Question

A convex glass lens is immersed in water. Compared to the power in air, its power in water will
A. Increase
B. Remain the same
C. Diminish for red light and increase for blue light
D. Diminish

Answer 1

As we know that, the power of a lens is inverse of its power, i.e., $P=\dfrac{1}{f}$. Also, the refractive indices of air and water are 1 and $\dfrac{4}{3}$ respectively. So we can see that The refractive index of the medium is being increased in the question. To answer this question remember the “Lens Maker’s Formula”.

Formula used:
For this question, we will use maker’s formula, i.e.
$\dfrac{1}{f}=(\dfrac{{{\mu }_{2}}}{{{\mu }_{1}}}-1)(\dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}})$

Complete step by step answer:
As given in the hint,
$P=\dfrac{1}{f}$
Where f is focal length of the given lens
Now, to answer the given question let us start by applying the Lens Maker’s formula
$\dfrac{1}{f}=(\dfrac{{{\mu }_{2}}}{{{\mu }_{1}}}-1)(\dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}})$
Where, μ$_1$ is the refractive index of the glass, and μ$_2$ is refractive index of the medium in which the lens is being placed (in this case water)
And R$_1$ and R$_2$ are the radius of curvature of both the surfaces of the lens
Now, the new power of lens to be $P=\dfrac{1}{f'}$
So, in accordance with the maker’s formula,
$\dfrac{1}{f'}=(\dfrac{{{\mu }_{2}}}{{{\mu }_{1}}}-1)(\dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}})$
Now, as we know that refractive index of water is greater than that of air
So, the term $(\dfrac{{{\mu }_{2}}}{{{\mu }_{1}}}-1)$ decreases
Therefore,
$P' > P$
Where P is power of lens when placed in the air
Therefore, we can say that, the power of lens will decrease when it is immersed in the water
Option D is the correct one.

Note:
We could easily solve such questions by remembering the basic conclusion from the Lens makers formula, i.e., the power of the lens is inversely proportional to the refractive index of the medium. In other words, the power of the lens decreases as the refractive index of the medium increases.