Question

The distance between an electric lamp and a screen is $d=1m$. A convergent lens of focal length $f=21cm$ is placed between the lamp and the lens such that a sharp image of the lamp filaments is formed on the screen.

• A. The positions of the lens from the lamp for which sharp images are formed on the screen are $35cm$ and $65cm$
• B. The position of the lens from the lamp for which sharp images are formed on the screen are $30cm$ and $70cm$
• C. Magnitude of the difference in magnification is $40/21$
• D. The size of the lamp filament for which there are two sharp images of $4.5cm$ and $2cm$, is $3cm$

Answer 1

Answer: (B), (C)

Magnitude of the difference in magnification is $40/21$
The position of the lens from the lamp for which sharp images are formed on the screen are $30cm$ and $70cm$

$\frac{1}{100-x}=\frac{1}{-x}=\frac{1}{21}$
$\Rightarrow x$, The distance of object from the lens is $30cm,70cm$
$m_1=\frac{70}{30},m_2=\frac{30}{-70}\therefore |m_1-m_2|=\frac{40}{21}$.