Question

An equiconvex lens of focal length $15cm$ is cut into two halves. The focal length of each half is $40cm$.
A.True
B.False

Answer 1

Let us analyze the radius of each side of the equiconvex lens. Now, we cut the lens into half vertically and find a new radius of each side. To find the focal length of the lens, we can use the lens maker formula. First, we have to find the relation between focal length and radius before cutting the lens and use it to find the focal length of the lens after cutting the lens.
Formula used:
We have to use the relation between the focal length and the radius of curvature as given by the formula given below:
$\Rightarrow \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where,
$f$ is the focal length.
${R_1}\,\& \,{R_2}$ are the radius of curvature of the lens of both the curved surface.
$\mu$ is the refractive index.

Complete answer:
A lens is made up of glass having two curved surfaces or one curved and one plane surface. It produces an image by the refraction of light at its two curved bounding surfaces. A convex lens is a type to lens having a thicker centre and thinner edges. We know that it is a converging lens as they converge parallel beams of light at a point called the principal focus.
An equiconvex lens is a normal convex lens and is symmetrical on both sides. So, let us assume the radius of curvature of both sides of the lens be ${R_1}$ and ${R_2}$ respectively. As they are symmetric on both sides, we get the radius of curvature to be equal but as for the first part of the lens, the radius of curvature is measured in the direction of propagation of light and for the second part, the radius of curvature is measured against the direction of propagation of light. Therefore we get the relation between two radii of curvature to be:
$\Rightarrow {R_1} = - {R_2}$
In the question, it is given that before cutting the lens, the focal length of the lens is 15 cm. We can use the lens maker’s formula as given below:
$\Rightarrow \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ --- 1
Where,
$f$ is the focal length.
${R_1}\,\& \,{R_2}$are the radius of curvature of the lens of both the curved surface.
$\mu$is the refractive index.

Let us assume, ${R_1} = R$and ${R_2} = - R$ $\mu - 1 = z$ . Putting these values in the lens maker’s formula given by equation (1), we get:
$\Rightarrow \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
$\Rightarrow \dfrac{1}{{15}} = \left( z \right)\left( {\dfrac{1}{R} - \dfrac{1}{{\left( { - R} \right)}}} \right)$
Now, we add the terms in the bracket, take the variable to one side and calculate its value as follows:
$\Rightarrow \dfrac{1}{{15}} = \left( z \right)\left( {\dfrac{1}{R} + \dfrac{1}{R}} \right)$
$\Rightarrow \dfrac{1}{{15}} = \left( z \right)\left( {\dfrac{2}{R}} \right)$
$\Rightarrow \dfrac{z}{R} = \dfrac{1}{{30}}\,\,\,\,\,\,\,......\left( 2 \right)$
Now, we cut the lens vertically. So, the equiconvex lens becomes two plano-convex lens i.e. one side is curved having a radius $R$ and the other side acts as a plane surface and we know that the radius of curvature of the plane surface is infinity. Let us assume the focal length to be $f$ . Again we use the lens maker’s formula given by equation (1) and putting all the values we get,
$\, \Rightarrow \dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
$\Rightarrow \dfrac{1}{f} = \left( z \right)\left( {\dfrac{1}{R} - \dfrac{1}{{\left( \infty \right)}}} \right)$
$\Rightarrow \,\dfrac{1}{f} = \dfrac{z}{R}\,\,\,\,\,\,\,\,......(3)$
Now, we use the value of $\dfrac{z}{R}$ from equation (2) and put it in equation (3) we get,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{30}}$
Focal length after cutting the lens is $30\,cm$and not $40\,cm$ as given in the question and hence, it is a false statement.

Therefore, option (b) is correct.

Note:
If the lens is cut into half horizontally instead of vertically, it would not have any effect on the value of focal length that is the focal length of both the full and half lens would be the same.
The only change that will occur is the intensity of the light. As the lens is half, half of the light rays will get converged at focus and the intensity of the light would be halved as compared to the full lens.