A point source of light is located at a distance of 2f from a...

Using the lens formula, $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

Given the object distance $u = -2f$, we have:

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

$\dfrac{1}{v} + \dfrac{1}{2f} = \dfrac{1}{f}$

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

$v = 2f$

The rays from the point source converge at the image point, which is at a distance of $2f$ on the opposite side of the lens. Since all the refracted rays converge at this point, the light energy is concentrated in a minimum area. Therefore, the intensity of light will be maximum at a distance of $2f$.