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 = [tex]tan^{-1}(\dfrac{y}{L})[/tex]
θ = [tex]tan^{-1}(\dfrac{0.0182}{2.38})[/tex]
θ = 0.438°
b) width of slit d is given by
[tex]d = \dfrac{m\lambda}{sin\theta}[/tex]
[tex]d = \dfrac{2\times 574}{sin\times 0.438}[/tex]
d = 1.5 x 10⁻⁴
