Question

A convex lens with a focal length of 0.2 m and made of glass $\left( {\mu _g^{} = 1.5} \right)$ is immersed in water $\left( {\mu _w^{} = 1.33} \right)$. Find the change in the focal length of the lens.
A) 6.8cm
B) 5.8cm
C) 0.58m
D) 7.8cm

Answer 1

Hint: 1. The focal length of a lens depends upon the refractive index of the material of the lens and the radii of curvatures of the two surfaces.
2. The Refractive index of a lens material is different in different mediums i.e. it will have different values of refractive index in water and air.
3. The lens is manufactured using the lens makers formula

Complete step by step solution:
This question can be solved using the lens makers formula as it relates the focal length, refractive index, and radius of curvature.
Lens makers formula
\[\dfrac{1}{f} = \left( {\mu _g^{} - {\mu _m}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
Where,
$f$: Focal length
$\mu _g^{}$: Refractive index of glass in the air
\[\mu _m^{}\]: Refractive index of glass in medium (here air and water are two mediums)
The Refractive index of air is $\left( {{\mu _a} = 1} \right)$
${R_1}$​ and ${R_2}$ are the radius of curvature
Step 1: Apply the lens makers formula for the lens in air
\[\Rightarrow \dfrac{1}{{{f_a}}} = \left( {\mu _g^{} - {\mu _a}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
Substituting the values of the refractive index of glass\[\left( {\mu _g^{} = 1.5} \right)\], the refractive index of air \[\left( {\mu _a^{} = 1} \right)\] , and focal length of the lens in the air $\left( {{f_a} = 0.2m} \right)$ in the above equation we get,
\[
\Rightarrow \dfrac{1}{{0.2}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \\
   \Rightarrow \left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) = 10m \\
 \]
\[\]
Again apply lens makers formula but this time as the lens is immersed in water so we will be using
Refractive index of glass in water i.e. \[\mu _w^{} = 1.33\]
Substituting other values we get
\[
  {\dfrac{1}{f}_w} = \left( {\mu _g^{} - {\mu _w}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \\
   \Rightarrow {\dfrac{1}{f}_w} = \left( {1.5 - 1.33} \right)\left( {10} \right) \\
   \Rightarrow {\dfrac{1}{f}_w} = 1.7 \\
   \Rightarrow {f_w} = \left( {\dfrac{1}{{1.7}}} \right) \\
   \Rightarrow {f_w} = 0.588m \\
 \] The focal length of the lens changes from 0.2m to 0.58 m when it is immersed in water.

Final answer is (C), 0.588m.

The focal length of the lens changes from 0.2m to 0.58 m when it is immersed in water.

Note: 1. In the calculation, the units of the different terms should be taken care of.
2. The medium used on both sides of the lens should always be the same.
3. light doesn't need a medium through which to travel.
4. The refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in the medium. It has no units.
5. Speed of light is different in a different medium.
6. Lenses of different focal lengths are used for various optical instruments.