InterviewSolution
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InterviewSolution
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The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit isA. (a) `1:4:9`B. (b) `1:2:3`C. (c) `1:(4)/(9pi^2):(4)/(25pi^2)`D. (d) `1:(1)/(pi^2):(9)/(pi^2)` |
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Answer» Correct Answer - C `I=I_0[(sin alpha)/(alpha)]^2`, where `alpha=varphi/2` For `n^(th)` secondary maxima `dsintheta=((2n+1)/(2))lambda` `impliesalpha=varphi/2=pi/2[dsintheta]=((2n+1)/(2))pi` `:. I=I_0[(sin((2n+1)/(2))pi)/(((2n+1)/(n))pi)]^2=(I_0)/({((2n+1))/(2)pi}^2)` So `I_0:I_1:I_2=I_0:(4)/(9pi^2)I_0:(4)/(25pi^2)I_0` `=1:(4)/(9pi^2):(4)/(25pi^2)` |
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