Physics - Wave Optics Question with Solution | TestHub

Given:
$d=2\text{mm}=2\times 10^{-3}\text{m}$
$D=2\text{m}$
${\lambda }_1=6000\text{A˚}=6\times 10^{-7}\text{m}$
${\lambda }_2=5000\text{A˚}=5\times 10^{-7}\text{m}$

Let the $n_1$-th bright fringe of ${\lambda }_1$ coincide with the $n_2$-th bright fringe of ${\lambda }_2$. The position of the $n$-th bright fringe from the central maximum is given by $y=\frac{n\lambda D}{d}$.

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

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

$\frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{5000}{6000}=\frac{5}{6}$

For the minimum distance from the central maximum, we take the smallest integer values for $n_1$ and $n_2$, which are $n_1=5$ and $n_2=6$.

Substituting $n_1=5$ into the position formula:
$y=\frac{5\times 6\times 10^{-7}\times 2}{2\times 10^{-3}}$

$y=30\times 10^{-4}\text{m}$

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

Answer: $3\text{mm}$