Question

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

• A. $1.56\text{mm}$
• B. $7.8\text{mm}$
• C. $9.75\text{mm}$
• D. $15.6\text{mm}$

Answer 1

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.