Question

A double convex lens is made of glass which has a refractive index of $1.55$ for violet rays and $1.50$ for red rays. If the focal length of violet rays is $20cm$, the focal length for red rays is:
A) $9cm$
B) $18cm$
C) $20cm$
D) $22cm$

Answer 1

To answer this question, we have to use the Lens Maker’s formula as it relates the focal length with the refractive index of the material. As violet rays and red have different wavelengths, their refractive index will also be different as given in the problem. We will use the Lens Maker’s formula for both the red and violet rays. Now, to eliminate the unknowns in the formula, we have to divide the two equations. This will give us an equation having the focal length of both the rays and their refractive index. Solving the equations will lead to the answer.

Formula used:
We have to use the relation between the focal length and the radius of curvature as given by the formula given below:
$\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where,
$f$ is the focal length.
${R_1}\,\& \,{R_2}$ are the radius of curvature of the lens of both the curved surfaces.
$\mu$ is the refractive index.

Complete answer:
Let the radius of curvature of the first sphere be ${R_1}$ and of the second sphere be ${R_2}$. Similarly, Let the focal lengths and refractive index for violet rays and red rays be ${f_v},{n_v}$ and ${f_r},{n_r}$ respectively.
We will use the lens maker’s formula for violet rays to get,
$\dfrac{1}{{{f_v}}} = \left( {{n_v} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)......(1)$
Similarly, using the lens maker’s formula for red rays we get,
$\dfrac{1}{{{f_r}}} = \left( {{n_r} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)......(2)$
As the radius of curvature of both the spheres is associated with the lens, therefore it will be the same for both the rays.
Therefore, dividing equation (1) by equation (2) we get the following equation,
$\dfrac{{\left( {\dfrac{1}{{{f_v}}}} \right)}}{{\left( {\dfrac{1}{{{f_r}}}} \right)}} = \dfrac{{\left( {{n_v} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}{{\left( {{n_r} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)}}$
On canceling the common terms we get,
$\Rightarrow \dfrac{{\left( {\dfrac{1}{{{f_v}}}} \right)}}{{\left( {\dfrac{1}{{{f_r}}}} \right)}} = \dfrac{{\left( {{n_v} - 1} \right)}}{{\left( {{n_r} - 1} \right)}}$
$\Rightarrow \dfrac{{{f_r}}}{{{f_v}}} = \dfrac{{\left( {{n_v} - 1} \right)}}{{\left( {{n_r} - 1} \right)}}$
$\Rightarrow {f_r} = {f_v} \times \dfrac{{\left( {{n_v} - 1} \right)}}{{\left( {{n_r} - 1} \right)}}......(3)$
Now, in the question, the values that are given are:
${f_v} = 20cm$
${n_v} = 1.55$
${n_r} = 1.50$
Putting the given values in the question in equation (3), we get,
$\Rightarrow {f_r} = 20 \times \dfrac{{\left( {1.55 - 1} \right)}}{{\left( {1.50 - 1} \right)}}$
$\Rightarrow {f_r} = 20 \times \dfrac{{0.55}}{{0.50}}$
$\Rightarrow {f_r} = 22cm$
Therefore, the focal length of red rays is $22cm$ and hence option (D) is correct.

Note: As a double convex lens is made up of two spherical surfaces, therefore it has two radii of curvature. The radius of curvature of the first sphere $({R_1})$ is positive as it is measured in the direction of propagation of light and the radius of curvature of the second sphere $({R_2})$is negative as it is measured in the opposite direction of the propagation of light. In a convex lens, the focal length is positive as it is measured in the direction of propagation of light.
For using the lens maker’s formula, we must make sure that the lens is thin as this would make sure that the separation between the two refracting surfaces will be small. Also, the lens maker’s formula can only be used if the medium on either side of the lens is the same.