Question

In Young's double slit experiment, the slits are $2\text{mm}$ apart and are illuminated by photons of two wavelengths ${\lambda }_1=12000\stackrel{\circ }{\text{A}}$ and ${\lambda }_2=10000\stackrel{\circ }{\text{A}}$ . 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. $6\text{mm}$
• B. $3\text{mm}$
• C. $4\text{mm}$
• D. $8\text{mm}$

Answer 1

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

Given,
$n_1=5,D=2\text{m},d=2\text{mm}=2\times {10}^{-3}\text{m}$
${\lambda }_1=12000\stackrel{\circ }{\text{A}}=12000\times {10}^{-10}\text{m}=12\times {10}^{-7}\text{m}$
${\lambda }_2=10000\stackrel{\circ }{\text{A}}=10000\times {10}^{-10}\text{m}=10\times {10}^{-7}\text{m}$

Let $n_1$ bright fringe of ${\lambda }_1$ coincides with $n_2$ bright fringe of ${\lambda }_2$. Then

$\frac{n_1{\lambda }_{1}D}{d}=\frac{n_2{\lambda }_{2}D}{d}$

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

$\frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{1000}{12000}=\frac{5}{6}$

Let $x$ be given distance.

$\therefore x=\frac{n_1{\lambda }_{1}D}{d}$

$x=\frac{5\times 12\times {10}^{-7}\times 2}{2\times {10}^{-3}}=6\times {10}^{-3}=6\text{mm}$