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In a single slit diffraction with lambda = 500 nm and a lens of diameter 0.1 mm then width of central maxima, obtain on screen at a distance of l m will be ...... |
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Answer» Solution :The WIDTH of central maximum means distance between first MINIMUM on both side of it. `2 SIN v_(1)=(2lambdaD)/(2)` `=(2xx5xx10^(-5)xx100)/(10^(-2))=10xx10^(-1) cm=10 mm` |
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