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In a Young's double slit experiment, $m$ th order and $n$th order of bright fringes are formed at point $P$ on a distant screen, if monochromatic source of wavelength $400 \mathrm{~nm}$ and $600 \mathrm{~nm}$ are used respectively. The minimum value of $m$ and $n$ are respectively,
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The correct answer is:
3,2
Let $m^{\text {th }}$ bright fringe of wavelength $\lambda_{r \pi}$ and $n^{\text {th }}$ bright fringe of wave length $\lambda_n$ formed at a distance $x_n$ from central maxima at same point $P$, i.e., they coincide, therefore, $x_n=x_n$.
$$
\begin{aligned}
\Rightarrow \quad \frac{m \lambda_{n m} D}{a} & =\frac{n \lambda_n D}{a} \\
\Rightarrow \quad \frac{m}{n} & =\frac{\lambda_n}{\lambda_n}
\end{aligned}
$$
Putting the given values, we get
$$
\begin{aligned}
& =\frac{600 \times 10^{-9}}{400 \times 10^{-9}} \\
\text { [Given, } \lambda_m & =400 \times 10^{-9} \mathrm{~m}, \lambda_n=600 \times 10^{-9} \mathrm{~m} \text { ] } \\
\therefore \quad & \frac{m}{n}=\frac{3}{2}
\end{aligned}
$$
So, least integral values of $m$ and $n$ satisfying above requirement are $m=3$ and $n=2$.
$$
\begin{aligned}
\Rightarrow \quad \frac{m \lambda_{n m} D}{a} & =\frac{n \lambda_n D}{a} \\
\Rightarrow \quad \frac{m}{n} & =\frac{\lambda_n}{\lambda_n}
\end{aligned}
$$
Putting the given values, we get
$$
\begin{aligned}
& =\frac{600 \times 10^{-9}}{400 \times 10^{-9}} \\
\text { [Given, } \lambda_m & =400 \times 10^{-9} \mathrm{~m}, \lambda_n=600 \times 10^{-9} \mathrm{~m} \text { ] } \\
\therefore \quad & \frac{m}{n}=\frac{3}{2}
\end{aligned}
$$
So, least integral values of $m$ and $n$ satisfying above requirement are $m=3$ and $n=2$.
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