Question

A lens has a power of $+ 5\,{\text{diopter}}$ in the air. What will be its power if completely immersed in water? The refractive index of the lens in air is 1.5. (${\mu _w} = \dfrac{4}{3}$)
A.$+ \dfrac{5}{4}\,{\text{D}}$
B. $+ \dfrac{{21}}{4}\,{\text{D}}$
C. $+ \dfrac{{10}}{3}\,{\text{D}}$
D. $+ 5\,{\text{D}}$

Answer 1

Use the lens maker’s formula. This formula gives the relation between the power of the lens and refractive indices of the lens and the medium.

Formula used:
The lens maker’s formula is
$D = \left( {\dfrac{\mu }{{{\mu _m}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (1)
Here, $D$ is the power of the lens, ${R_1}$ and ${R_2}$ are the radii of curvature of the lens, $\mu$ is the refractive index of the lens and

Complete step by step answer:
The refractive indices of the lens, air and water are $1.5$, $1$ and $\dfrac{4}{3}$ respectively.
The power of the lens in the air is $+ 5\,{\text{diopter}}$.
Calculate the power of the lens in water.
Rewrite equation (1) for the power of the lens in the air.
${D_{la}} = \left( {\dfrac{\mu }{{{\mu _a}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Here, ${D_{la}}$ is the power of the lens in the air, $\mu$ is the refractive index of the lens and ${\mu _a}$ is the refractive index of air.
Substitute $+ 5\,{\text{diopter}}$ for ${D_{la}}$, $1.5$ for $\mu$ and $1$ for ${\mu _a}$ in the above equation.
$\Rightarrow + 5\,{\text{diopter}} = \left( {\dfrac{{1.5}}{1} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
$\Rightarrow + 5\,{\text{diopter}} = \left( {0.5} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
$\Rightarrow \left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) = 10$ …… (2)
Rewrite equation (1) for the power of the lens in the water.
$\Rightarrow {D_{lw}} = \left( {\dfrac{\mu }{{{\mu _w}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Here, ${D_{lw}}$ is the power of the lens in the water, $\mu$ is the refractive index of the lens and ${\mu _w}$ is the refractive index of water.
Substitute $1.5$ for $\mu$ and $\dfrac{4}{3}$ for ${\mu _a}$ in the above equation.
$\Rightarrow {D_{lw}} = \left( {\dfrac{{1.5}}{{\dfrac{4}{3}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
$\Rightarrow \left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) = \dfrac{{{D_{lw}}}}{{0.125}}$ …… (3)
Divide equation (3) by equation (2).
$\Rightarrow \dfrac{{\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}{{\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}} = \dfrac{{0.125}}{{10}}$
$\Rightarrow 10 = \dfrac{{{D_{lw}}}}{{0.125}}$
$\Rightarrow {D_{lw}} = + \dfrac{5}{4}\,{\text{D}}$
Therefore, the power of the lens when completely immersed in water is$+ \dfrac{5}{4}\,{\text{D}}$.
Hence, the correct option is A.

Note: Remember that the radii of curvature ${R_1}$ and ${R_2}$of the lens are constant whatever may be the medium and sign should be taken as positive if we measured the distance along the direction of incident rays and vice-versa.