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Intensity at A due to source is I. Without concave mirror, then find out the intensity of A after placing concave mirror. |
Answer» Solution : `P=IxxP` `=Ixx4pi (60)^(2)` Intensity at `P=P/("AREA")` `=(4pi (60)^(2) I)/(4pi (30)^(2))` `=4I` Now `Delta PAB ~ Delta ACD` `rArr (AP)/(PB)=(AC)/(CD)` `rArr CD=30/60xxR=R/2` Energy at area of R radius `=4 I=pi R^(2)` Now energy will fall on the screen but at an Area of radius `R/2`. So intensity from mirror `=(4IxxpiR^(2))/(pi(R//2)^(2))=16 I` Total Intensity `=16 I+I=17 I` |
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