Question

A thin glass (refractive lens 1.5) lens has optical power of -8D in air. Its optical power in a liquid medium with refractive index 1.6 will be:

Answer 1

Hint:Firstly, we will use the expression of power in terms of focal length and then use the lens maker’s formula which shows the reciprocal of focal length in terms of refractive indices and radius of curvature. Then we find the ratio of the power of the lens in air to that of medium and hence find an expression in terms of refractive indices.

Formula used:
Expression of the power of lens;
\[P = \dfrac{1}{f}\]
Here, $P$ is the power of the lens and $f$ is the focal length.
Lens maker’s formula;
\[\dfrac{1}{f} = (\dfrac{{{n_2}}}{{{n_1}}} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}})\]
Here, $f$ is the focal length, ${n_g}$ is the refractive index of the lens and ${n_a}$ is the refractive index of the material from where the light is incident. ${R_1}$ and ${R_1}$ is the radius of curvature.

Complete step by step answer:
We have a thin glass with a refractive index of 1.5 and an optical power of -8D in air. We have to find the optical power of the lens in a liquid medium of refractive index of 1.6. We know that the power of a lens is given by the formula;
\[P = \dfrac{1}{f}\]
First we need to find the focal length which we will do using the lens maker’s formula as follows:
\[\dfrac{1}{f} = (\dfrac{{{n_2}}}{{{n_1}}} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}})\]
Here, let us assume the refractive index of glass as ${n_2}$ and refractive index of air as ${n_2}$. Thus we have the power for air medium ${P_a}$ as ;
\[
{P_a} = (\dfrac{{{n_g}}}{{{n_a}}} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) \\
\Rightarrow {P_a} = (\dfrac{{1.5}}{1} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) \\
\]
For the other medium, the power is Pm and refractive index nm. Thus we have;
\[
{P_m} = (\dfrac{{{n_m}}}{{{n_g}}} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) \\
\Rightarrow {P_m} = (\dfrac{{1.5}}{{1.6}} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) \\
\]
Dividing the two powers, we get;
\[\dfrac{{{P_m}}}{{{P_a}}} = \dfrac{{(\dfrac{{1.5}}{{1.6}} - 1)}}{{(\dfrac{{1.5}}{1} - 1)}}\]
\[
\Rightarrow {P_m} = - 0.125 \times - 8 \\
\therefore {P_m} = 1D \\ \]
Hence, option B is the correct answer.

Note: Here, we are given the values for the refractive indices and the power of lenses in air, but we are not provided with the expression that is directly related between the two. Hence we derive the required formula through a combination of two different formulas that we already know. Also, the focal length should be in meters to get power in dioptre.