Monochromatic light of wavelength 574 nm is incident on a narrow slit. on a screen 2.38 m away, the distance between the second diffraction minimum and the central maximum is 1.82 cm. (a) calculate the angle of diffraction θ of the second minimum. (b) find the width of the slit.

Answer

given,

Wavelength of the light =  574 nm

distance from the screen = 2.38 m

distance between  the second diffraction minimum and the central maximum =  1.82 cm  = 0.0182 m

a) Angle of diffraction for the second minimal

θ =

θ =

θ = 0.438°

b) width of slit d is given by

        d = 1.5 x 10⁻⁴

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explanation: