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A point source of light is placed 60 cm ...

A point source of light is placed 60 cm away from screen. Intensity detected at point P is `I` . Now a diverging lens of focal length 20cm is placed 20cm away from S between S and P. The lens transmits `75%` of light incident on it. Find the new value intensity at P.

Text Solution

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`u=-20,f=-20rArrv=-10`
Let P=Power of source
`I=(P)/(4pi(60)^(2))`
Energy receibed by lens,
`E_(2)=(P)/(4pi(20)^2)A_1`
`:. I_(2)=(0.75E_(2))/(A_(2))`
From similar triangles `(A_(2))/(A_(1))=25`
`:. I_(2)=0.27I`
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