Question

An object of height 4.25 mm is placed at a distance of 10 cm from a convex lens of power +5 D. Find (i) focal length of the lens, and (ii) size of the image.

Answer 1

Object height, h = 4.25 mm = 0.425 cm
Object distance, u = -10 cm
Power, P = +5 D
Focal length, f = ?
Image distance, v = ?
Image height, h' =?

Power, $P=\frac{1}{f}f=\frac{1}{P}=\frac{1}{5}=0.2m=20cm$

Using the lens formula, we get:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}-\frac{1}{-10}=\frac{1}{20}\Rightarrow \frac{1}{v}+\frac{1}{10}=\frac{1}{20}\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{10}\Rightarrow \frac{1}{v}=-\frac{1}{20}v=-20cm$

Now magnification, $m=\frac{h^{\prime }}{h}=\frac{v}{u}$

Substituting the values we get:

$\frac{h^{\prime }}{0.425}=\frac{-20}{-10}{h}^{\prime }=2\times 0.425=0.85cm=8.5mm$

Thus the image is 8.5 mm long; It is also erect and virtual.