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

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The positions of the lens from the lamp for which sharp images are formed on the screen are 30 cm and 70 cm

The positions of the lens from the lamp for which sharp images are formed on the screen are 35 cm and 65 cm

Magnitude of the difference in magnification is 40/21

The size of the lamp filament for which there are two sharp images of 4.5 cm and 2 cm, is 3 cm

answer is B, C.

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Detailed Solution

$\frac{1}{100 - x} = \frac{1}{- x} = \frac{1}{21}$

$\Rightarrow x,$ The distance of object from the lens is 30 cm, 70 cm

$m_{1} = \frac{70}{- 30}, m_{2} = \frac{30}{- 70} ∴ | m_{1} - m_{2} | = \frac{40}{21}$

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