Question

In a Young's double slit experiment, the slits are $2\text{mm}$ apart and are illuminated by photons of two wavelengths ${\lambda }_1=12000\mathring{\text{A}}$ and ${\lambda }_2=10000\mathring{\text{A}}$. At what minimum distance from the common central bright fringe on the screen, $2\text{m}$ from the slit, will a bright fringe from one interference pattern coincide, with a bright fringe from the other?

• A. $3\text{mm}$
• B. $8\text{mm}$
• C. $6\text{mm}$
• D. $4\text{mm}$

Answer 1

The correct option is C $6\text{mm}$

Let $n_1$ and $n_2$ are the orders of the bright bands of the wavelengths ${\lambda }_1$ and ${\lambda }_2$ respectively, which coincide.

$\Rightarrow n_1{\lambda }_1=n_2{\lambda }_2$

Therefore,

$\frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{10000}{12000}=\frac{5}{6}$


So, the minimum value $n_1$ and $n_2$ can have is $5$ and $6$ respectively.

Consider the interference pattern of ${\lambda }_1$.

$y_{\text{min}}=\frac{n_1{\lambda }_{1}D}{d}=\frac{5(12000\times {10}^{-10})\left(2\right)}{2\times {10}^{-3}}$

$=6\times {10}^{-3}\text{m}=6\text{mm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.