Question

In Young’s double-slit experiment, suppose the separation between the two slits is d  = 0.320 mm. If a beam of 500 nm light strike the slits and produces an interference pattern. How many maxima will there be in the angular range  − 45.0° < θ < 45.0° ?

Answer 1

On the viewing screen, light intensity is a maximum when the two waves interfere constructively. This occurs when

d sin θ = nλ,    n = 0, ±1, ±2...

where λ is the wavelength of the light. At θ = 45.0°, 4 = 3.20 x 10^-4 m and  λ = 500 x 10^-9m, we get n = d sin θ/λ = 452.5

Thus, there are 452 maxima in the range 0 < θ = 45.5°. By symmetry, there are also 452 maxima in the range  -45.0°< θ < 0. Including the one for n = 0 straight ahead, the total number of maxima is N = 452 + 452 + 1 = 905