Question

An object 1.4 cm high when placed in front of a convex lens at a distance of 6 cm, forms a virtual image at a distance of 24 cm from the lens. Calculate (a) the focal length of the lens (b) the size of the image.

Answer 1

Height of the object (h_o) = 1.4 cm

Distance of the object from the lens (u) = −6 cm [∵ u is always −ive]

Distance of the image from the lens (v) = −24 cm [∵ Image is virtual]

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

Size of the image (h_i) = ? [to be calculated]

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

\[\Rightarrow \ \frac{1}{-24} - \frac{1}{-6} = \frac{1}{f}\]

\[ \Rightarrow \ \frac{1}{f} = -\frac{1}{24} + \frac{1}{6} \]

\[\Rightarrow \ \frac{1}{f} = \frac{-1 + 4}{24} \]

\[\Rightarrow \ \frac{1}{f}= \frac{3}{24}\]

\[ \Rightarrow \ f = \frac{24}{3}\]

f = 8 cm

Thus, the focal length of the lens is 8 cm.

(b) Applying, \[ \frac{h_{i}}{h_{o}} = \frac{v}{u} \]

\[\Rightarrow \ \frac{h_{i}}{1.4} = \frac{24}{6}\]

\[ \Rightarrow \ h_{i} = 1.4 \times 4  \]

\[\Rightarrow \ h_{i} = 5.6 \ \text{cm}\]

Thus, the size of the image is 5.6 cm.