Question

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.

Answer 1

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`

y₁=0.66375 m

For 596 nm

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

y₂=0.6705 m

Separation between the positions of first maxima = y₂− y₁ = 0.00675 m