The distance between an object and a screen is $100 \mathrm{~cm}$. A lens can produce real image of the object on the screen for two different positions between the screen and the object. The distance between these two positions is $40 \mathrm{~cm}$. If the
power of the lens is close to $\left(\frac{N}{100}\right) D$ where $N$ is an integer, the value of $N$ is_________
$(476.19)$
Given,
Distance between an object and screen, $D=100 \mathrm{~cm}$
Distance between the two position of lens, $d=40 \mathrm{~cm}$
Focal length of lens,
$f=\frac{D^{2}-d^{2}}{4 D}=\frac{100^{2}-40^{2}}{4(100)}=\frac{(100+40)(100-40)}{4(100)}=21 \mathrm{~cm}$
Power, $P=\frac{1}{f}=\frac{100}{21}=\frac{N}{100}$
$\therefore N=476.19 .$
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