InterviewSolution
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InterviewSolution
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Light of wavelength 500 nm Pass through a slit of 0.2 mm wide. The diffraction pattern is formed on a screen 60cm away. Determine the. (i) angular spread of centeral maximum (ii) the distance between the central maximum and the second minimum. = |
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Answer» Solution :`lambda=500nmxx500xx10^(-9)m,a=0.2xx10^(-3)m,D=60cm=60xx10^(-2)m` (i) Equation for diffraction minimumis, `a sin 0 = n lambda` The central maximum is spread up to the first minimum. Hence, n = 1 Rewriting, `sin 0 (lambda)/(a) (or) = sin^(-1) ((lambda)/(a))`  `theta = 0.0025 rad` (ii) To find the value of `y_(1)` for maximum, which is spread up to first minimum with `(n = 1) is, a sin theta = lambda`  As `theta` is very small, `sinthetaapproxtantheta=(y_(1))/(D)` `a(y_(1))/(D)=lambda "rewriting", lambda_(1)=(yD)/(a)` Substituting, `y_(1)=(500xx10^(-9)xx60xx10^(-2))/(0.2xx10^(-3))=1.5xx10^(-3)=1.5 mm` To find the value of `y_(2)` for SECOND minimum with `(n=2)is,asintheta=2lambda` `a(y_(2))/(D)=2lambda "rewriting", y_(2)=(2lambdaD)/(a)` Substituting, `y_(2)=(2xx500xx10^(-9)xx60xx10^(-2))/(0.2xx10^(-3))=3xx10^(-3)=3mm` The distance between the central maximum and second minimum is, `y_(2) - y_(1) y_(2) - y_(1) = 3mm - 1.5 mm` NOTE: The above calculation SHOWS that in the differactionpatten caused by single slit, the width of each maximum is equalwith centralmaximum as the double that of others. But the bright and dark FRINGES are not of equal width. |
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