Question

A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal axes coinciding. A beam of light travelling parallel to the principal axis and having a beam diameter 5.0 mm, is incident on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the emergent beam.

Answer 1

Given,
Focal length of the convex lens, f_d = 20 cm
Focal length of the concave lens, f_c = 10 cm
Beam diameter of the incident light, d = 5.0 mm
Distance between both the lenses is 10 cm.

As per the question, the incident beam of light is parallel to the principal axis.
Let it be incident on the convex lens.
Now, let B be the focus of the convex lens where the image by the convex lens should be formed.
For the concave lens,
Object distance (u) = + 10 cm (Virtual object is on the right of concave lens.)
Focal length, f_c = − 10 cm
Using the lens formula,
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{-10}+\frac{1}{+10}=\infty \\ \Rightarrow v=\infty \end{gathered}$$
Thus, after refraction in the concave lens, the emergent beam becomes parallel.
As shown, in triangles XYB and PQB,

$$\begin{gathered} \frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{RB}}{\mathrm{ZB}}=\frac{10}{20}=\frac{1}{2} \\ \mathrm{PQ}=\frac{1}{2}\times 5=2.5\mathrm{mm} \end{gathered}$$

Thus, the beam diameter of the emergent light is 2.5 mm.
Similarly, we can prove that if the beam of light is incident on the side of the concave lens, the beam diameter (d) of the emergent light will be 1 cm.