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
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]
\[ \Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u}\]
\[\frac{1}{v} = \frac{1}{10} - \frac{1}{15}\]
\[ \Rightarrow v = 30 \text{ cm }\]
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
\[\frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1}\]
\[ \Rightarrow \frac{1}{v_1} = \frac{1}{5} - \frac{1}{10}\]
\[ \Rightarrow v_1 = 10 \text{ cm }\] 
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 \[m = \frac{h_i}{h_0} = \frac{v}{u}\]
\[ \Rightarrow h_i = - \frac{5 \times 30}{15}\]
\[ \Rightarrow h_i = - 10 mm\] 
Magnification by the second lens: \[\frac{h_{final}}{h_i} = \frac{v}{u}\]
\[ \Rightarrow \frac{10}{- 10} = \frac{h_{final}}{- 10}\]
\[ \Rightarrow h_{final} = 10 \text{ mm }\]
Thus, the image will be erect and real.
(c) Size of the final image is 10 mm.