Question

A convex refractive index $\dfrac{3}{2}$ has a power of $2.5\,{\text{D}}$ in air. If it is placed in a liquid of refractive index 2, then the new power of the lens is:
A. $- 1.25\,{\text{D}}$
B. $- 1.5\,{\text{D}}$
C. $1.25\,{\text{D}}$
D. $1.5\,{\text{D}}$

Answer 1

Use the Lens Maker’s formula. This formula gives the relation between the power of the lens in the medium, refractive index of the material of the lens, refractive index of the medium and radii of curvature of the lens. Rewrite this equation for power of lens in air medium and power of lens in water medium. Take division of these two equations and determine the new power of lens in water.

Formula used:
The Lens Maker’s formula is given by
$P = \left( {\dfrac{{{\mu _{material}}}}{{{\mu _{medium}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (1)
Here, $P$ is power of the lens, ${\mu _{material}}$ is the refractive index of lens material, ${\mu _{medium}}$ is the refractive index of the medium and ${R_1}$ and ${R_2}$ are radii of curvature of the lens.

Complete step by step answer:
We have given that the refractive index of the convex lens is $\dfrac{3}{2}$ and its power is $2.5\,{\text{D}}$ when it is in the air.
${\mu _{material}} = \dfrac{3}{2}$
$\Rightarrow{P_{air}} = 2.5\,{\text{D}}$
We know that the refractive index of the air medium is 1 and the refractive index of the water medium is given as 2.
${\mu _{air}} = 1$
$\Rightarrow{\mu _{water}} = 2$
We have asked to determine the power of the lens when the lens is in the water medium.

We can write equation (1) for the power ${P_{air}}$ of lens when the lens is in the air medium as
${P_{air}} = \left( {\dfrac{{{\mu _{material}}}}{{{\mu _{air}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (2)
We can write the equation (1) for the power ${P_{water}}$ of lens when the lens is in the water medium as
${P_{water}} = \left( {\dfrac{{{\mu _{material}}}}{{{\mu _{water}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (3)

Let now divide equation (3) by equation (2).
$\dfrac{{{P_{water}}}}{{{P_{air}}}} = \dfrac{{\left( {\dfrac{{{\mu _{material}}}}{{{\mu _{water}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}{{\left( {\dfrac{{{\mu _{material}}}}{{{\mu _{air}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}$
$\Rightarrow \dfrac{{{P_{water}}}}{{{P_{air}}}} = \dfrac{{\left( {\dfrac{{{\mu _{material}}}}{{{\mu _{water}}}} - 1} \right)}}{{\left( {\dfrac{{{\mu _{material}}}}{{{\mu _{air}}}} - 1} \right)}}$
Rearrange the above equation for ${P_{water}}$.
$\Rightarrow {P_{water}} = \dfrac{{\left( {\dfrac{{{\mu _{material}}}}{{{\mu _{water}}}} - 1} \right){P_{air}}}}{{\left( {\dfrac{{{\mu _{material}}}}{{{\mu _{air}}}} - 1} \right)}}$

Substitute $\dfrac{3}{2}$ for ${\mu _{material}}$, $2$ for ${\mu _{water}}$, $1$ for ${\mu _{air}}$ and $2.5\,{\text{D}}$ for ${P_{air}}$ in the above equation.
$\Rightarrow {P_{water}} = \dfrac{{\left( {\dfrac{{\dfrac{3}{2}}}{2} - 1} \right)\left( {2.5\,{\text{D}}} \right)}}{{\left( {\dfrac{{\dfrac{3}{2}}}{1} - 1} \right)}}$
$\Rightarrow {P_{water}} = \dfrac{{\left( { - \dfrac{1}{4}} \right)\left( {2.5\,{\text{D}}} \right)}}{{\left( {\dfrac{1}{2}} \right)}}$
$\therefore {P_{water}} = - 1.25\,{\text{D}}$
Therefore, the new power of lens in the water medium is $- 1.25\,{\text{D}}$.

Hence, the correct option is A.

Note: The students may think that we have written the same radii of curvature of the convex lens for both the equations for power of lens in air medium and power of lend in water medium because although the medium is changed, the construction of the convex lens remains the same. Hence, the radii of curvature remains the same in any medium.