In Young's double- slit experiment, slits are separated by $0.5\text{mm}$ and the screen is placed $150\text{cm}$ away. A beam of light consisting of two wavelengths, $650\text{nm}$ and $520\text{nm}$, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide are

The correct option is B $7.8\text{mm}$
Given :-
$d=0.5\text{mm};D=150\text{cm}$
${\lambda }_1=650\text{nm};{\lambda }_2=520\text{nm}$

Let $y$ be the distance from the central maximum to the point where the bright fringes due to both the wavelengths coincides.

Now, for ${\lambda }_1$ :-

$y_1=\frac{m{\lambda }_{1}D}{d}$

For ${\lambda }_2$:-

$y_2=\frac{n{\lambda }_{2}D}{d}$

At interference,

$y_1=y_2$

$\Rightarrow \frac{m{\lambda }_{1}D}{d}=\frac{n{\lambda }_{2}D}{d}$

$\Rightarrow \frac{m}{n}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{520}{650}$

$\Rightarrow \frac{m}{n}=\frac{4}{5}$

It means, with respect to the central maximum, $4^{th}$ bright fringe of ${\lambda }_1$ coincides with $5^{th}$ bright fringe of ${\lambda }_2$.

Now, $y=\frac{m{\lambda }_{1}D}{d}$

$\Rightarrow y=\frac{4\times 650\times {10}^{-9}\times 150\times {10}^{-2}}{0.5\times {10}^{-3}}$

$\therefore y=7.8\text{mm}$

Hence, option $\left(\text{B}\right)$ is correct.