Question

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.

Answer 1

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:

$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f_c} \\ \Rightarrow \frac{1}{f_{\mathrm{c}}}=\frac{1}{30}-\frac{1}{(-15)} \\ \Rightarrow \frac{1}{f_{\mathrm{c}}}=\frac{1}{30}+\frac{1}{15}=\frac{3}{30} \\ \Rightarrow f_{\mathrm{c}}=10\mathrm{cm} \end{gathered}$$
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:
$v_{\mathrm{f}}$ = + (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:

$$\begin{gathered} \frac{1}{v_{\mathrm{f}}}-\frac{1}{v}=\frac{1}{f_{\mathrm{d}}} \\ \Rightarrow \frac{1}{f_{\mathrm{d}}}=\frac{1}{60}-\frac{1}{30} \\ \Rightarrow \frac{1}{f_d}=\frac{30-60}{60\times 30}=\frac{-30}{60\times 30} \\ \Rightarrow f_d=-60\mathrm{cm} \end{gathered}$$
Hence, the focal length (f_c ) of convex lens is 10 cm and that of the concave lens (f_d ) is 60 cm.