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

$2f$

The correct answer is Option (2) → $2f$

Given: Point source at distance $u = 2f$ from a converging lens of focal length $f$

Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \text{ implies }\frac{1}{v} = \frac{1}{f} + \frac{1}{2f} = \frac{3}{2f} \text{ implies }v = \frac{2f}{3}$

To maximize intensity at the image, the source should be at the focal point of the lens ($u = f$) so that parallel rays are formed, converging to a bright image after focusing. At $u = 2f$, the image is formed at $v = 2f$, which is the lens's conjugate point.

Maximum intensity occurs when the object is at $2f$ and the image is also formed at $2f$ (real, inverted, and magnification = 1). This is the case for a point source placed at twice the focal length, forming a bright image of maximum intensity.

Image at distance $v = 2f$ on the other side of the lens