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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.
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`
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
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