Question

A convex lens has $20cm$ focal length in air. What is focal length in water? (Refractive index of air-water $ = 1.33$, refractive index for air-glass $ = 1.5$)

Answer 1

Hint: Here we are asked to find the focal length of the convex lens when it is in water. And we can solve this by using the formula of focal length of the convex lens. First we will write the focal length of the convex lens when it is in air then write the focal length of the convex lens when it is in water. Now taking the ratio of the two terms we will get the solution to this problem.

Complete step by step answer:
As per the problem we have a convex lens having $20\,cm$ focal length in air. We need to calculate the focal length of the convex lens in water.
We know, refractive index of air-water $ = 1.33$
Refractive index for air-glass $ = 1.5$
We know the formula for focal length of the convex lens in air as,
$\dfrac{1}{{{f_{air}}}} = \left( {\dfrac{{{\mu _g}}}{{{\mu _{air}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \ldots \ldots \left( 1 \right)$
Where, ${f_{air}}$ is the final length of lens in air, ${\mu _g}$ is the refractive index of the glass, ${\mu _{air}}$ is the refractive index of air and ${R_1}$, ${R_2}$ are the radii of the convex lens.

The formula for focal length of the convex lens in water is,
$\dfrac{1}{{{f_{water}}}} = \left( {\dfrac{{{\mu _g}}}{{{\mu _{water}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \ldots \ldots \left( 2 \right)$
Where, ${f_{water}}$ is the final length of lens in air, ${\mu _g}$ is the refractive index of the glass, ${\mu _{water}}$ is the refractive index of air and ${R_1}$, ${R_2}$ are the radii of the convex lens.
Now taking the ratio of the above two equation we will get,
$\dfrac{{{f_{water}}}}{{{f_{air}}}} = \dfrac{{\left( {\dfrac{{{\mu _g}}}{{{\mu _{air}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}{{\left( {\dfrac{{{\mu _g}}}{{{\mu _{water}}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}$

Cancelling the common term we will get,
$\dfrac{{{f_{water}}}}{{{f_{air}}}} = \dfrac{{\left( {\dfrac{{{\mu _g}}}{{{\mu _{air}}}} - 1} \right)}}{{\left( {\dfrac{{{\mu _g}}}{{{\mu _{water}}}} - 1} \right)}}$
Refractive index of air-water $ = 1.33$
Refractive index for air-glass $ = 1.5$
Refractive index of water $ = 1$
${f_{air}}$ is the final length of lens in air $ = 20\,cm$
Putting the known value we will get,
$\dfrac{{{f_{water}}}}{{20 = }} = \dfrac{{\left( {\dfrac{{1.5}}{1} - 1} \right)}}{{\left( {\dfrac{{1.5}}{{1.33}} - 1} \right)}}$
On further solving we will get,
$\therefore {f_{water}} = 78.23\,cm$

Hence, the focal length in water is $78.23\,cm$.

Note: Remember that focal length of any kind of lens depends on its matter in which it is placed due to which it varies from matter to matter as the refractive index varies. We can also note that focal length is directly proportional to the refractive index of the matter. As in this problem, the initial focal length of the convex lens increases.