Two wavelength of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture `2xx10^(-6)`m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the position of first maxima of the diffraction pattern obtained in the two cases.

Explanation

For `lamda_(1)=590nm`
Location of 1 st maxima, `y_(1)=(2n+1)(Dlamda_(1))/(2a)`
If n = 1 `rArry_(1)=(3Dlamda_(1))/(2a)`
For `lamda_(2)=596nm`
Location of 2nd maxima,
`y_(2)=(2n+1)(Dlamda_(2))/(2a)`
If n = 1 `rArry_(2)=(3Dlamda_(2))/(2a)`
`therefore` Path difference = `y_(2)-y_(1)=(3D)/(2a)(lamda_(2)-lamda_(1))`
= `(3xx1.5)/(2xx2xx10^(-6))(596-590)xx10^(-9)`
= `6.75xx10^(-3)`m