Question

An object 4 cm high is placed at a distance of 40 cm from the optical centre of a concave lens of focal length 20 cm. Calculate (a) the position of the image (b) the size of the image. Also state the characteristics of the image.

Answer 1

Size of the object h_o = 4 cm

Distance of the object from the lens (u) = −40 cm  [u is a always −ive]

Focal length of the lens (f) = −20 cm [f for concave lens is −ive]

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

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

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

\[\Rightarrow \ \frac{1}{v} - \frac{1}{-40} = \frac{1}{-20}\]

\[ \Rightarrow \ \frac{1}{v} = -\frac{1}{20} - \frac{1}{40}  \]

\[\Rightarrow \ \frac{1}{v} = \frac{-2 - 1}{40}\]

\[\Rightarrow \ \frac{1}{v} = -\frac{3}{40}\]

\[ \Rightarrow \ v = -\frac{40}{3} \]

\[\Rightarrow \ v = - \ 13.33 \ \text{cm}\]

Thus, the image is formed at a distance of 13.33 cm on the same side as the object.

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

\[\Rightarrow \ \frac{h_{i}}{4} = \frac{40}{3 \times 40}\]

\[ \Rightarrow \ h_{i} = \frac{4}{3} \]

\[ \Rightarrow \ h_{i} = 1.33 \ \text{cm} \]

Thus, the size of the image is 1.33 cm.

The image is virtual, erect and diminished.