If a slit of width ' $x$ ' was illuminated by red light having wavelength $6500 Å$, the first minima was…
If a slit of width ' $x$ ' was illuminated by red light having wavelength $6500 Å$, the first minima was obtained at $\theta=30^{\circ}$. Then the value of ' $x$ ' is
$1.4 \times 10^{-4} \mu \mathrm{~m}$
$1.2 \times 10^{-5} \mathrm{~m}$
$1.3 \mu \mathrm{~m}$
$1.2 \mu \mathrm{~m}$
Solution
In diffraction, for first minima,
$\begin{aligned}
& x \sin \theta=1 \times \lambda=6500 \times 10^{-10} \\
& \Rightarrow \quad x=\frac{6500 \times 10^{-10}}{\sin 30^{\circ}}=1.3 \mu \mathrm{~m}
\end{aligned}$