1.

Two beams of light of intensity I_(1), and I_(2)interfere to give an interference pattern. If the ratio of maximum intensity to that of minimum intensity is (16)/(4) then I_(1):I_(2)= .......

Answer»

`1:9`
`4:1`
`1:4`
`9:1`

Solution :`(I_(max))/(I_(min))=((a_(1)+a_(2))/(a_(1)-a_(2)))^(2)`
`(16)/(4)=((a_(1)+a_(2))/(a_(1)-a_(2)))^(2)`
`=2(a_(1)+a_(2))/(a_(1)-a_(2))^(2)`
`2a_(1)-2a_(2)=a_(1)+a_(2)`
`:.a_(1)=3a_(2)`
`:.(a_(1))/(a_(2))=(3)/(1)`
`:.(a_(1)^(2))/(a_(2)^(2))=(9)/(1)`
`:.(I_(1))/(I_(2))=(9)/(1) "" [ a_(1)^(2) PROP I_(1),a_(2)^(2) prop I_(2)`Taking K=1 in both]


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