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If distance between source and screen increases by 2%, then intensity obtained at screen will be .......

Answer»

<P> INCREASED by 4%
increased by 2%
decreased by 2%
decreased by 4%

Solution :Intensity of light `I=(P)/(A)` [ P is same]
`:. I prop (I)/(r^(2))`
`:. (I_(2))/(I_(1))=((r_(1))/(r_(2)))^(2)` but `r_(1)=r`
`2% "of" r_(2)=r_(1)+r_(1)`
`=((r)/(1.02r))^(2) "" =r+rxx0.02`
`=1.02r `
`=((1)/(1.02))^(2)`
`=0.96`
`:. I_(2)=0.96 I_(1) ""I_(1) gt I_(2)`
`(I_(1)-I_(2))/(I_(1))xx100=(1-096I_(1))/(I_(1))xx100%`
`=0.04xx100%=` decreasesed by 4


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