$A$ 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?

(N/A) Given:
$\lambda_{1} = 650\; nm$
$\lambda_{2} = 520\; nm$
Let $D$ be the distance of the screen from the slits and $d$ be the distance between the slits.
$(a)$ The distance of the $n^{th}$ bright fringe from the central maximum is $x_n = n \lambda \frac{D}{d}$.
For the third bright fringe $(n=3)$ with $\lambda_{1} = 650\; nm$:
$x_3 = 3 \times 650 \times \frac{D}{d} = 1950 \frac{D}{d}\; nm$.
$(b)$ The bright fringes coincide when $n_1 \lambda_1 = n_2 \lambda_2$.
$n_1 (650) = n_2 (520)$
$\frac{n_1}{n_2} = \frac{520}{650} = \frac{4}{5}$.
The fringes coincide at the $4^{th}$ bright fringe of $\lambda_1$ and the $5^{th}$ bright fringe of $\lambda_2$.
The least distance is $x = 4 \times 650 \times \frac{D}{d} = 2600 \frac{D}{d}\; nm$.