Question

The focal length of a thin lens made from the material of refractive index \[1.5\] is \[20cm\]. When it is placed in a liquid of refractive index \[4/3\], its focal length will be \[\_\_\_\_\_cm.\]
(A) \[80\]
(B) \[45\]
(C) \[60\]
(D) \[78\]

Answer 1

Hint: Since we have to find the variation in the focal length of the lens and the data that has been provided to us only includes the refractive indices of the lens and the medium in which the lens is placed, our best bet would be to apply lens maker formula; once for the lens in air and then for the lens submerged in the liquid. We can find the focal length of the submerged lens by comparison of the two expressions.

Formula Used:
\[\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)\]

Complete step by step answer:
Let the refractive index of the liquid be \[\mu \] and the refractive index of the lens be \[\eta \]. The refractive index of air is ideally taken to be \[1\]. Since there are two refractive surfaces of the lens, we will have two curvatures \[{{R}_{1}}\] and \[{{R}_{2}}\] such that \[{{R}_{1}}={{R}_{2}}=R\] (say).
The focal length of the lens in air \[f=20cm\]
Let the focal length of the lens in the liquid be \[{{f}_{L}}\] .
When the lens is placed in air, we can apply lens maker formula as
\[\begin{align}
  & \dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{R}+\dfrac{1}{R} \right) \\
 & \Rightarrow \dfrac{1}{f}=\left( \mu -1 \right)\dfrac{2}{R}---equation(1) \\
\end{align}\]
Now, when the lens is immersed in the liquid, we will apply lens maker formula as
\[\begin{align}
  & \dfrac{1}{{{f}_{L}}}=\left( \dfrac{\mu }{\eta }-1 \right)\left( \dfrac{1}{R}+\dfrac{1}{R} \right) \\
 & \Rightarrow \dfrac{1}{{{f}_{L}}}=\left( \dfrac{\mu }{\eta }-1 \right)\dfrac{2}{R}---equation(2) \\
\end{align}\]
Taking the ratio of \[equation(1)\] and \[equation(2)\], we get
\[\begin{align}
  & \dfrac{{{f}_{L}}}{f}=\dfrac{\left( \mu -1 \right)}{\left( \dfrac{\mu }{\eta }-1 \right)} \\
 & \Rightarrow {{f}_{L}}=\dfrac{\left( \mu -1 \right)}{\left( \dfrac{\mu }{\eta }-1 \right)}\times f \\
\end{align}\]
We know that \[\mu =1.5\] and \[\eta =\dfrac{4}{3}\] (given in the question)
So substituting these values, we get
 \[\begin{align}
  & {{f}_{L}}=\dfrac{\left( 1.5-1 \right)}{\left( \dfrac{1.5}{4/3}-1 \right)}\times 20cm \\
 & \Rightarrow {{f}_{L}}=\dfrac{\left( 1.5-1 \right)}{\left( 1.125-1 \right)}\times 20cm \\
 & \Rightarrow {{f}_{L}}=\dfrac{\left( 0.5 \right)}{\left( 0.125 \right)}\times 20cm \\
 & \Rightarrow {{f}_{L}}=4\times 20cm=80cm \\
\end{align}\]
We can see that the correct option is (A).

Note: At first sight, you might think that we cannot apply lens maker formula as the radius of curvature of the lens is not known to us, but since we had two situations, by comparing them we ruled out the common factor, that is, the radius of curvature and we did not need its value. The same can be done in a lot of questions when some data is left out. If we have two or more cases, we always start by assuming some value for the unknown term and then rule it out during solution.