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InterviewSolution
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A spherial light bulb with a diameter of`3.0cm` radiateds equally in all the directions with a power of `4.5piW.`(a find the light intensity of the surface of the bulb.(b) Find the light set up with its axis pointing towards the bulb.The lens has a circular face witb a diameter of `15.0cm`and a focal length of `30.0cm`Find the diamerter of the image of the bulb formed on a screen kept at the location of hte image .(d)Find the light intensity at the image. |
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Answer» Correct Answer - `(a)5000w//m^(2) (b)0.02W//m^(2)` ©0.125cm(d)`288W//m^(2)` (a)`I=(P)/(4piR^(2))` `I=(4.5pi)/(4pi(1.5xx10^(-2))^(2))rArr I=(4.5xx10^(4))/(4xx2.25)"watt"//m^(2)` `I=5000"watt"//m^(2)` (b) `I_(1)=(P)/(4pir^(2)) rArr I_(1)=(4.5pi)/(4pi(15//2)^(2))` `I_(1)=(4.5xx10^(4))/(4pixx(225)/(4)) "Watt"//m^(2) rArr I_(1)=(4.5)/(225)"Watt"//m^(2)` `I_(1)=0.02"Watt"//m^(2)` (c) `(h_(I))/(h^(@))=(v)/(u)=(f)/(f+u) rArr` since `h_(@)=d//2=1.5cm` hence`(h_(1))/(1.5)=(30)/(30+(-750)) rArr h_(1) =-(1)/(16)` diameter of the image=`2h_(1)` diameter of the image=`2xx(1)/(16)cm rArr` diameter of the image=`0.125cm` Light intensity at image`I_(2)=("energy received by the lens per second")/("area of image")` energy received per second=intensity at lens x area of lens `I_(1)xxpi(d_("lens")//2)^(2)` area of image=`pi(d_("image")//2)^(2)` `I_(2)=(I_(1)xxpi(d_("lens")//2)^(2))/(pi(d_("image")//2)^(2))rArr I_(2)=I_(1)xx(d_(("lens"))/(d_("image")^(2)))^(2) rArr I_(2)=0.02xx((15)/(0.125))^(2)` `=0.02xx(120)^(2)` `I_(2)=288 "Watt"//m^(2)` |
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