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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. |
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Answer» SOLUTION :`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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