Two convex lenses of focal length f and 2 f are arranged to have their foci at the same point as shown in the figure. A beam of parallel light of diameter 1 cm is incident into the first lens. The beam emerging from the second lens is

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The parallel beam converges at the focus of the first lens (f). This point acts as the object for the second lens. Since the foci of both lenses coincide, the object for the second lens is at its focus. Therefore, the beam emerging from the second lens will be parallel.

Let $d_{1}$ be the diameter of the beam incident on the first lens, and $d_{2}$ be the diameter of the beam emerging from the second lens.

Using similar triangles, we can relate the diameters to the focal lengths:

$\frac{d _{2}}{f _{2}} = \frac{d _{1}}{f _{1}}$

Given $d_{1} = 1$ cm, $f_{1} = f$ , and $f_{2} = 2 f$ , we have:

$\frac{d _{2}}{2 f} = \frac{1}{f}$

$d_{2} = \frac{2 f}{f} = 2 c m$

Therefore, the beam emerging from the second lens is parallel with a diameter of 2 cm.

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