Question

A solid transparent sphere has a small opaque dot at its centre. When observed from the outside, the apparent position of the dot will be
A. closer to the eye than the actual position
B. farther away from the eye than its actual position.
C. same as its actual position
D. independent of the refractive index of the sphere

Answer 1

The above problem can be resolved by applying the fundamental concept of lenses regarding its shape and size, which directly affects the position of image formed by the mirror. Moreover, the mathematical relation for the lens equation is used that relates the object distance as well as image distance with the refractive indices of the media, where the lens is kept.

Complete step by step answer:
Given: A solid transparent sphere has a small opaque dot at its centre.
Let the refractive index of the inner and outer medium be ${\mu _1}$ and ${\mu _2}$.
Assuming the solid transparent sphere, as the spherical lens of radius R. Applying the lens formula as,
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
Here, v and u are the image distance and the object distance respectively. Considering the outer medium to be air ( ${\mu _2} = 1$).
The formula is given as,
$\dfrac{1}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{1 - {\mu _1}}}{R}$
As, the object distance (u) is equivalent to radius of curvature of lens ( R), u = R.
Then the final expression is given as,

\dfrac{1}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{1 - {\mu _1}}}{R}\\\ \dfrac{1}{v} - \dfrac{{{\mu _1}}}{R} = \dfrac{{1 - {\mu _1}}}{R}\\\ \dfrac{1}{v} = \dfrac{1}{{ - R}} \end{array}$$ The above result shows that image formation will happen at the same place and therefore it is independent of the value of the refractive index of the medium. **So, the correct answer is “Option D”.** **Note:** In order to achieve the correct solution for the given condition, it is necessary to remember the concept under the lens formula. And one must have the clear idea about the position of the object, to predict the approximate position for the formation of the image of the object.