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A slit of width $a$ is illuminated by red light of wavelength $6500 Å$. If the first diffraction minimum falls at $30^{\circ}$, then the value of $a$ is
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Verified Answer
The correct answer is:
$1.3$ micron
Given, $\lambda=6500 Å=6500 \times 10^{-10} \mathrm{~m}, \theta=30^{\circ}$ According to the diffraction pattern of single slit,
$a \sin \theta=n \lambda$
where, $n=$ order of secondary minima $=1$
$\Rightarrow a \sin 30^{\circ}=1 \times\left(6500 \times 10^{-10}\right)$
$\Rightarrow \quad \frac{a}{2}=6500 \times 10^{-10}$
or
$\begin{aligned}
a &=13000 \times 10^{-10} \\
&=1.3 \times 10^{-6} \mathrm{~m}=1.3 \text { micron }
\end{aligned}$
$a \sin \theta=n \lambda$
where, $n=$ order of secondary minima $=1$
$\Rightarrow a \sin 30^{\circ}=1 \times\left(6500 \times 10^{-10}\right)$
$\Rightarrow \quad \frac{a}{2}=6500 \times 10^{-10}$
or
$\begin{aligned}
a &=13000 \times 10^{-10} \\
&=1.3 \times 10^{-6} \mathrm{~m}=1.3 \text { micron }
\end{aligned}$
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