Question

The image of the needle placed 45 cm from a lens is formed on a screen placed 90 cm on the other side of the lens. The displacement of the image, if the needle is moved by 5.0 cm away from the lens is

• A. 10 cm, toward the lens
• B. 15 cm, away from the lens
• C. 15 cm, towards the lens
• D. 10 cm, away from the lens

Answer 1

The correct option is C 15 cm, towards the lens

Here, u = -45 cm, v = 90 cm
$\therefore$ $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{90}+\frac{1}{45}=\frac{1}{30}\Rightarrow f=30cm$
when the needle is move 5 cm away from the lens,
u=-(45 + 5) = -50 cm
$\therefore$ $\frac{1}{v^{\prime }}=\frac{1}{f}+\frac{1}{u^{\prime }}=\frac{1}{30}+\frac{1}{(-50)}=\frac{2}{150}=\Rightarrow v^{\prime }=75cm$
$\therefore$ Displacement of image = v-v' =90-75= 15 cm, towards the lens