A point object is placed at a distance of 15 cm from a convex lens. The image is formed on the other side at a distance of 30 cm from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by 30 cm. Calculate the focal lengths of the two lenses.

#### Solution

Given,

Distance between point object and convex lens, *u* = 15 cm

Distance between the image of the point object and convex lens, *v* = 30 cm

Let* **f*_{c} be the focal length of the convex lens.

Then, using lens formula, we have:

\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f_c}\]

\[ \Rightarrow \frac{1}{f_c} = \frac{1}{30} - \frac{1}{( - 15)}\]

\[ \Rightarrow \frac{1}{f_c} = \frac{1}{30} + \frac{1}{15} = \frac{3}{30}\]

\[ \Rightarrow f_c = 10 \text{ cm }\]

Now, as per the question, the concave lens is placed in contact with the convex lens. So the image is shifted by a distance of 30 cm.

Again, let *v*_{f} be the final image distance from concave lens, then:

\[v_f\] = + (30 + 30) = + 60 cm

Object distance from the concave lens, *v *= 30 cm

If *f*_{d} is the focal length of concave lens then

Using lens formula, we have:

\[\frac{1}{v_f} - \frac{1}{v} = \frac{1}{f_d}\]

\[ \Rightarrow \frac{1}{f_d} = \frac{1}{60} - \frac{1}{30}\]

\[ \Rightarrow \frac{1}{f_d} = \frac{30 - 60}{60 \times 30} = \frac{- 30}{60 \times 30}\]

\[ \Rightarrow f_d = - 60 \text{ cm }\]

Hence, the focal length (*f*_{c}_{ }) of convex lens is 10 cm and that of the concave lens (*f*_{d} ) is 60 cm.