Question

Light consisting of a plane waves of wavelength, ${\lambda }_1=8\times 10^{-5}cm$ and ${\lambda }_2=6\times 10^{-5}cm$ generates an interference pattern in Young's double slit experiment. If $n_1$ denotes the $n_1$ th dark fringe due to light of wavelength ${\lambda }_1$ which coincides with $n_2$ th bright fringe due to light of wavelength ${\lambda }_2$, then

• A. $n_1=3,n_2=1$
• B. $n_1=4,n_2=5$
• C. $n_1=1,n_2=2$
• D. $n_1=3,n_2=2$

Answer 1

Answer: (C)

Given, ${\lambda }_1=8\times 10^{-5}cm$ and ${\lambda }_2=6\times 10^{-5}cm$
Position $n^{\text{th }}$, dark fringe (due to light of wavelength ${\lambda }_1$ ) is given by,
$x_{n_1}=\left(2n_1\times 1\right)\frac{D{\lambda }_1}{2d}$
Position of $n_2^{\text{th}}$ bright fringe due to light of wavelength ${\lambda }_2$ is given by
$x_{n_2}=\frac{n_{2}D{\lambda }_2}{d}$
Since both the fringes are coincide to each other.
Hence,
$\left(2n_1+1\right)\frac{D{\lambda }_1}{2d}=\frac{n_{2}D{\lambda }_2}{d}$
$\frac{2n_1+1}{n_2}=\frac{2{\lambda }_2}{{\lambda }_1}=\frac{2\times 6\times 10^{-5}}{8\times 10^{-5}}$
$\frac{2n_1+1}{n_2}=\frac{3}{2}$
$4n_1+2=3n_2$
$4n_1-3n_2=-2$
$4n_1-3n_2=-2...(i)$
when $n_1=1$ and $n_2=2$, then Eq. (i) is satisfied. Hence, for $n_1=1$ and $n_2=2$, given fringes coincide to each other.