
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
134.7k+ views
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.
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.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Diffraction of Light - Young’s Single Slit Experiment

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
