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A screen is placed at `50cm` from a single slit, which is illuminated with `600nm` light. If separation between the first and third minima in the diffraction pattern is `3.0mm`, then width of the slit is:A. (a) `0.4mm`B. (b) `0.1mm`C. (c) `0.3mm`D. (d) `0.2mm` |
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Answer» Correct Answer - D `y_3-y_1=(3lambdax)/(alpha)-(lambdax)/(alpha)` `implies3xx10^-3=(2xx600xx10^-9xx0.5)/(alpha)` `=a=0.2nm` |
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