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.

Explanation

(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`