Question

An object of height 2 cm is placed at a distance of 8 cm from a convex lens of focal length 10 cm. Calculate (a) position of the image (b) the size of the image. State the characteristics of the image.

Answer 1

Height of the object (h_o) = 2 cm

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

Focal length of the lens (f) = +10 cm [f is always +ive for a convex lens]

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

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

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

\[\Rightarrow \ \frac{1}{v} - \frac{1}{-8} = \frac{1}{10}\]

\[ \Rightarrow \ \frac{1}{v} = \frac{1}{10} - \frac{1}{8}\]

\[\Rightarrow \ \frac{1}{v} = \frac{4 - 5}{40}\]

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

\[\Rightarrow \ v = - \ 40 \ \text{cm}\]

Thus, the image is formed at a distance of 40 cm from the lens, on the same side as the object.

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

\[\Rightarrow \ \frac{h_{i}}{2} = \frac{40}{8}\]

\[ \Rightarrow \ h_{i} = 10 \ \text{cm} \]

Thus, the height of the image is 10 cm.

The image is virtual, erect and enlarged.