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Light of wavelength `6328Å` is incident normally on slit having a width of `0.2mm`The width of the central maximum measured form minimum to minimum of diffraction pattens on a screen `9.0meter`away will be about-
A. `0.36^(@)`
B. `0.18^(@)`
C. `0.72^(@)`
D. `0.09^(@)`

1 Answer

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Best answer
Correct Answer - A
`asintheta=nlamda`
Since width of central maxima is twice the distance of `1^(st)` dark bands on both sides,
`2sintheta=(2xx1xxlamda)/(a)=(2xx6.328xx10^(-7))/(2xx10^(-4))=0.006328`
`:.2theta=sin^(-1)(0.006328)=0.36^(@)`

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