Question

450 nm of light falls on a single slit of width 0.30 mm. What is the angular width of the central diffraction peak

Answer 1

Angular width is 3 x 10^-3

Let D be the distance between source and screed d the distance between coherent source then for central diffraction maxima,

where λ is wavelength

Given:

λ = 450 nm = 450×10^−9m

d = 0.3x10^−3m, D = 1m

W = 2 x 450×10^−9/0.3x10^−3*1

To Find:

Angular width

Solution: The width of the central maxima is nothing but the difference between the positions of the first two minima. Hence we will use the expression for the position of minima and accordingly obtain the expression of the width of central maxima and secondary maxima

θ = W/D

θ = 2 x 450×10^−9/0.3x10^−3*1/1 = 3 x 10^-3

Hence, angular width is 3 x 10^-3

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