Question

The two surfaces of a concave lens, made of glass of refractive index 1.5 have the same radii of curvature $R$ . It is now immersed in a refractive index $1.75$ , then the lens:
(A) Becomes a convergent lens of focal length $3.5R$
(B) Becomes a convergent lens of focal length $3.0R$
(C) Changes as a diverging lens of focal length $3.5R$
(D) Changes as a diverging lens of focal length $3.0R$

Answer 1

First of all, we will find out the refractive index of the lens with respect to the medium. After that we will use the lens maker’s formula, with appropriate signs of both the radii. We will manipulate accordingly and obtain the result.

Complete step by step solution:
We will go through the situation given in the question.We know from the laws of optics that when put in a medium with a higher refractive index, a lens reverses its behaviour. As we know in theory, we also need to prove it in support of our answer.Let us proceed to solve the problem. In order to proceed further, let us assume the lens as medium $1$ and the other medium in which the lens was dipped as medium $2$.We will now find out the refractive index of the first medium with respect to the second medium, which is given by the following calculation given below:
${n_{12}} = \dfrac{{{n_1}}}{{{n_2}}}$ …… (1)
Where,
${n_{12}}$ indicates the refractive index of medium one with respect to medium two.
${n_1}$ indicates the refractive index of the medium one i.e. the material of the lens.
${n_2}$ indicates the refractive index of the other medium in which the lens was dipped.
Now, we substitute the required values in the equation (1) and we get:
${n_{12}} = \dfrac{{{n_1}}}{{{n_2}}} \\\ \Rightarrow {n_{12}} = \dfrac{{1.5}}{{1.75}} \\\ \Rightarrow {n_{12}} = \dfrac{6}{7}$
Now, we will apply the lens maker’s formula in order to find the focus of the new system:
$\dfrac{1}{f} = \left( {{n_{12}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ …… (2)
Where,
$f$ indicates the new focal length.
${R_1}$ and ${R_2}$ indicate the radii of curvature of the two surfaces of the lens.
We know, according to the sign convention that:
${R_1} = - R$ and
${R_2} = R$
Now, we will substitute the required values in the equation (2) and we get:
$\dfrac{1}{f} = \left( {{n_{12}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \\\ \Rightarrow \dfrac{1}{f} = \left( {\dfrac{6}{7} - 1} \right)\left( {\dfrac{1}{{ - R}} - \dfrac{1}{R}} \right) \\\ \Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{7} \times \left( { - \dfrac{2}{R}} \right) \\\ \Rightarrow \dfrac{1}{f} = \dfrac{2}{{7R}} \\\ \Rightarrow f = \dfrac{{7R}}{2} \\\ \therefore f = + 3.5R$
Therefore, the focal length is found to be $+ 3.5R$ .We have found the focal length positive, which means it becomes convergent.

The correct option is A.

Note: While solving the problem, it is important to remember that a concave lens typically diverges light, but it converges parallel light rays to a focal point when inserted in a medium with a higher refractive index and vice-versa. It is also important to note that, both the radii of the two surfaces will have a specific sign convention. Left surface has a negative radius while the right surface has a positive radius according to the sign convention. If we ignore the sign convention, the focal length will come out as zero, which is wrong.