Question

# A thin lens of focal length f and its aperture diameter d , forms a real image of intensity I. Now the central part of the aperture upto diameter (d2) is blocked by an opaque paper. If the new focal length is  xf and new image intensity is yI4,then xy is

Solution

## Initially energy/sec =I×π(d2)2=πd2I4 Now energy/sec =I[π(d2)2−π(d4)2] =Iπd2[316] So, Now  Intensity final  intensity initial = I πd23/16Iπd2/4=34 So, new intensity would be 3I4. Focus will not change. ∴x=1,y=3 ⇒xy=3 Physics

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