A point source of light is placed at a distance twice the focal length of a converging lens. If the focal length of the lens is \( f \), the intensity of the image of light source on the other side of the lens will be maximum at
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2f \)
The light is most concentrated where the lens brings the rays from the source together, that is at the position of the real image. Use the lens formula with the Cartesian sign convention (distances measured from the optical centre, positive in the direction of the incident light).
\( u = -2f \)
\( \text{focal length} = +f \)
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \;\Rightarrow\; \frac{1}{v} = \frac{1}{f} - \frac{1}{2f} = \frac{1}{2f} \)
\( \displaystyle v = +2f \)
So the image, and the maximum intensity, is at \( 2f \) on the other side of the lens.