Question

A convex glass lens is immersed in water. How will its power change in water?
A. increases
B. decreases
C. remain unchanged
D. increases for red colour and decreases for blue colour

Answer 1

Hint: The power of lens is defined as the reciprocal of the focal length of the lens. The lens maker’s formula can be used to solve this question. In this equation, the power of the lens is equivalent to the product of the change in refractive indices of the media and the difference of the reciprocal off the radii of curvature of both the surfaces. These all will help you to solve this question.

Complete Answer:
The power of the lens is given by the equation,
$P=\dfrac{1}{F}$
Where $F$be the focal length of the lens.
According to the lens maker’s equation, we can write that,
$P=\dfrac{1}{F}=\left( \dfrac{{{n}_{2}}}{{{n}_{1}}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
Where ${{n}_{2}}$be the refractive index of the lens, ${{n}_{1}}$ be the refractive index of the surrounding medium,${{R}_{1}}$ and ${{R}_{2}}$ be the radii of curvature of the lens. $P=\dfrac{1}{F}$$P=\dfrac{1}{F}$$P=\dfrac{1}{F}$
As the lens is dipped in water, the refractive index of the surrounding medium has been changed. That is previously it was air and now it is water. Therefore the refractive index will be increased as,
${{n}_{1}}=1\to \dfrac{4}{3}$
Therefore the ratio of the refractive index term in the equation gets decreased,
$\dfrac{{{n}_{2}}}{{{n}_{1}}}$ will get decreased, and also \[\left( \dfrac{{{n}_{2}}}{{{n}_{1}}}-1 \right)\] will be decreased. This will cause the reciprocal of the focal length \[\left( \dfrac{1}{F} \right)\] to decrease. And thus the power of the lens gets decreased.

Therefore the correct answer is given as option B.

Note:
Lens power is determined in dioptres (D). The convex lenses are having positive focal lengths, therefore they will be having a positive power value in air. The concave lenses have a negative focal length, so they will be having negative power value in air.