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

(a) 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, $n=3$

$\therefore x=3 \times 650 \frac{D}{d}=1950\left(\frac{D}{d}\right) \mathrm{nm}$

(b) Let the $n^{\text {th }}$ bright fringe due to wavelength $\lambda_{2}$ and $(n-1)^{\text {th }}$ bright fringe due to wavelength $\lambda_{1}$ coincide on the screen. We can equate the conditions for bright fringes as:

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

$520 n=650 n-650$

$650=130 n$

$\therefore n=5$

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

$x=n \lambda_{2} \frac{D}{d}$

$=5 \times 520 \frac{D}{d}=2600 \frac{D}{d} \mathrm{~nm}$

Note: The value of d and D are not given in the question.