Question

.A double-slit experiment is performed with light of wavelength 500 nm and bright interference fringes spaced 1.5 mm apart are observed on the viewing screen. What would the spacing between the fringes be if the wavelength were changed to 640 nm?

Answer 1

Answer:

spacing is 0.8307 mm

Explanation:

given data

wavelength = 500 nm

interference fringes N = 1.5 mm

wavelength = 640 nm

to find out

spacing between the fringes

solution

we know bright fringe spacing in double slit is express as

w × sinθ = N × wavelength

so angle is

sinθ = N × wavelength / w

and spacing is express as

spacing = d sinθ

and that is = N × (d/w) × wavelength

so

spacing = 1.5 × 360 / 650

so spacing is 0.8307 mm