Question

A converging lens of focal length 15 cm and a converging mirror of focal length 10 cm are placed 50 cm apart. If a pin of length 2.0 cm is placed 30 cm from the lens farther away from the mirror, where will the final image form and what will be the size of the final image?

Answer 1

Given,
Convex lens of focal length f_l = 15 cm
Concave mirror of focal length f_m = 10 cm
Distance between mirror and lens = 50 cm
Length of the pin (object length) h₀ = 2.0 cm



As per the question
The pin (object) is placed at a distance of 30 cm from the lens on the principle axis.
Using lens formula,
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f_{\mathrm{l}}} \\ \Rightarrow v=\frac{u{f}_{\mathrm{l}}}{u+f_{\mathrm{l}}} \end{gathered}$$

Since, u = −30 cm and f_l = 15 cm
$\mathrm{So},v=\frac{(-30)\times 15}{(-30+15)}=\frac{-450}{-15}=30\mathrm{cm}$
From the figure it can be seen that image of the object (AB) is real and inverted (A'B') and it is of the same size as the object. This image (A'B') is at a distance of 20 cm from the concave mirror, which is formed at the centre of curvature of the mirror. Thus, mirror will form the image (A'B') at the same place as (A''B'') and will be of the same size. Now, due to the refraction from the lens, the final image (A''B'') will be formed at AB and will be of the same size as the object (AB).