Question

A thin converging lens of focal length $f = 25\, cm$ forms the image of an object on a screen placed at a distance of $75\, cm$ from the lens. The screen is moved closer to the lens by a distance of $25\, cm$ . The distance through which the object has to be shifted, so that its image on the screen in sharp again is

• A. 37.5 cm
• B. 16.25 cm
• C. 12.5 cm
• D. 13.5 cm

Answer 1

Answer: (C)

12.5 cm

Answer 2

According to the first condition $f =25 \,cm , v=75 \,cm$ $u =?$ $\frac{1}{f} =\frac{1}{v}-\frac{1}{u}$ $\frac{1}{25} =\frac{1}{75}-\frac{1}{u}$ $\frac{1}{u} =\frac{1}{75}-\frac{1}{25}$ $\frac{1}{u} =\frac{1-3}{75}$ $u =-\frac{75}{2}=-37.5 \,cm$ According to the second condition $v_{1}=50 \, cm , f =25 \, cm , u_{1}=?$ $\frac{1}{f} =\frac{1}{v_{1}}-\frac{1}{u_{1}}$ $\frac{1}{25} =\frac{1}{50}-\frac{1}{u_{1}}$ $\Rightarrow \frac{1}{u_{1}}=\frac{1}{50}-\frac{1}{25}$ $\Rightarrow \frac{1}{u_{1}}=\frac{1-2}{50}$ $u_{1}=-50 \,cm$ So, the screen is sharp again is $\Delta u =u_{1}-u$ $=50-37.5$ $=12.5 \, cm$