Question

A double convex lens (\[\mu = 1.5\]) in air has its focal length equal to 20 cm. When immersed in water (\[\mu = 4/3\]) its focal length would be
A. 20 cm
B. 80/9 cm
C. 360/9 cm
D. 80 cm

Answer 1

Hint: Use the lens maker’s formula to express the focal length of the lens for both water medium and air medium. Take the ratio of the focal lengths keeping the radii of curvature the same for both the cases as it is a constant quantity.

Formula used:
\[\dfrac{1}{f} = \left( {\dfrac{{{\mu _2}}}{{{\mu _1}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
Here, f is the focal length of the lens, \[{\mu _2}\] is the refractive index of the lens, \[{\mu _1}\] is the refractive index of the medium, \[{R_1}\] and \[{R_2}\] is the radius of curvatures of both surfaces of the lens.

Complete step by step answer:We know the lens maker’s formula for thin lens is,
\[\dfrac{1}{f} = \left( {\dfrac{{{\mu _2}}}{{{\mu _1}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]

Here, f is the focal length of the lens, \[{\mu _2}\] is the refractive index of the lens, \[{\mu _1}\] is the refractive index of the medium, \[{R_1}\] and \[{R_2}\] is the radius of curvatures of both surfaces of the lens.

We can express the focal length of the lens in air and water as follows,
\[\dfrac{1}{{{f_{air}}}} = \left( {\dfrac{{{\mu _{lens}}}}{{{\mu _{air}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\] ……. (1)
\[\dfrac{1}{{{f_{water}}}} = \left( {\dfrac{{{\mu _{lens}}}}{{{\mu _{water}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\] ….… (2)

Divide equation (1) by equation (2).
\[\dfrac{{\dfrac{1}{{{f_{air}}}}}}{{\dfrac{1}{{{f_{water}}}}}} = \dfrac{{\left( {\dfrac{{{\mu _{lens}}}}{{{\mu _{air}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}{{\left( {\dfrac{{{\mu _{lens}}}}{{{\mu _{water}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}\]
\[ \Rightarrow \dfrac{{{f_{water}}}}{{{f_{air}}}} = \dfrac{{\left( {\dfrac{{{\mu _{lens}}}}{{{\mu _{air}}}} - 1} \right)}}{{\left( {\dfrac{{{\mu _{lens}}}}{{{\mu _{water}}}} - 1} \right)}}\]
\[ \Rightarrow \dfrac{{{f_{water}}}}{{{f_{air}}}} = \dfrac{{\left( {\dfrac{{{\mu _{lens}} - {\mu _{air}}}}{{{\mu _{air}}}}} \right)}}{{\left( {\dfrac{{{\mu _{lens}} - {\mu _{water}}}}{{{\mu _{water}}}}} \right)}}\]
\[ \Rightarrow {f_{water}} = \left( {\dfrac{{{\mu _{lens}} - {\mu _{air}}}}{{{\mu _{lens}} - {\mu _{water}}}}} \right)\left( {\dfrac{{{\mu _{water}}}}{{{\mu _{air}}}}} \right){f_{air}}\]

Substitute 1 for \[{\mu _{air}}\], 1.5 for \[{\mu _{lens}}\], 4/3 for \[{\mu _{water}}\] and 20 cm for \[{f_{air}}\] in the above equation.
\[{f_{water}} = \left( {\dfrac{{1.5 - 1}}{{1.5 - \dfrac{4}{3}}}} \right)\left( {\dfrac{{\dfrac{4}{3}}}{1}} \right)\left( {20\,cm} \right)\]
\[ \Rightarrow {f_{water}} = \left( 4 \right)\left( {20\,cm} \right)\]
\[ \Rightarrow {f_{water}} = 80\,cm\]

So, the correct answer is option (D).

Note:Lens maker’s formula is used to construct the lens of desired focal length. We have got the positive focal length of the lens, which means the lens still acts as a convex lens. When the refractive index of the medium is greater than that of the lens, the focal length becomes negative. The focal length is negative for a concave lens. Therefore, when we place the convex lens in a medium of refractive index greater than the refractive index of the lens, the convex lens behaves as a concave lens.