Question

An object is placed at a distance of 30 cm from a converging lens of focal length 15 cm. A normal eye (near point 25 cm, far point infinity) is placed close to the lens on the other side. (a) Can the eye see the object clearly? (b) What should be the minimum separation between the lens and the eye so that the eye can clearly see the object? (c) Can a diverging lens, placed in contact with the converging lens, help in seeing the object clearly when the eye is close to the lens?

Answer 1

Object distance, u = $-$30 cm
Focal length, f = 15 cm
Image distance, v = ?
The lens formula is given by
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}-\frac{1}{-30}=\frac{1}{15} \\ \Rightarrow \frac{1}{v}=\frac{1}{30} \\ \Rightarrow v=+30\mathrm{cm} \end{gathered}$$
(on the opposite side of the object)

(a) No, the eye placed close to the lens cannot see the object clearly.

(b) The eye should be 30 cm away from the lens to see the object clearly.

(c) The diverging lens will always form an image at a large distance from the eye; this image cannot be seen through the human eye.