Question

A double-slit pattern is observed on a screen 4m from the slits. Given that the light is incident normally and has a wavelength of 480nm, what is the minimum slit seperation for a point 4.4mm from the centre of the middle bright fringe to be
a) minimum?
b) maximum?

Answer 1

a) you have a first maximum at approx 8.8 mm for a distance of 4 m. At that point you have two wave fronts adding up in amplitude. This means the light leaving the two slits differ exactly one wavelenght of 480nm
if x is the distance between the slits,
\[x = \left( 4m \over 8,8mm \right).480nm\]
is 0.2181818 mm

Answer 2

for b) the first maximum occurs here at 4.4mm so formula becomes
\[x = \left( 4m \over 4.4mm \right).480nm = 4.3636.10^{-4} m = 0.43636 mm\]