Q144 / 160 · TS EAMCET 2019 (06 May Shift 1) · Physics
Light consisting of a plane waves of wavelength, $\lambda_1=8 \times 10^{-5} \mathrm{~cm}$ and $\lambda_2=6 \times 10^{-5} \mathrm{~cm}$ 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$n_1=4, n_2=5$
- Correct answerC$n_1=1, n_2=2$
- D$n_1=3, n_2=2$
Solution
Given, $\lambda_1=8 \times 10^{-5} \mathrm{~cm}$ and $\lambda_2=6 \times 10^{-5} \mathrm{~cm}$
Position $n^{\text {h }}$, 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,
$\begin{aligned}\left(2 n_1+1\right) \frac{D \lambda_1}{2 d} & =\frac{n_2 D \lambda_2}{d} \\ \frac{2 n_1+1}{n_2} & =\frac{2 \lambda_2}{\lambda_1}=\frac{2 \times 6 \times 10^{-5}}{8 \times 10^{-5}} \\ \frac{2 n_1+1}{n_2} & =\frac{3}{2} \\ 4 n_1+2 & =3 n_2 \\ 4 n_1-3 n_2 & =-2 \\ 4 n_1-3 n_2 & =-2\end{aligned}$
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.
Hence, the correct answer is option: 'C'.