Question

A bulb is located on a wall. Its image is to be obtained on a parallel wall with the help of a convex lens. If the distance parallel walls is $d$ then required focal length of lens placed in between the walls is:
A) Only $\dfrac{d}{4}$
B) Only $\dfrac{d}{2}$
C) More than $\dfrac{d}{4}$ but less than $\dfrac{d}{2}$
D) Less than or equal to $\dfrac{d}{4}$

Answer 1

Hint: The two walls are at a distance $d$ away from each other. We need to obtain the image of the bulb on the second wall using a convex lens, the lens is placed between the walls. To obtain a real image with a convex lens, the minimum distance between the screen and the object must be four times the focal length of the lens.

Complete step by step solution:
We are given that:
The distance between the walls, $d$
The lens is placed between the walls, distance from source (bulb), $\dfrac{d}{2}$ --equation $1$. The image is to be obtained on the parallel wall which is same distance from the lens as the distance between the lens and the bulb thus, we have $\left| v \right| = \left| u \right|$ where $v$ is the distance of the image from the lens and $u$ is the distance of source from image as the distance between lens and image is equal to the distance between the source.
But from equation $1$ , we have
Distance between lens and wall is $\dfrac{d}{2}$ , thus we get $\left| v \right| = \left| u \right| = \dfrac{d}{2}$
For a convex lens, image is formed only when the minimum distance between the source (bulb) and screen is $4f$ where $f$ is the focal length of the convex lens.
But the distance between the screen (wall) and the bulb is $d$ , we have
$d = 4f$
$ \Rightarrow f = \dfrac{d}{4}$
The required focal length of lens placed in between the walls is $\dfrac{d}{4}$.
Therefore, option A is the correct answer.

Note: For a convex lens, real image is obtained when the distance between the source and the screen is four times the focal length of the screen. In the given problem the magnification $m$ will be given as $m = \dfrac{v}{u}$ . In this problem, as the real image is formed $u,v$ will have opposite signs as they are on the opposite side of the lens.