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A single slit of a width $a$ is illuminated by a monochromatic light of wavelength $600 \mathrm{~nm}$. The value of ' $a$ ' for which first minimum appears at $\theta$ $=30^{\circ}$ on the screen will be :
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The correct answer is:
$1.2 \mu \mathrm{m}$
Wavelength of monochromatic light,
$\lambda=600 \times 10^{-9} \mathrm{~m}$
As fir first minima a $\sin \theta=\lambda$
$\begin{aligned}
& \Rightarrow \mathrm{a} \sin 30^{\circ}=600 \times 10^{-9} \\
& \Rightarrow \mathrm{a}=1200 \times 10^{-9} \mathrm{~m}=1.2 \mu \mathrm{m}
\end{aligned}$
$\lambda=600 \times 10^{-9} \mathrm{~m}$
As fir first minima a $\sin \theta=\lambda$
$\begin{aligned}
& \Rightarrow \mathrm{a} \sin 30^{\circ}=600 \times 10^{-9} \\
& \Rightarrow \mathrm{a}=1200 \times 10^{-9} \mathrm{~m}=1.2 \mu \mathrm{m}
\end{aligned}$
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