JEE Main 2017 — Wave Optics Question with Solution

Given, 
Wavelength of first light is, ${\lambda }_1=650 \mathrm{nm}$
Wavelength of second light is, ${\lambda }_2=520 \mathrm{nm}$
Screen distance is, $D=150 \mathrm{cm}$
Slit width is, $d=0.5 \mathrm{mm}$

For first wavelength, location of fringe from central line,
$y_1=\frac{n_1{\lambda }_{1}D}{d}$ for bright fringes

For second wavelength, location of fringe from central line,
$y_2=\frac{n_2{\lambda }_{2}D}{d}$ for bright fringes

Condition of coincide of fringe is, 
$y_1=y_2$
$\Rightarrow n_1{\lambda }_1=n_2{\lambda }_2$
$\Rightarrow n_1\times 650=n_2\times 520$
$\Rightarrow \frac{n_1}{n_2}=\frac{4}{5}$

For the minimum value of $n_1$ and $n_2$
$n_1=4$ and $n_2=5$
${\left(y_1\right)}_{min}=4\times 650 \mathrm{nm}\times \frac{1.50 \mathrm{m}}{0.5\times {10}^{-3} \mathrm{m}}$
$=7800\times {10}^{-6} \mathrm{m}$
$=7.8 \mathrm{mm}$