Question

A thin convex lens made from crown glass $\left( {\mu = \dfrac{3}{2}} \right)$ has focal length $f$. When it is measured in two different liquids having refractive indices $\dfrac{4}{3}$ and $\dfrac{5}{3}$, it has the focal lengths ${f_1}$ and ${f_2}$ respectively. The correct relation between the focal lengths is:
A) ${f_2} > f$ and ${f_1}$ becomes negative
B) ${f_1}$ and ${f_2}$ become negative
C) ${f_1} = {f_2} < f$
D) ${f_1} > f$ and ${f_2}$ becomes negative

Answer 1

Hint: In order to solve this question you have to know all the concepts related to lenses and apply the lens maker formula for both the liquids giving in the question. Then from that lens maker formula find the focal length for both the liquids and then compare.

Formula used:
The lens maker formula is given by,
$\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where, $f$ is the focal length
$\mu $ is the refractive index
${R_1}$ and ${R_2}$ are the radius of aperture.

Complete step by step solution:
According to the lens maker formula
$\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where, $f$ is the focal length
$\mu $ is the refractive index
${R_1}$ and ${R_2}$ are the radius of aperture.
Now taking the ratio of focal length in first liquid to the focal length in air
$\dfrac{{{f_1}}}{f} = \dfrac{{(\mu - 1)}}{{\left( {\dfrac{\mu }{{{\mu _m}}} - 1} \right)}}$
On putting the values of refractive indices given, we have
$ \Rightarrow \dfrac{{{f_1}}}{f} = \dfrac{{\left( {\dfrac{3}{2} - 1} \right)}}{{\left( {\dfrac{{\dfrac{3}{2}}}{{\dfrac{4}{3}}} - 1} \right)}}$
On further solving, we get
$ \Rightarrow {f_1} = 4f$

Now taking the ratio of focal length in second liquid to the focal length in air, we have
$\dfrac{{{f_2}}}{f} = \dfrac{{(\mu - 1)}}{{\left( {\dfrac{\mu }{{{\mu _m}}} - 1} \right)}}$
On putting the values of refractive indices given, we have
$ \Rightarrow \dfrac{{{f_2}}}{f} = \dfrac{{\left( {\dfrac{3}{2} - 1} \right)}}{{\left( {\dfrac{{\dfrac{3}{2}}}{{\dfrac{5}{3}}} - 1} \right)}} = - 5$
On the above equation, we have
$ \Rightarrow {f_2} < 0$
Hence, from the above solution we conclude that ${f_1} > f$ and ${f_2}$ becomes negative

Therefore, the correct option is (D).

Note: The focal length of the lens depends upon refractive index of lens with respect to the medium. The focal length of a lens is very much affected when immersed in water, after immersing the lens in the water, the focal length of the lens increases. The focal length of a lens is defined mainly by two properties of a lens that is the material's index of refraction and the curvature of the lens' surfaces.