Question

A 5 mm high pin is placed at a distance of 15 cm from a convex lens of focal length 10 cm. A second lens of focal length 5 cm is placed 40 cm from the first lens and 55 cm from the pin. Find (a) the position of the final image, (b) its nature and (c) its size.

Answer 1

Given:
Length of the high pin = 5.00 mm
Focal length of the first convex lens, f = 10 cm
Distance between the first lens and the pin = 15 cm
Focal length of the second convex lens, f₁ = 5 cm
Distance between the first lens and the second lens = 40 cm
Distance between the second lens and the pin = 55 cm

(a) Image formed by the first lens:
Here,
Object distance, u = − 15 cm
Focal length, f = 10 cm
The lens formula is given by

$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\ \frac{1}{v}=\frac{1}{10}-\frac{1}{15} \\ \Rightarrow v=30\mathrm{cm} \end{gathered}$$

Now,
This will be object for the second lens.
∴ Object distance for the second lens, $u_1$ = − (40 − 30)
$\Rightarrow u_1$ = − 10 cm
Focal length of the second lens, f₁ = 5 cm
The lens formula is given by

$$\begin{gathered} \frac{1}{v_1}=\frac{1}{f_1}+\frac{1}{u_1} \\ \Rightarrow \frac{1}{v_1}=\frac{1}{5}-\frac{1}{10} \\ \Rightarrow v_1=10\mathrm{cm} \end{gathered}$$
Therefore, the final position of the image is 10 cm right from the second lens.

(b) Magnification $\left(m\right)$ by the first lens is given by
$$\begin{gathered} m=\frac{h_i}{h_0}=\frac{v}{u} \\ \Rightarrow h_i=-\frac{5\times 30}{15} \\ \Rightarrow h_i=-10\mathrm{mm} \end{gathered}$$
Magnification by the second lens:

$$\begin{gathered} \frac{h_{\mathrm{final}}}{h_i}=\frac{v}{u} \\ \Rightarrow \frac{10}{-10}=\frac{h_{\mathrm{final}}}{-10} \\ \Rightarrow h_{\mathrm{final}}=10\mathrm{mm} \end{gathered}$$
Thus, the image will be erect and real.
(c) Size of the final image is 10 mm.