InterviewSolution
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InterviewSolution
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Two wavelength of sodium light `590 nm` and `596 nm` are used, in turn, to study the diffraction taking placed at a single slit of aperture `2 xx 10^(-4)m`. The distance between the slit and screen is `1.5 m`. Calculate the separation between the positions of first maxima of diffraction pattern obtained in the two cases. |
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Answer» (a) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of the two slits (b) Given that: Wavelength of the light beam, `lambda_(1)=590 nm=5.9xx10^(-7)m` Wavelength of another light beam, `lambda-(2)=596nm=5.96xx10^(-7)m` Distance the slits from the screen `=D=1.5m` Distance between the two slits `alpha=2xx10^(-4)m` For the first secondary maxima, `sintheta=(3lambda_(1))/(2alpha)=(chi_(1))/(D)` `chi_(1)=(3lambda_(1)D)/(2alpha) and chi_(2)=(3lambda_(2)D)/(2alpha)` `therefore` Separation between the positions of first secondary maxima of two sodium lines `chi_(1)-chi_(2)(3D)/(2alpha)(lambda_(2)-lambda_(1))` `=(3xx1.5)/(2xx2xx10^(-4))(5.96xx10^(-7)-5.9xx10^(-7))` `=6.75xx10^(-5)m` |
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