In Fraunhofer diffraction by a single slit the width of the slit is 0.01 cm. If the wavelength of the light incident normally on the slit is 6000 A, the angular distance of the second maximum from the mid-line of the central maximum is

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(2) 0.015 rad. Given: a=0.01 cm=10⁻⁴ m, λ=6000 A=6×10⁻⁷ m, and the second maximum is wanted. Idea: the minima of a single slit lie at a sin θ=nλ, and the secondary maxima fall roughly midway between consecutive minima, at a sin θ=(n+1/2)λ. For the second secondary maximum, n=2: sin θ=(2.5λ)/a, and the angle is small enough that θ≈sin θ. θ=(2.5×6×10⁻⁷)/(10⁻⁴). θ=1.5×10⁻². θ=0.015 rad.

In Fraunhofer diffraction by a single slit the width of the slit is 0.01 cm. If the wavelength of the light incident normally on the slit is 6000 A, the angular distance of the second maximum from the mid-line of the central maximum is

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