Question

Two slits separated by 1 mm in Young's double slit experiment are illuminated by the violet light of the wavelength 400 nm. The interference fringes are obtained on the screen placed at 1 m from the slits. Find the fringe width. If the violet light is replaced by the red light of the wavelength 700 nm, find the percentage change in fringe width.

Answer 1

Slit separation, d = 1 mm = 1 x 10^-3 m

wavelength \(\lambda\) = 6.5 x 10^-7 m

screen distance D = 1m

Distance b/w 3^rd dark fringe and 5^th bright fringe

Dark fringe y_n = \(\frac{(2n-1)\lambda D}{2d}\)

Bright fringe y_n = \(\frac{n\lambda D}d\)

y₅ - y₃ = \(\frac{5\times6.5\times10^{-7}\times1}{10^{-3}}\) - \(\frac{(2\times3-1)\times6.5\times10^{-7}}{2\times10^{-3}}\)

 = 32.5 x 10^-4 - 16.25 x 10^-4

= 16.25 x 10^-4 m

The distance b/w the third dark fringe and fifth bright fringe is 1.625 mm.