Let "y' be the linear distance between the centre of screen and the point at which the bright fringes due to both wavelength coincides. Let ${\mathrm{n}}_1^{\text{th}}$ number of bright fringe with wavelength ${\mathrm{\lambda }}_1$ coincides with ${\text{n}}_2^{\text{m}}$ number of bright fringe with wavelength ${\mathrm{\lambda }}_2$ .
$\therefore {\text{n}}_1{\mathrm{\beta }}_1={\text{n}}_2{\mathrm{\beta }}_2$
$\Rightarrow {\text{n}}_1\frac{{\mathrm{\lambda }}_1\text{D}}{\text{d}}={\text{n}}_2\frac{{\mathrm{\lambda }}_2\text{D}}{\text{d}}$
$\Rightarrow {\mathrm{n}}_1{\mathrm{\lambda }}_1={\mathrm{n}}_2{\mathrm{\lambda }}_2$ …(i)
Also at first position, the ${\mathrm{n}}^{\text{th}}$ bright fringe of one will coincide with ${(\text{n}+1)}^{\text{th}}$ bright of other.
If ${\mathrm{\lambda }}_2<{\mathrm{\lambda }}_1$
So then ${\mathrm{n}}_2>{\mathrm{n}}_1$
$\Rightarrow {\text{n}}_2={\text{n}}_1+1$ …(ii)
Using equation (ii) and equation (i),
${\mathrm{n}}_1{\mathrm{\lambda }}_1=\left({\mathrm{n}}_1+1\right){\mathrm{\lambda }}_2$
$\Rightarrow \left({\mathrm{n}}_1\right)\left(650\times {10}^{-9}\right)=\left({\mathrm{n}}_1+1\right)\left(520\times {10}^{-9}\right)$
$\Rightarrow 65{\mathrm{n}}_1=52{\mathrm{n}}_1+52$
$\Rightarrow 13{\mathrm{n}}_1=52$
$\Rightarrow {\mathrm{n}}_1=4$
Thus, $\mathrm{y}={\mathrm{n}}_1{\mathrm{\beta }}_1$
$={\mathrm{n}}_1\frac{{\mathrm{\lambda }}_1\mathrm{D}}{\mathrm{d}}$
$=4\left[\frac{\left(6.5\times {10}^{-7}\right)\left(1.5\right)}{3\times {10}^{-3}}\right]$
$=1.3\times {10}^{-\mathrm{3 }}\text{m}$
$=\mathrm{1.3 }\text{mm}$