Question

A double-slit experiment is performed with light of wavelength 640 nm. The bright interference fringes are spaced 1.6 mm apart on the viewing screen.What will the fringe spacing be if the light is changed to a wavelength of 360nm?

Answer 1

Answer:

1.44*10^-3m

Explanation:

Given that distance BTW two bright fringes is

DetaY = lambda* L/d

So for second wavelength

Deta Y2= Lambda 2* L/d

=lambda 2 x deta y1/ lambda1

So substituting

= 360 x 10^-9 x (1.6*10^-3/640*10^-9)

1.44*10^ -3m