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

Answer: (C)

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,
$u = - (45 + 5) = - 50 \,cm$
$\therefore \frac{1}{v'} = \frac{1}{f} + \frac{1}{u'} $
$= \frac{1}{30} + \frac{1}{(-50)} = \frac{2}{150} $
$\Rightarrow v' = 75\,cm$
$\therefore $ Displacement of image
$= v - v' = 90 - 75 = 15 \,cm$, towards the lens