
Here,
Refractive index of air from where incident light start travel $\left( {{\mu _1}} \right)$ = 1
Refractive index of sphere where incident light strikes $\left( {{\mu _2}} \right) = \mu $ (given)
Distance of object from the sphere (u) = -u (to be calculated; to the left side of the sphere, hence negative)
Distance of object from the sphere (v) = $\infty $ (for the image to be real)
Radius of sphere (R) = R (given)
Substituting these values, we get:
$
  \dfrac{\mu }{\infty } + \dfrac{1}{{\left( { - u} \right)}} = \dfrac{{\mu - 1}}{R} \\
   \Rightarrow \dfrac{1}{u} = \dfrac{{\mu - 1}}{R}\left(
  \because \dfrac{1}{\infty } = 0 \right) \\
  0 \times \mu = 0 \\
  \therefore u = \dfrac{R}{{\mu - 1}} \\
 $
Therefore, the distance of a point object from the surface of the sphere to form a real image at the same distance from the second surface of the sphere is $u = \dfrac{R}{{\mu - 1}}$.

Hence, the correct option is (C).

Note: Refractive index is basically the ratio of speed of light in medium to the speed of light in air. So when we take the refractive index of air with respect to air, it comes out to be 1. Refractive index has no dimensions and units as we have both speeds in numerator and denominator. Refraction takes place when light travels from one medium to another because in different mediums, the speed of light is different.