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In double slit experiment , the distance between two slits is `0.6mm` and these are illuminated with light of wavelength `4800 Å`. The angular width of dark fringe on the screen at a distance 120 cm from slits will beA. `8xx10^(-4)` radB. `6xx10^(-4)` radC. `4xx10^(-4)` radD. `16xx10^(-4)` rad |
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Answer» Correct Answer - A Given, `d=0.6mm=0.6xx10^(-3)m` `lambda=4800"Å"=4.8xx10^(-7)m,n=1` Angular fringe width of first minima `(2x)/(D)=2(2n-1)(lambda)/(2d)=(2xx1-1)(lambda)/(d)` `therefore (2x)/(D)=((2-1)xx4.8xx10^(-7))/(0.6xx10^(-3))=8xx10^(-4)` rad |
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