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Light of wavelength $550 \mathrm{~nm}$ falls normally on a slit of width $22.0 \times 10^{-5} \mathrm{~cm}$. The angular position of the second minima from the central maximum will be (in radians)
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Verified Answer
The correct answer is:
$\frac{\pi}{8}$
$\frac{\pi}{8}$
If angular position of $2^{\text {nd }}$ maxima from central maxima is $\theta$ then
$$
\begin{aligned}
&\sin \theta=\frac{(2 \mathrm{n}-1) \lambda}{2 \mathrm{a}}=\frac{3 \lambda}{2 \mathrm{a}}=\frac{3 \times 550 \times 10^{-9}}{2 \times 22 \times 10^{-7}} \\
&\therefore \quad \theta=\frac{\pi}{8} \mathrm{rad}
\end{aligned}
$$
$$
\begin{aligned}
&\sin \theta=\frac{(2 \mathrm{n}-1) \lambda}{2 \mathrm{a}}=\frac{3 \lambda}{2 \mathrm{a}}=\frac{3 \times 550 \times 10^{-9}}{2 \times 22 \times 10^{-7}} \\
&\therefore \quad \theta=\frac{\pi}{8} \mathrm{rad}
\end{aligned}
$$
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