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The distance between the first and the sixth minima in the diffraction pattern of a single slit is 0.5 mm. The screen is 0.5 m away from the slit. If the wavelength of light used is 5000 Å. Then the slit width will beA. `2.5mm`B. `5 mm`C. `1.25 mm`D. `1 mm` |
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Answer» Correct Answer - A `x_(6)-x_(1)=(5Dlamda)/(a)` `:.a=(5Dlamda)/(x_(6)xxx_(1))=(5xx0.5xx5xx10^(-7))/(0.5xx10^(-3))` `=25xx10^(-4)m` |
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