Question

A convex lens forms a 2.5 times magnified virtual image when an object is placed at a distance of 8 cm from the lens. Calculate (a) the distance of the image from the lens (b) the focal length of lens.

Answer 1

Magnification (m) = 2.5

Distance of the object from the lens (u) = −8 cm [∵ u is always −ive]

Distance of the image from the lens (v) = ? [to be calculated]

Focal length of the lens (f) = ? [to be calculated]

(a) Applying, \[ m = \frac{v}{u} \]

\[\Rightarrow \ 2.5 = \frac{v}{8}\]

v = 2.5 × 8

∴ v = 20 cm

As the image is virtual, therefore v = −20 cm.

Thus, the image is formed at a distance of 20 cm from the lens, on the same side as the object.

(b) Applying, \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]

\[\Rightarrow \ \frac{1}{-20} - \frac{1}{-8} = \frac{1}{f}\]

\[ \Rightarrow \ \frac{1}{f} = -\frac{1}{20} + \frac{1}{8} \]

\[\Rightarrow \ \frac{1}{f} = \frac{-2 + 5}{40}\]

\[\Rightarrow \ \frac{1}{f} = \frac{3}{40}\]

\[\Rightarrow  f = \frac{40}{3}\]

∴ f = 13.3 cm

Thus, the focal length of the lens is 13.3 cm.