Question

A thin glass (refractive index 1.5) lens has optical power of $- 5D$ in air. Its optical power in a liquid medium with refractive index 1.6 will be
A. 25 D
B. $- 25D$
C. 1 D
D. None of these

Answer 1

The focal length of the lens changes when we place it in the medium of different refractive index. Recall the formula for the focal length of the lens placed in the medium of refractive index ${\mu _m}$, and express the focal length in air and in liquid medium. Taking the ratio of the focal lengths, you will get the focal length of the glass in liquid medium in terms of focal length in the air medium. The optical power of the lens is inversely proportional to the focal length.

Formula used:
$\dfrac{1}{{{f_m}}} = \left( {\dfrac{{{\mu _g}}}{{{\mu _m}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Here, ${\mu _g}$ is the refractive index of the glass, ${R_1}$ and ${R_2}$ are the radii of curvatures of the two surfaces of the lens.

Complete step by step answer:
We know that when we place a lens in the medium of different refractive index, it behaves differently. The focal length of the lens changes after we place it in a different medium. We have the formula for the focal length of the lens placed in the medium of refractive index ${\mu _m}$.
$\dfrac{1}{{{f_m}}} = \left( {\dfrac{{{\mu _g}}}{{{\mu _m}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Here, ${\mu _g}$ is the refractive index of the glass, ${R_1}$ and ${R_2}$ are the radii of curvatures of the two surfaces of the lens.

Let us express the refractive index of the glass when it placed in the air as follows,
$\dfrac{1}{{{f_a}}} = \left( {\dfrac{{{\mu _g}}}{{{\mu _a}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (1)
Let us express the refractive index of the glass when it placed in the liquid medium as follows,
$\dfrac{1}{{{f_m}}} = \left( {\dfrac{{{\mu _g}}}{{{\mu _m}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (2)
Dividing equation (1) by equation (2), we get,
$\dfrac{{{f_m}}}{{{f_a}}} = \dfrac{{\dfrac{{{\mu _g}}}{{{\mu _a}}} - 1}}{{\dfrac{{{\mu _g}}}{{{\mu _m}}} - 1}}$
$\Rightarrow \dfrac{{{f_m}}}{{{f_a}}} = \left( {\dfrac{{{\mu _m}}}{{{\mu _a}}}} \right)\left( {\dfrac{{{\mu _g} - {\mu _a}}}{{{\mu _g} - {\mu _m}}}} \right)$

Substituting ${\mu _m} = 1.6$, ${\mu _a} = 1$ and ${\mu _g} = 1.5$ in the above equation, we get,
$\dfrac{{{f_m}}}{{{f_a}}} = \left( {\dfrac{{1.6}}{1}} \right)\left( {\dfrac{{1.5 - 1}}{{1.5 - 1.6}}} \right)$
$\Rightarrow \dfrac{{{f_m}}}{{{f_a}}} = - 8$ …… (3)
The relation between optical power and focal length of the lens placed in the medium of refractive index $\mu$. Let us express the optical power of the glass in the air as follows,
${P_a} = \dfrac{{{\mu _a}}}{{{f_a}}}$

Substituting ${P_a} = - 5$ and ${\mu _a} = 1$ in the above equation, we get,
$- 5 = \dfrac{1}{{{f_a}}}$
$\Rightarrow {f_a} = - \dfrac{1}{5}$
Substituting ${f_a} = - \dfrac{1}{5}$ in equation (3), we get,
${f_m} = \dfrac{8}{5}$
Let us express the optical power of the glass in the liquid medium as follows,
${P_m} = \dfrac{{{\mu _m}}}{{{f_m}}}$
Substituting ${\mu _m} = 1.6$ and ${f_m} = \dfrac{8}{5}$ in the above equation, we get,
${P_m} = \dfrac{{1.6}}{{8/5}}$
$\therefore {P_m} = 1\,{\text{D}}$

So, the correct answer is option C.

Note: Students must be able to identify the nature of the glass by looking at the focal length of the glass in the liquid medium. Since the focal length is positive, the glass must behave as a convex lens. The radii of curvatures of the glass remain the same in both the cases since the structure of the glass does not change.