Question

The focal length of a lens is in the ratio $13:8$ when it is immersed in two different liquids of refractive indices $1.3$ and $1.2$ respectively. The refractive index of the material of the lens is
(A) $1.25$
(B) $1.56$
(C) $1.5$
(D) $0.5$

Answer 1

Hint
The ratio of the focal lengths of the lens are given in the question. Thus, using the values of the focal lengths in the lens material formula, we will find the refractive index of the material of the lens.
Formulae used:
$\Rightarrow \dfrac{1}{f} = \left( {\dfrac{{{n_2} - {n_1}}}{{{n_1}}}} \right)\left[ {\dfrac{1}{{R{}_1}} - \dfrac{1}{{{R_2}}}} \right]$
Where $f$ is the focal length, ${n_1},{n_2}$ are the refractive indices and ${R_1},{R_2}$ are the radii of curvature.

Complete step by step answer
The lens maker’s formula defines the relation between the focal length of a lens, radii of curvature of its two surfaces and the refractive index.
The formula for calculating the focal length given by the lens maker’s formula as,
$\Rightarrow \dfrac{1}{f} = \left( {\dfrac{{{n_2} - {n_1}}}{{{n_1}}}} \right)\left[ {\dfrac{1}{{R{}_1}} - \dfrac{1}{{{R_2}}}} \right]$
For the given case, we will consider the refractive indices of medium and that of the lens as $\Rightarrow {n_{medium}}$ and ${n_{lens}}$ respectively .
So, we have,
$\Rightarrow \dfrac{1}{f} = \left( {\dfrac{{{n_{lens}} - {n_{medium}}}}{{{n_{medium}}}}} \right)\left[ {\dfrac{1}{{R{}_1}} - \dfrac{1}{{{R_2}}}} \right]$
Now by substituting the given value of the refractive index of the first medium, which is $1.3$ , we can find the value of the focal length of the lens.
So, we have,
$\Rightarrow \dfrac{1}{{{f_1}}} = \left( {\dfrac{{{n_{lens}} - 1.3}}{{1.3}}} \right)\left[ {\dfrac{1}{{R{}_1}} - \dfrac{1}{{{R_2}}}} \right]$ …… (1)
Again, by substituting the given value of the refractive index of the second medium, which is $1.2$ , we can find the value of the focal length of the lens as,
$\Rightarrow \dfrac{1}{{{f_2}}} = \left( {\dfrac{{{n_{lens}} - 1.2}}{{1.2}}} \right)\left[ {\dfrac{1}{{R{}_1}} - \dfrac{1}{{{R_2}}}} \right]$ …… (2)
As we have obtained the expressions for the two different focal lengths, let us divide the equations (1) and (2) to carry out the further calculation as,
\Rightarrow \dfrac{{{\raise0.7ex\hbox{ 1 } \\!\mathord{\left/ {\vphantom {1 {{f_1}}}}\right.} \\!\lower0.7ex\hbox{ {{f_1}} }}}}{{{\raise0.7ex\hbox{ 1 } \\!\mathord{\left/ {\vphantom {1 {{f_2}}}}\right.} \\!\lower0.7ex\hbox{ {{f_2}} }}}} = \dfrac{{\left( {\dfrac{{{n_{lens}} - 1.3}}{{1.3}}} \right)\left[ {\dfrac{1}{{R{}_1}} - \dfrac{1}{{{R_2}}}} \right]}}{{\left( {\dfrac{{{n_{lens}} - 1.2}}{{1.2}}} \right)\left[ {\dfrac{1}{{R{}_1}} - \dfrac{1}{{{R_2}}}} \right]}}
Now the term $\left[ {\dfrac{1}{{R{}_1}} - \dfrac{1}{{{R_2}}}} \right]$ gets cancelled from numerator and denominator and we can simplify this equation as,
$\Rightarrow \dfrac{{{f_2}}}{{{f^1}}} = \left( {\dfrac{{{n_{lens}} - 1.3}}{{{n_{lens}} - 1.2}}} \right) \times \dfrac{{1.2}}{{1.3}}$
As we are already given with the ratio of the focal lengths, so we will substitute the same in the above equation.
The ratio of the focal length is given to be as follows.
$\Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{13}}{8}$
So the reciprocal is,
$\Rightarrow \dfrac{{{f_2}}}{{{f_1}}} = \dfrac{8}{{13}}$
Now substitute this value of the ratio of focal lengths in the above equation.
So, we get,
$\Rightarrow \dfrac{8}{{13}} = \left( {\dfrac{{{n_{lens}} - 1.3}}{{{n_{lens}} - 1.2}}} \right) \times \dfrac{{1.2}}{{1.3}}$
Continue the further calculation by doing cross multiplication, we get
$\Rightarrow 10.4({n_{lens}} - 1.2) = 15.6({n_{lens}} - 1.3)$
On opening the brackets we get
$\Rightarrow 10.4{n_{lens}} - 12.48 = 15.6{n_{lens}} - 20.28$
Rearranging the terms we can the refractive index of the material of the lens to one side as,
$\Rightarrow 20.28 - 12.48 = 15.6{n_{lens}} - 10.4{n_{lens}}$
On doing the calculation we have,
$\Rightarrow 7.8 = 5.2{n_{lens}}$
Therefore we get,
$\Rightarrow {n_{lens}} = 1.5$
$\therefore$ The refractive index of the lens is 1.5, thus, option (C) is correct.

Note
The focal length of any optical system measures how strongly the lens converges or diverges the light that is passing through it. So from this problem we can see that by changing the medium, we can change the focal length, that is the converging or the diverging power of the lens.