Question

A point object O is placed on the principal axis of a convex lens of focal length f = 20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. An eye is placed 60 cm to right of the lens and a distance h below the principal axis. The maximum value of h to see the image is

Options

• 0

• 2.5 cm

• 5 cm

• 10 cm.

Answer 1

2.5 cm

As the focal length of the lens is 20 cm and object distance is 40 cm from the lens, the image is formed at the centre of curvature at the right side of the lens.

From right angled triangle ABC,
\[\tan\alpha = \frac{AB}{BC}\]
\[\tan\alpha = \frac{5}{40}\]
\[\alpha = \tan^{- 1} \left( \frac{5}{40} \right)\]
and from right angled triangle, we have:
\[\tan\alpha = \frac{AB}{BC}\]
\[\tan\alpha = \frac{5}{40}\]
\[\alpha = \tan^{- 1} \left( \frac{5}{40} \right)\]
and from right angled triangle, we have:
\[\tan\alpha = \frac{AB}{BC}\]
\[\tan\alpha = \frac{5}{40}\]
\[\alpha = \tan^{- 1} \left( \frac{5}{40} \right)\]
Putting the value of angle alpha, we get:
\[\tan\left( \tan^{- 1} \left( \frac{5}{40} \right) \right) = \frac{h}{20}\]
\[\frac{5}{40} = \frac{h}{20}\]
\[ \Rightarrow h = 2 . 5 cm\]