Two slits in an opaque barrier each have a width of 0.020 mm and are separated by 0.050 mm. When coherent monochromatic light passes through the slits the number of interference maxima within the central diffraction maximum: Group of answer choices is 1 is 4 is 2 is 5 cannot be determined unless the wavelength is given

The answer is 5

Explanation:

The maximum interference is:

m * λ = d * sinθi

Where m = 0,1,2,3, ...

The first minimum diffraction is:

λ = a * sinθd

|sinθi| < sinθd

Where

(|m| * λ) / d < λ/a

|m| < d/a = 2.5

|m|max = 2

It can be concluded that coherent monochromatic light passes through the slits, therefore the maximum number of interference is 5.