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

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 = $tan^{-1}(\dfrac{y}{L})$

θ = $tan^{-1}(\dfrac{0.0182}{2.38})$

θ = 0.438°

b) width of slit d is given by

 $d = \dfrac{m\lambda}{sin\theta}$

 $d = \dfrac{2\times 574}{sin\times 0.438}$

        d = 1.5 x 10⁻⁴