Ray Optics

JEE Main level, numerical answer, medium

188 of 295

Question 188 of 295

The distance between an object and a screen is 100cm100cm. 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 40cm40cm. If the power of the lens is close to (N100)D\left(\dfrac{N}{100}\right)D where NN is an integer, the value of NN is _________

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Answer: 476

f=D2−d24D=1002−4024(100)=(100+40)(100−40)4(100)=21cmf=\dfrac{{D}^{2}-{d}^{2}}{4D}=\dfrac{{100}^{2}-{40}^{2}}{4(100)}=\dfrac{(100+40)(100-40)}{4(100)}=21cm
P=1f=10021=N100P=\dfrac{1}{f}=\dfrac{100}{21}=\dfrac{N}{100}
N=476.19N=476.19

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