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प्रश्न
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.
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उत्तर
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λ1 = 590 nm = 590 × 10−9 m
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λ2 = 596 nm = 596 × 10−9 m
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Slit width a = 2 × 10−6 m
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Distance to screen D = 1.5 m
For λ1 = 590 × 10−9:
`y = (1.43lamdaD)/a`
`y_1 = (1.43xx590xx10^-9 xx 1.5)/(2xx10^-6)`
`= (1.43 xx 590 xx 1.5)/2 xx 10^-3`
`= (1.43xx885)/2 xx 10^-3`
`1264.55/2 xx 10^-3 = 632.275 mm`
For λ2 = 596 × 10−9:
`y_2 = (1.43 xx 596 xx 10^-9 xx 1.5)/(2 xx 10^-6)`
`= (1.43 xx 596 xx 1.5)/2 xx 10^-3`
`= 1277.22/2 xx 10^-3 = 638.61 xx 10^-3`
= 0.63861m = 638.61mm
Δy = y2 − y1 = 638.61 mm − 632.275 mm = 6.335 mm
Separation = 6.34 mm
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