A point source of light S is placed on the axis of a lens of focal length 20 cm as shown. A screen is placed normal to the axis of lens at a distance x from it. Then INCORRECT statement.

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Intensity variation at the centre of the screen from (x = 90 cm to x = 100 cm) and (x = 100 cm and x = 110 cm) is same.

As x is increased from zero, intensity continuously decreases.

Radius of the bright circle obtained on the screen is equal to 4/5 cm for x = 20 cm.

As x is increased from zero, intensity first increases then decreases.

answer is A.

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Detailed Solution

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} \Rightarrow \frac{1}{v} - \frac{1}{(- 25)} = \frac{1}{20}$

$\frac{1}{v} = \frac{1}{20} - \frac{1}{25} = \frac{1}{100} \Rightarrow V = 100 cm$
From O to I intensity increases and then decreases at x = 90 cm and 110 cm intensity is same (d)
Radius at x = 20 cm is equal to radius of lens.

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