A beam of light consisting of two wavelengths,

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:

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