Question

An object is placed 30 cm away from a convex lens of focal length 10 cm and a sharp image is formed on a screen. Now a concave lens is placed in contact with the convex lens. The screen now has to be moved by 45 cm to get a sharp image again. The magnitude of focal length of the concave len is (in cm)

• A. 72
• B. 60
• C. 36
• D. 20

Answer 1

The correct option is D 20

An object is placed 30 cm in front of convex lens of focal length 10 cm :

Given : $u_1=-30$ cm $f_1=10$ cm

From lens formula, $\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$

$\therefore$ $\frac{1}{v_1}-\frac{1}{-30}=\frac{1}{10}$ $⟹v=15$ cm

Thus image is formed at a distance of 15 cm behind the lens.

Now a concave lens of focal length $f_2$ is placed in contact with convex lens. so the screen has to be shifted by 45 cm further away.

Now the new image distance $v_2=15+45=60$ cm

Let the focal length of combination of lens be $F$.

Using lens formula, $\frac{1}{v_2}-\frac{1}{u_1}=\frac{1}{F}$

$\therefore$ $\frac{1}{60}-\frac{1}{-30}=\frac{1}{F}$ $⟹F=20$ cm

Focal length of combination of lenses $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$

$\therefore$ $\frac{1}{20}=\frac{1}{10}+\frac{1}{f_2}$

$⟹f_2=-20$ cm (minus sign comes as the focal length of concave lens is negative)