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

Hint: 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.