Question 45 of 191

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is $30^{\circ}$ in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is _______ $\mu \mathrm{m}$.

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Answer: 6

$\begin{aligned} & \theta_1=\sin ^{-1}\left(\frac{2 \lambda}{a}\right) \\ & \theta_2=\sin ^{-1}\left(\frac{3 \lambda}{a}\right) \\ & \because \quad \theta_1+\theta_2=30^{\circ}\end{aligned}$
$\begin{aligned} & \Rightarrow \sin ^{-1}\left(\frac{2 \lambda}{\mathrm{a}}\right)+\sin ^{-1}\left(\frac{3 \lambda}{\mathrm{a}}\right)=\frac{\pi}{6} \\ & \Rightarrow \frac{2 \lambda}{\mathrm{a}} \sqrt{1-\left(\frac{3 \lambda}{\mathrm{a}}\right)^2}+\frac{3 \lambda}{\mathrm{a}} \sqrt{1+\left(\frac{2 \lambda}{\mathrm{a}}\right)^2}=\sin \frac{\pi}{6}\end{aligned}$
Here $\lambda=628 \mathrm{~nm}$
After solving
$\mathrm{A}=6.07 \mu \mathrm{~m}$
Approximate Method :
$\begin{aligned} & \theta=\theta_1+\theta_2 \\ & \Rightarrow \frac{\pi}{6}=\frac{2 \lambda}{a}+\frac{3 \lambda}{a} \\ & \Rightarrow \frac{\pi}{6}=\frac{5}{a}(628 \mathrm{~nm}) \\ & \Rightarrow a=6 \mu \mathrm{~m}\end{aligned}$