In a Young's double slit experiment, slits are separated by 0 . 5mm...

Given,
Wavelength of first light is, ${\lambda}_{1} = 650nm$
Wavelength of second light is, ${\lambda}_{2} = 520nm$
Screen distance is, $D = 150cm$
Slit width is, $d = 0 . 5mm$

For first wavelength, location of fringe from central line,
${y}_{1} = \dfrac{{n}_{1}{\lambda}_{1}D}{d}$ for bright fringes

For second wavelength, location of fringe from central line,
${y}_{2} = \dfrac{{n}_{2}{\lambda}_{2}D}{d}$ for bright fringes

Condition of coincide of fringe is,
${y}_{1} = {y}_{2}$
$⇒ {n}_{1}{\lambda}_{1} = {n}_{2}{\lambda}_{2}$
$⇒ {n}_{1} \times 650 = {n}_{2} \times 520$
$⇒ \dfrac{{n}_{1}}{{n}_{2}} = \dfrac{4}{5}$

For the minimum value of ${n}_{1}$ and ${n}_{2}$
${n}_{1} = 4$ and ${n}_{2} = 5$
{\left({y}_{1}\right)}_\min = 4 \times 650nm \times \dfrac{1 . 50m}{0.5 \times {10}^{- 3}m}
$= 7800 \times {10}^{- 6}m$
$= 7.8mm$