A diverging lens of focal length 20 cm and a converging lens of focal length 30 cm are placed 15 cm apart with their principal axes coinciding. Where should an object be placed on the principal axis so that its image is formed at infinity?

#### Solution

Given,

Focal length of convex lens, *f*_{c} = 30 cm

Focal length of concave lens, *f*_{d} = 15 cm

Distance between both the lenses,* d* = 15 cm

Let (*f*) be the equivalent focal length of both the lenses.\[\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}\]

\[ = \frac{1}{30} + \left( \frac{1}{20} \right) - \left( \frac{5}{30( - 20)} \right)\]

\[ = \frac{1}{120}\]

\[\Rightarrow f=120 \text{ cm }\]

As focal length is positive, so it will be a converging lens.

Let '*d*_{1}' be the distance from diverging lens, so that the emergent beam is parallel to the principal axis and the image will be formed at infinity.

\[d_1 = \frac{df}{f_c} = \frac{15 \times 120}{30} = 60 \text{ cm }\]

It should be placed 60 cm left to the diverging lens. The object should be placed

(120 − 60) = 60 cm from the diverging lens

Let *d*_{2} be the distance from the converging lens. Then,

\[d_2 = \frac{df}{f_d} = \frac{15 \times 120}{20}\]*d*_{2} = 90 cm

Thus, it should be placed (120 + 90) cm = 210 cm right to converging lens.