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

Answer:

at 2f

Explanation:

hope it helped

Answer 2

The intensity on the other side of the lens is maximum at a distance is 2 f.

Explanation:

As the object is set at 2 f.

The target image is formed at a distance of 2 f from the converging lens.

It can be shown from the formula of the lens too:  

                     $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

The distance of object (u)  

The distance of image (v)  

f is focal length .    

Here is , u = − 2 f and f = f

When we put the respective values we get the following :

                   $\frac{1}{v}-\frac{1}{-2 f}=\frac{1}{f}$

                   $\frac{1}{v}=\frac{1}{f}-\frac{1}{2 f}$

                   $\frac{1}{v}=\frac{2 f-f}{2 f}$

                   $\frac{1}{v}=\frac{1}{2 f}$                                              

So, picture gap v = 2 f.

Thus, the option (c) 2 f is the correct answer.