JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Geometrical Optics

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The distance between an object and a screen is 100 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 cm. If the power of the lens is close to

(\frac{N}{100}) D

where N is an integer, the value of N is

Answer: 476

Solution

This is the displacement (Bessel) method, in which a single lens forms a real image for two positions separated by

d = 40 cm

while the object-to-screen distance is

D = 100 cm

. The two conjugate distances are

u = - 30 cm

and

v = 70 cm

, so the lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

gives

\frac{1}{70} + \frac{1}{30} = \frac{1}{f}

, hence

f = 21 cm = 0.21 m

. The power is

P = \frac{1}{0.21} = \frac{100}{21} \approx 4.76 D

, and writing

P = \frac{N}{100}

gives

N \approx 476

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Integer answer

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JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Geometrical Optics

Solution

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