Question

A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will be-
(A). Remains unchanged
(B). Becomes small but not zero
(C). Becomes zero
(D). Becomes infinite

Answer 1

Hint: Image formation in lenses takes place by refraction. For refraction to take place, light has to travel between different mediums. The image formation in a lens is also determined by the focal length of the lens which depends on the surrounding medium, medium of the lens and the radii of curvature of the lens.

Formulae used:
$\dfrac{1}{f}=(\dfrac{{{\mu }_{1}}}{{{\mu }_{2}}}-1)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$

Complete step-by-step solution:
A convex lens is made of two curved surfaces bound together such that it is thicker in the middle and thinner at the edges. It can form both real and virtual images for different positions of the object. When real images are formed by it, by convention, its focal length and image distance is taken as positive and object distance is taken as negative.

The focal length of a lens depends on different factors like radius of curvature of the lens and the refractive indices of the lens as well as the surrounding medium.

The formula for lens is-
$\dfrac{1}{f}=(\dfrac{{{\mu }_{1}}}{{{\mu }_{2}}}-1)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$ - (1)
Here,
$f$ is the focal length
${{\mu }_{1}}$ is the refractive index of the surrounding medium
${{\mu }_{2}}$ is the refractive index of the medium in the lens
${{R}_{1}}$ is the radius of of the first curved surface
${{R}_{2}}$ is the radius of the second curved surface

Given that the refractive index of the surrounding medium is equal to the refractive index of the material of the lens.
Therefore,
$\dfrac{{{\mu }_{2}}}{{{\mu }_{1}}}=1$ - (2)

We substitute eq (1) in eq (2), to get,
$\begin{align}
  & \dfrac{1}{f}=(\dfrac{{{\mu }_{1}}}{{{\mu }_{2}}}-1)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right) \\
 & \Rightarrow \dfrac{1}{f}=(1-1)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right) \\
 & \Rightarrow \dfrac{1}{f}=0 \\
 & \therefore f=\infty \\
\end{align}$
The focal length becomes infinity, $f=\infty $.

Therefore, the correct option is (D).

Note:
Refraction is the phenomenon in which, when light travels from one medium to another, it deviates from its path and hence bending of light takes place. Here, since the refractive indices are the same, the densities of the mediums are same and hence no bending of light takes place. When travelling from a denser medium to a rarer medium, light bends away from the normal. On the other hand, when travelling from rarer medium to denser medium, it bends towards the normal.