Question

A convex lens A of focal length $20cm$ and a concave lens B of focal length $5cm$ are kept along the same axis with the distance $d$ between them. If a parallel beam of light falling on A leaves B as a parallel beam, then the distance $d$ in $cm$ will be
(A). $15$
(B). $25$
(C). $30$
(D). $50$

Answer 1

You can start the solution by calculating the distance at which the convex lens produces the image of the parallel ray of light coming from infinity by using the lens, i.e. $\dfrac{1}{{{f_1}}} = \dfrac{1}{v} + \dfrac{1}{u}$ . We know that this image will acts as a virtual object for the concave lens, so again use the equation $\dfrac{1}{{{f_1}}} = \dfrac{1}{v} + \dfrac{1}{u}$ to calculate the distance of the virtual object from the concave lens. Then find the difference between the distance of the image formed by the convex lens and the virtual object of the concave lens to reach the solution.

Complete step-by-step answer :

For the convex lens, let’s assume that the distance of the object and the image formed from the convex lens be ${u_1}$ and ${v_1}$ . Let’s also assume that the focal length of the convex lens is ${f_1}$ .
For the concave lens let’s assume that the distance of the image formed by the convex lens and image formed by the concave lens form the concave lens be ${u_2}$ and ${v_2}$ . Let’s also assume that the focal length of the concave lens is ${f_2}$ .
Given, ${u_1} = \infty$ and ${v_2} = \infty$
Using the lens formula for the convex lens
$\dfrac{1}{{{f_1}}} = \dfrac{1}{v} + \dfrac{1}{u}$
$\dfrac{1}{{20}} = \dfrac{1}{{{v_1}}} + \dfrac{1}{\infty }$
${v_1} = 20cm$
So the image formed by the convex lens is at a distance of $20cm$ right from the convex lens. This will be a real image.
Using the lens formula for the concave lens
$\dfrac{1}{{\left( { - 5} \right)}} = \dfrac{1}{\infty } - \dfrac{1}{{{u_2}}}$
${u_2} = 5cm$
So the image formed by the convex lens (which in this case will act as the object for the concave mirror) is at a distance of $5cm$ right from the concave lens. This means that the object is virtual.
As we can see in the diagram that, $d = {v_1} - {u_2}$
$d = 20 - 15 = 15cm$
Hence, option A is the correct choice.

Note : You could have also solved this problem theoretically. We know that the convex lens would produce the image of a parallel ray coming from infinity at the focus ( $20cm$ away from it). We also know that a concave lens only forms a real image when the object is virtual. Also, the image formed is at infinity so the virtual object should be at the focus ( $5cm$ away from it). You could have used this information to reach the solution.