Question

A concave lens with unequal radii of curvature made of glass $\left( {{\mu _g} = 1.5} \right)$ has a focal length of $40\,cm$. If it is immersed in a liquid of refractive index $\left( {{\mu _1} = 2} \right)$, then
A. It behaves like a convex lens of $80\,cm$ focal length
B. It behaves like a convex lens of $20\,cm$ focal length
C. Its focal length becomes $60\,cm$
D. Nothing can be said

Answer 1

Apply the lens maker’s formula to find the focal length and then find the resultant focal length when immersed in a liquid by applying the refractive index formula.

Complete step by step solution:
Let ${R_1}$ and ${R_2}$ be the radii of the curvature of the two surfaces of the lens. If f is its focal length in air, then by lens maker formula, we have
$- \dfrac{1}{f} = \dfrac{1}{2}\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right) \Rightarrow \dfrac{1}{f} = - \dfrac{1}{2}\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right)$

But it is given that$f = 40\,cm$, as the focal length of a concave lens is negative. Thus
$- \dfrac{1}{2}\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}} \right) = - 40 \Rightarrow \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} = 80$
As $f{\text{ }} > {\text{ }}0$ , it behaves like a convex lens.

Hence option (A) is correct.

When it is kept in a liquid of refractive index $\left( {{\mu _1} = 2} \right)$ its focal length is given by
$\dfrac{1}{f} = \dfrac{{1 - 2}}{{ - {R_1}}} + \dfrac{{2 - 1}}{{{R_2}}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} \Rightarrow \dfrac{1}{f} = 80$

Note: Lens maker’s formula is a relation that connects focal length of a lens to radii of curvature of the two surfaces of the lens and the refractive index of the material of the lens. It is useful to design lenses of desired focal length using suitable material and surfaces of suitable radii of curvature.
In case of spherical mirrors, there will be no change in the focal length of the mirrors when immersed in water. This is because the focal length of mirrors does not depend on the external medium in which it is kept.In case of lenses, refraction takes place. When a lens is immersed in water, the refractive index decreases. Hence its focal length increases. Therefore, When a concave lens is immersed in water its focal length increases as focal length of a lens is directly proportional to the refractive index of the medium and when it is immersed in a denser medium its focal length increases.