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In a single slit diffraction pattern, the distance between first minima on the right and first minima on the left of central maximum is 4mm. The screeon on which the pattern is displaced is 2m from the slit and wavelength of light used is 6000Å. Calculate width of the slit and width of the central maximum. |
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Answer» SOLUTION :GIVEN: `lambda=6000Å`, `D=2m` CONSIDER that `X_(1)` and `X_(2)` be the distance between FIRST minima on the right and left respectively of central maximum. `X_(1)+X_(2)=4mm` `X_(1)+X_(2)=(2lambdaD)/(d)` `4xx10^(-3)=(2xx6xx10^(-7)xx2)/(d)` `d=6xx10^(-4)m` `d=0.6mm` `:. ` Width of central maximum, `W=X_(1)+X_(2)` `=4xx10^(-3)m` `=4mm`. |
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