Question

A point source of light is placed at a distance of 2 f from a converging lens of focal length f. The intensity on the other side of the lens is maximum at a distance
(a) f
(b) between f and 2 f
(c) 2 f
(d) more than 2 f.

Answer 1

(c) 2 f

Since the object is placed at 2 f, the image of the object will be formed at distance of 2 f from a converging lens.
It can also be shown from the lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Here, u = − 2 f and f = f
On putting the respective values we get:
$$\begin{gathered} \frac{1}{v}-\frac{1}{-2f}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{2f} \\ =\frac{1}{2f} \end{gathered}$$
Therefore, image distance v = 2 f