Question

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 1

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 =
`tan^(-1)((y)/(L))`

θ =
`tan^(-1)((0.0182)/(2.38))`

θ = 0.438°

b) width of slit d is given by

`d = (m\lambda)/(sin\theta)`

`d = (2* 574)/(sin* 0.438)`

d = 1.5 x 10⁻⁴