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In the diffraction due to a single-slit experiment, the aperture of the slit is 3mm. If monochromatic light of wavelength 620 mm is incident normally on the slit, calculate the separation between the first order minima and the 3rd order maxima on one side of the screen. the distance between the slit and the screen is 1.5 m. |
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Answer» Solution :Here aperture of the slit `a=3mm=3xx10^(-3)m`, wavelength of light `lamda=620nm=620xx10^(-9)m` and DISTANCE of screen from the slit, `D=1.5m` `therefore` Distance of first ORDER minima from central maxima point on the screen `x=(lamdaD)/(a)` and distance of 3rd order maxima from central maxima point `x.=((2xx3+1)lamdaD)/(2a)=(7lamdaD)/(2a)` `therefore` Separation between first order minima and THIRD order maxima `Deltax=x.-x=(7lamdaD)/(2a)-(lamdaD)/(a)=(5lamdaD)/(2a)=(5xx6200xx10^(-9)xx1.5)/(2xx3xx10^(-3))=7.75xx10^(-3)m=7.75mm` |
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