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Question

# The angular width of central maxima in the Fraunhofer diffraction pattern is measured. The slit is illuminated by the light of wavelength 6000˚A If the slit is illuminated by a light of another wavelength, angular width decreases by 30%. The wavelength of light used is :

A
3500˚A
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B
4200˚A
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C
4700˚A
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D
6000˚A
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Solution

## The correct option is B 4200˚Adsinθ=nλn = 1, we havedsinθ=λIf angle is small, then sinθ=θdθ=λHalf angular width θ=λdFull angular width 2θ=2λdω′=2λdλ′λ=ω′ω⇒λ′=λ′ω′omegaλ′=6000×0.7=4200˚A

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