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

Here, $u=-45 cm , v=90 cm$

$\therefore \frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{90}+\frac{1}{45}=\frac{1}{30}\Rightarrow f=30 cm$

When the needle is moved 5 cm away from the lens, 

$u^{\text{'}}=-\left(45+5\right)=-50 cm$

$\therefore \frac{1}{v^{\text{'}}}=\frac{1}{f}+\frac{1}{u^{\text{'}}}=\frac{1}{30}+\frac{1}{\left(-50\right)}=\frac{2}{150}\Rightarrow v^{\text{'}}=75 cm$

$\therefore$ Displacement of image $=v-v\text{'} =90 -75 = 15 cm$, towards the lens