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Question
If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
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Solution
We know that intensity is directly proportional to the square of an amplitude
`I prop a^2`
if `I_1 = I/2`
if intensity reduced to 50%, the amplitude will be `a/sqrta2`
then `r = sqrt2`
`I_"max"/I_"min" = (r+1)^2/(r-1)^2 = (sqrt2 + 1)^2/(sqrt2 - 1)^2`
`I_"max"/I_"min" = ((2.414)/(0.414))^2 = (5.83)^2`
`I_"max"/I_"min"` = 33.98 ≅ 34
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