Question

When a convex lens is dipped in water, then it becomes
A. more converging
B. less converging
C. remains the same
D. none of these

Answer 1

When the convex lens is dipped in water, use the formula $\dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\left( \dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R} \right)$ with convection, for the refraction of light on both spherical surfaces of the lens. Then find the relation between the positions of the object (u) and is image (v). Place the object at infinity in both the mediums that is air water and calculate values of v in both cases. Compare the values of v and find out what happens to the lens.

Formula used:
$\dfrac{1}{v}-\dfrac{1}{u}=\left( {{\mu }_{l}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
$\dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\left( \dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R} \right)$

Complete step-by-step solution:
Suppose a convex lens is placed in air. An object is placed in front of the lens. Then the relation between the positions of the object and its image is given as $\dfrac{1}{v}-\dfrac{1}{u}=\left( {{\mu }_{l}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$ …..(i), (according to the sign convection).
Here u and v are the positions of the object and its image respectively. ${{\mu }_{l}}$ is the refractive index of the lens. ${{R}_{1}}$ is the radius of the spherical surface on which the rays are incident and ${{R}_{2}}$ is the radius of the other spherical surface. The values of ${{R}_{1}}$ and ${{R}_{2}}$ are also according to sign convection.
Let now dip the lens in water. Now the relation between u and v will not be the same as in the above case. Let us find the relation between v and u when the lens is in water. For this, we will use the formula of refraction through a spherical surface.
i.e. $\dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\left( \dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R} \right)$ … (ii).
Here, ${{\mu }_{1}}$ is the refractive index of the medium in which the rays are incident at the surface and the ${{\mu }_{2}}$ is the refractive index of the medium in which the rays enter after refraction at the surface.
R is the radius of the spherical surface.
Let place an object in the same position as that in air.
Let us apply the formula (ii) to the refraction at the surface of radius ${{R}_{1}}$.
Let the image from due this refraction be at position v’.
And, ${{\mu }_{1}}$ is the refractive index of water (${{\mu }_{w}}$) and ${{\mu }_{2}}$ is the refractive index of a lens (${{\mu }_{l}}$).
$\dfrac{{{\mu }_{l}}}{v'}-\dfrac{{{\mu }_{w}}}{u}=\left( \dfrac{{{\mu }_{l}}-{{\mu }_{w}}}{{{R}_{1}}} \right)$ …. (iii).
Then when the light rays refract at the other surface, ${{\mu }_{1}}$=${{\mu }_{l}}$ and ${{\mu }_{2}}$=${{\mu }_{w}}$ and R=u=v’ let the position of final image be v. Hence, we get,
$\dfrac{{{\mu }_{w}}}{v}-\dfrac{{{\mu }_{l}}}{v'}=\left( \dfrac{{{\mu }_{w}}-{{\mu }_{l}}}{{{R}_{2}}} \right)$ .. (iv).
Now add (iii) and (iv).
$\Rightarrow \dfrac{{{\mu }_{l}}}{v'}-\dfrac{{{\mu }_{w}}}{u}+\dfrac{{{\mu }_{w}}}{v}-\dfrac{{{\mu }_{l}}}{v'}=\left( \dfrac{{{\mu }_{l}}-{{\mu }_{w}}}{{{R}_{1}}} \right)+\left( \dfrac{{{\mu }_{w}}-{{\mu }_{l}}}{{{R}_{2}}} \right)$
$\Rightarrow {{\mu }_{w}}\left( \dfrac{1}{v}-\dfrac{1}{u} \right)=\left( {{\mu }_{l}}-{{\mu }_{w}} \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$.
$\Rightarrow \left( \dfrac{1}{v}-\dfrac{1}{u} \right)=\dfrac{\left( {{\mu }_{l}}-{{\mu }_{w}} \right)}{{{\mu }_{w}}}\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
$\Rightarrow \left( \dfrac{1}{v}-\dfrac{1}{u} \right)=\left( \dfrac{{{\mu }_{l}}}{{{\mu }_{w}}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$ …. (v)
Now, consider equation (i).
If we placed the object at an infinite distance i.e. $u=-\infty$, then
$\Rightarrow \left( \dfrac{1}{{{v}_{air}}}-\dfrac{1}{-\infty } \right)=\left( {{\mu }_{l}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
$\Rightarrow \left( \dfrac{1}{{{v}_{air}}}-0 \right)=\left( {{\mu }_{l}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
$\Rightarrow {{v}_{air}}=\dfrac{1}{\left( {{\mu }_{l}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)}$ … (vi).
Now, consider equation (v).
If we placed the object at an infinite distance i.e. $u=-\infty$, then
$\Rightarrow \left( \dfrac{1}{{{v}_{water}}}-\dfrac{1}{-\infty } \right)=\left( \dfrac{{{\mu }_{l}}}{{{\mu }_{w}}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
$\Rightarrow \left( \dfrac{1}{{{v}_{water}}}-0 \right)=\left( \dfrac{{{\mu }_{l}}}{{{\mu }_{w}}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$
$\Rightarrow {{v}_{water}}=\dfrac{1}{\left( \dfrac{{{\mu }_{l}}}{{{\mu }_{w}}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)}$.
We can understand that $\left( \dfrac{{{\mu }_{l}}}{{{\mu }_{w}}}-1 \right)$ < $\left( {{\mu }_{l}}-1 \right)$.
Therefore, ${{v}_{water}}$ > ${{v}_{air}}$.
This means that the lens is less converging in water than air.
Hence, the correct option is B.

Note: We can also answer this question by calculating and comparing the focal length of the lens in air and water.
The formula for focal length is given as $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$, where f is the focal length of the lens.
If we substitute the $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ in (i) and (v), we find that the focal length of the lens in the water is greater than that of in air.
Since, the focal length of the lens increases, we can say that it becomes less converging.