Question

A convex lens made up of glass $\left( {{\mu _g} = 1.5} \right)$ when placed in air has a focal length $4{\text{ cm}}$, if it is immersed in water $\left( {{\mu _w} = 1.33} \right)$ then what will be its focal length in water
A. $4{\text{ cm}}$
B. ${\text{8 cm}}$
C. ${\text{16 cm}}$
D. ${\text{32 cm}}$

Answer 1

Hint A convex lens is that type of lens which is thinner at edges and thicker at middle and it is also called converging lens as it converges the light rays coming from infinity to its focus.
The Radius of curvature of the lens doesn’t change when the surrounding medium changes. So, you can apply the lens maker formula to relate the focal lengths when the lens is kept in air and water.
The lens maker formula is given by $\dfrac{1}{f} = \left[ {\dfrac{{{n_1}}}{{{n_2}}} - 1} \right]\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ where $f$ is the focal length of the lens, ${n_1}$ and ${n_2}$ are the refractive index of the lens in consideration and the surrounding medium respectively and ${R_1}$ & ${R_2}$ are the radius of curvature of sphere 1 and sphere 2 of the lens.

Complete step by step answer
Let us first discuss a convex lens. It is that type of lens which is thinner at edges and thicker at middle and it is also called a converging lens as it converges the light rays coming from infinity to its focus.
As given in the question that the lens is immersed in water after being kept in air. We know that Radii of the curvature of the lens doesn’t change when the surrounding medium changes. So, we can apply the lens maker formula to relate the focal lengths when the lens is kept in air and water.
Lens maker formula is basically a relation between the focal length, refractive index of the lens and the surrounding medium and radii of curvature of the two spheres used in lenses. It is given by $\dfrac{1}{f} = \left[ {\dfrac{{{n_1}}}{{{n_2}}} - 1} \right]\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ where $f$ is the focal length of the lens, ${n_1}$ and ${n_2}$ are the refractive index of the lens in consideration and the surrounding medium respectively and ${R_1}$ & ${R_2}$ are the radius of curvature of sphere 1 and sphere 2 of the lens.
As given in the question, the lens made up of glass $\left( {{\mu _g} = 1.5} \right)$ when placed in air has a focal length $4{\text{ cm}}$. So, applying the lens maker formula in air using ${n_1} = {\mu _g} = 1.5$ and ${n_2} = 1$ (refractive index of air) we get
$\dfrac{1}{4} = \left( {\dfrac{{1.5}}{1} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …..(i)
Now, the lens is immersed in water $\left( {{\mu _w} = 1.33} \right)$. Let the focal length of the length in this case become $f$. So, applying lens maker formula in water using ${n_1} = {\mu _g} = 1.5$ and ${n_2} = {\mu _w} = 1.33$ we get
$\dfrac{1}{f} = \left( {\dfrac{{1.5}}{{1.33}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …..(ii)
Now by dividing equation (i) by (ii) we get
$\dfrac{f}{4} = \left( {0.5} \right)\left( {\dfrac{{1.33}}{{1.5 - 1.33}}} \right) = 0.5 \times 7.82 = 3.91$
On simplifying we have
$f = 4 \times 3.91 \approx 16$
So, the focal length of the lens when it is immersed in water is $16{\text{ cm}}$ .

Hence, option C is correct.

Note Lens maker’s formula is used by the Lens manufacturers to manufacture lenses of the focal length for desired purposes as different optical instruments required lenses of different focal lengths. The main fact of making this formula is that the focal length of a lens depends on the refractive index of the lens and its radii of curvature.