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In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is $6600 Å$, then wavelength of first maximum will be:
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The correct answer is:
$4400 Å$
$4400 Å$
In a single slit experiment, For diffraction maxima, $a \sin \theta=\left(2 n+1 \frac{Å}{2}\right.$ and for diffraction minima, $\mathrm{a} \sin \theta=\mathrm{n} \lambda$ According to question,
$$
\begin{aligned}
&\left(2 \times 1+1 \frac{\lambda}{2}=1 \times 6600\right. \\
&\left(\because \lambda_{\mathrm{R}}=6600 Å\right) \\
&\lambda=\frac{6600 \times 2}{3} \\
&\lambda=4400 Å
\end{aligned}
$$
$$
\begin{aligned}
&\left(2 \times 1+1 \frac{\lambda}{2}=1 \times 6600\right. \\
&\left(\because \lambda_{\mathrm{R}}=6600 Å\right) \\
&\lambda=\frac{6600 \times 2}{3} \\
&\lambda=4400 Å
\end{aligned}
$$
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