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,
$x_{n_{1}}=\left(2 n_{1} \times 1\right) \frac{D \lambda_{1}}{2 d}$
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(2 n_{1}+1\right) \frac{D \lambda_{1}}{2 d}=\frac{n_{2} D \lambda_{2}}{d}$
$\frac{2 n_{1}+1}{n_{2}}=\frac{2 \lambda_{2}}{\lambda_{1}}=\frac{2 \times 6 \times 10^{-5}}{8 \times 10^{-5}}$
$\frac{2 n_{1}+1}{n_{2}}=\frac{3}{2}$
$4 n_{1}+2=3 n_{2}$
$4 n_{1}-3 n_{2}=-2$
$4 n_{1}-3 n_{2}=-2\,\,\,...(i)$
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.