# A beam of light consisting of two wavelengths,

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?

Solution:

Wavelength of the light beam, $\lambda_{1}=650 \mathrm{~nm}$

Wavelength of another light beam, $\lambda_{2}=520 \mathrm{~nm}$

Distance of the slits from the screen = D

Distance between the two slits = d

(i) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,

$x=n \lambda_{1}\left(\frac{D}{d}\right)$

For third bright fringe, n=3

Therefore $\mathrm{x}=3 \times 650 \frac{D}{d}=1950 \frac{D}{d} n m$

(b) Let, the $n^{t h}$ bright fringe due to wavelength $\lambda_{2}$ and $(n-1)^{t h}$ bright fringe due to wavelength $\lambda_{2}$ coincide

on the screen. The value of n can be obtained by equating the conditions for bright fringes:

$n \lambda_{2}=(n-1) \lambda_{1}$

520n = 650n – 650

650=130n

Therefore n = 5

Hence, the least distance from the central maximum can be obtained by the relation:

$\mathbf{x}=n \lambda_{2} \frac{D}{d}=5 \times 520 \frac{D}{d}=2600 \frac{D}{d} \mathrm{~nm}$