Two wavelengths 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 2 × 10^{−6} m. The distance between the slit and the screen is 1·5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.

#### Solution

For first maxima of the diffraction pattern we know `sintheta=(3lambda)/(2a)`

where *a* is aperture of slit.

For small values of *θ,* sinθ≈ tanθ = `y/D`

Where* y* is the distance of first minima from central line and *D* is the distance between the slit and the screen.

So

`y= (3lambda)/(2a)D`

For 590 nm,

`y_1=(3xx590xx10^(-9))/(2xx2xx10^(-6))xx1.5`

y_{1}=0.66375 m

For 596 nm

`y_2=(3xx596xx10^(-9))/(2xx2xx10^(-6))xx1.5`

y_{2}=0.6705 m

Separation between the positions of first maxima = *y*_{2}*− **y*_{1} = 0.00675 m