Question

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

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

Answer 1

Answer: (C)

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