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In Fraumhroffer diffraction pattern due to a single slit, the screen is at distance of 100 cm from the slit and slit is illuminated by `lambda=5893Å`. The width of the slit is 0.1 mm. Then separation between central maxima and the first secondary minimaA. `0.5893 cm`B. `5.893 cm`C. `0.2946 cm`D. `2.946 cm ` |
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Answer» Correct Answer - A `x=(flambda)/(d)=(100xx5893xx10^(-8))/(0.01)` 0.5893 cm |
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