Question

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall $3m$ away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Answer 1

In order to solve this question, the concept of lens formula and the properties of convex lens is important. It should be kept in mind that the image forming on the wall is the real image of the electric bulb. The condition asked here is the condition of maxima.

Complete step by step answer:
Here it is given in the question that the distance between the object that is the electric bulb and the image is $3m$.
Let this distance be represented by variable $d$.
So here, $d = 3m$.
The lens used to create the image is a convex lens.
For the case of real image forming on the wall the minimum distance between the object and the wall (where the image is formed) should be $4f$ .
Here $f$ is the focal length of the lens.
Now the maximum focal length of the lens is given by:
${f_{max}} = \dfrac{d}{4}$
For real image to be obtained on the wall we have $d = 3m$
So, ${f_{\max }} = \dfrac{3}{4}$
On solving we get,
${f_{max}} = 0.75m$
So, the maximum possible focal length for the image to be obtained on the wall $3m$ away from the object is $0.75m$ .

Note: It is important to note that this question can be solved by using lens formula also. In that case first we have to assume the distance between the object and the lens as $x$ and then the distance between lens and image would become $3 - x$. The distance on the left hand side of the lens would be negative and the distance on the right hand side of the image is positive.