Question

A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Answer 1

Wavelength of the light beam, \(λ_1 = 650 nm\)
Wavelength of another light beam, \(λ_2 = 520 nm\)
Distance of the slits from the screen = D
Distance between the two slits = d
Distance of the \(n^{ th}\) bright fringe on the screen from the central maximum is given by the relation,
\(x=nλ_1(\frac{D}{d})\)
For third bright fringe, n = 3
\(∴ x= 3 ×650 \frac{D}{d} = 1950(\frac{D}{d}) nm\)
Let the \(n^{th}\) bright fringe due to wavelength \(λ_2\) and \((n-1)^{th}\) bright fringe due to wavelength \(λ_1\) coincide on the screen. We can equate the conditions for bright fringes as:
\(nλ_2 = (n-1)λ_1\)
520n = 650n-650
650 = 130n
∴ n = 5
Hence, the least distance from the central maximum can be obtained by the relation:
\(x = nλ_2\frac{D}{d}\)
\(= 5 ×520\frac{D}{d}=2600\frac{D}{d} nm\)
Note: The value of d and D are not given in the question.