Here $a-\angle \mathrm{mm}, D=1.2 \mathrm{~m}, \lambda_1=650 \mathrm{~nm}, \lambda_2=520 \mathrm{~nm}$
(a) Distance of third bright fringe from the central maximum for the wavelength $650 \mathrm{~nm}$.
$$
y_3=\frac{3 \lambda D}{d}=\frac{3\left(650 \times 10^{-9}\right) 1.2}{2 \times 10^{-3}}=1.17 \mathrm{~m}
$$
(b) Let at linear distance ' $y$ ' from center of screen the bright fringes due to both wavelength coincides. Let $n_1$ number of bright fringe with wavelength $\lambda_1$ coincides with $n_2$ number of bright fringe with wavelength $\lambda_2$.
We can write
$$
\begin{aligned}
&y=n_1 \beta_1=n_2 \beta_2 \\
&n_1=\frac{\lambda_1 D}{d}=n_2 \frac{D \lambda_2}{d} \text { or } n_1 \lambda_1=n_2 \lambda_2 \ldots \text { (i) }
\end{aligned}
$$
Also at first position of coincide the $\mathrm{n}^{\text {th }}$ bright of one will coincide with $(\mathrm{n}+1)^{\text {th }}$ bright fringe of other.
If $\lambda_2 < \lambda_1$
So then $n_2>n_1$
then $n_2=n_1+1$
Using equation (ii) in equation (i)
$n_1 \lambda_1=\left(n_1+1\right) \lambda_2$
$n_1(650) \times 10^{-9}=\left(n_1+1\right) 520 \times 10^{-9}$
$65 n_1=52 n_1+52$ or $12 n_1=52$ or $n_1=4$
So, the fourth bright fringe of wavelength $520 \mathrm{~nm}$ coincides with 5 th bright fringe of wavelength $650 \mathrm{~nm}$