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 2x${{10}^{-6}}$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.

For $λ_{ 1 }$ = 590 nm

Location of 1st maxima, $y_{ 1 }$ = (2n+1) D$λ_{ 1 }$ /2a

If n=1 ====> $y_{ 1 }$ = 3D $λ_{ 1 }$/2a

For $λ_{ 2}$ = 596nm

Location of 2nd maxima,$y_{ 2 }$= (2n+1) D $λ_{ 2}$/2a

If n=1 ====> $y_{ 2 }$ = 3D$λ_{ 2}$/2a

Path difference = $y_{ 2 }$ -$y_{ 1 }$ = 3D/2a($λ_{ 2}$ -$λ_{ 1 }$)

= 3x 1.5/2x2x${{10}^{-6}}$ (596-590) x ${{10}^{-9}}$

= 6.75 x ${{10}^{-3}}$ m