Question

What happens to the focal length of a convex lens when it is immersed in water? Refractive index of the material of the lens is greater than that of water.

Answer 1

Hint: We can solve this question by using the lens maker formula. First, we find the focal length of the lens when it is in the air using the lens maker formula. We then find the focal length of the lens when it is in water. By comparing the focal length in both the cases to get a relation between them we solve the problem.
Formula used: lens maker formula \[\dfrac{1}{f} = (\mu - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}})\]
Here,
Focal length of the lens is represented by $f$
Radius of curvature of the lens on both the surfaces is represented by ${R_1},{R_2}$ respectively
Refractive index of the lens with respect to the medium it is present in is represented by $\mu $

Step by step solution
The lens maker formula gives a relation between the focal length of the lens, refractive index of the lens and the radii of curvature of the lens.
$\dfrac{1}{f} = (\mu - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}})$
From the lens maker formula
$\dfrac{1}{{{f_a}}} = ({\mu _{l/a}} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}})$
This can also be written as
 $ \dfrac{1}{{{f_a}}} = (\dfrac{{{\mu _l}}}{{{\mu _a}}} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) $
$ \because {\mu _a} = 1 $
$ \dfrac{1}{{{f_a}}} = ({\mu _l} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) $….(1)
Here,
\[{\mu _{l/a}}\] is the refractive index of the lens with respect to air
\[{\mu _a}\] is the refractive index of air
${f_a}$ is the focal length of the lens in the air
When the lens is underwater the refractive index with respect to water becomes \[{\mu _{l/w}}\]
Substituting this in the lens maker formula we get
\[\dfrac{1}{{{f_w}}} = ({\mu _{l/w}} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) \]
\[\dfrac{1}{{{f_w}}} = (\dfrac{{{\mu _l}}}{{{\mu _w}}} - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) \]….. (2)
Here, \[\dfrac{{{\mu _l}}}{{{\mu _w}}}\]is the refractive index of lens with respect to water
${f_w}$ is the focal length of lens in water
From the question refractive index of lens is greater than refractive index of water
$ {\mu _l} > {\mu _w} $
$\therefore {\mu _l} > \dfrac{{{\mu _l}}}{{{\mu _w}}} $
Subtracting $1$ from both sides,
$\Rightarrow ({\mu _l} - 1) > (\dfrac{{{\mu _l}}}{{{\mu _w}}} - 1) $
$\Rightarrow f \propto \dfrac{1}{{\mu - 1}} $
Comparing equations (1) and (2),
$\dfrac{1}{{{f_a}}} > \dfrac{1}{{{f_w}}}$
We get
${f_w} > {f_a}$

Hence, the focal length of the lens increases when it is immersed in water.

Note: From this, we can say that the focal length is inversely proportional to the refractive index. It can be noted that ${\mu _l} > \dfrac{{{\mu _l}}}{{{\mu _w}}} $ this expression will not be true if ${\mu _w} < 1$. But it is known to us the refractive index of water is greater than 1 so our result is not affected.