Question

A lens of glass $\left( {\mu = 1.5} \right)$ of focal length $+ 10\,{\rm{cm}}$ is immersed in water $\left( {\mu = 1.33} \right)$. The new focal length is
A) $20\,{\rm{cm}}$
B) $40\,{\rm{cm}}$
C) $48\,{\rm{cm}}$
D) $12\,{\rm{cm}}$

Answer 1

Use the lens maker formula to calculate the unknown focal length. Before calculating the final focal length, split the situation into two parts, one part is when the glass is in the air and another situation is when the glass is kept inside the water.

Complete step by step solution:
Given,
The Refractive index of the glass is $\mu = 1.5$.
The focal length of the glass is $f = + 10\;{\rm{cm}}$.
The Refractive index of water is $\mu = 1.33$.
The lens maker formula can be written as,
$\dfrac{1}{f} = \left( {\dfrac{{{\mu _g}}}{{{\mu _a}}}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ ….. (1)
Here, ${\mu _g}$,${\mu _a},{R_1}$and ${R_2}$are the refractive index of glass, air, the radius of curvature of two surfaces respectively.
When the glass is kept in the air, then the ray of light will pass through the air first, then it will pass through glass. After substituting the value of $f$, this situation can be shown in terms of mathematics as follows from equation (1),
$\dfrac{1}{{10}} = \left( {\dfrac{{{\mu _g}}}{{{\mu _a}}}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ ….. (2)

And, when the glass is kept inside the water, then light ray passes through water first then it will pass through the glass material. Here, we can have,
$\dfrac{1}{f} = \left( {\dfrac{{{\mu _g}}}{{{\mu _w}}}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (3)

Dividing above equation (1) by equation (2),
$\dfrac{f}{{10}} = \left( {\dfrac{{\dfrac{{{\mu _g}}}{{{\mu _a}}} - 1}}{{\dfrac{{{\mu _g}}}{{{\mu _w}}} - 1}}} \right)$ ….. (4)

Substitute the values of ${\mu _g}$,${\mu _a},{R_1}$and ${R_2}$ equation (4), we can have,
$\begin{array}{l}\dfrac{f}{{12}} = \left( {\dfrac{{\dfrac{{1.5}}{1} - 1}}{{\dfrac{1}{{1.33}} - 1}}} \right)\\\f = 10 \times \left( {\dfrac{{\dfrac{{1.5}}{1} - 1}}{{\dfrac{{1.5}}{{1.33}} - 1}}} \right)\\\f = 38.40\;{\rm{cm}}\end{array}$

Hence, the required new focal length is $f = 38.40\,{\rm{cm}}$.

Note:
While using the lens maker formula, students should take care of the position of the refractive index of the different mediums. The student shouldn’t confuse whether a concave or convex lens is used in the given situation or not because the focal length is already given with sign.