Question

Optical axis of a thin equi-convex lens is the X-axis. The co-ordinate of a point object and its image are (-20 cm, 1 cm )and(25 cm, -2 cm) respectively :-

• A. the lens is located at = 5 cm
• B. the lens is located at = -9 cm
• C. the focal length of the lens is 10 cm
• D. the focal length of the lens is 15 cm

Answer 1

Answer: (C)

the focal length of the lens is 10 cm

Given,

The height of object is $h=1cm$

The height of image is $h^{\prime }=-2cm$

The magnification is given as

$m=\frac{h^{\prime }}{h}$

$m=\frac{-2}{1}=\frac{v}{u}$

$\left|\frac{v}{u}\right|=2\Rightarrow v=2u$

Now the total distance between the object and image is

$|v|+|u|=20+25\Rightarrow 2u+u=45$

$3u=45\Rightarrow u=15cm$

Hence $v=30cm$

Now using lens formula

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\Rightarrow \frac{1}{f}=\frac{1}{30}-\frac{1}{-15}$

$\frac{1}{f}=\frac{3}{30}\Rightarrow f=10cm$

Therefore the focal length of lens is $10cm$ .