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 _______ $\mu \mathrm{m}$.

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	heta_1=\sin ^{-1}\left(\frac{2 \lambda}{a}ight) \ & 	heta_2=\sin ^{-1}\left(\frac{3 \lambda}{a}ight) \ & \because \quad 	heta_1+	heta_2=30^{\circ}\end{aligned}$ $\begin{aligned} & \Rightarrow \sin ^{-1}\left(\frac{2 \lambda}{\mathrm{a}}ight)+\sin ^{-1}\left(\frac{3 \lambda}{\mathrm{a}}ight)=\frac{\pi}{6} \ & \Rightarrow \frac{2 \lambda}{\mathrm{a}} \sqrt{1-\left(\frac{3 \lambda}{\mathrm{a}}ight)^2}+\frac{3 \lambda}{\mathrm{a}} \sqrt{1+\left(\frac{2 \lambda}{\mathrm{a}}ight)^2}=\sin \frac{\pi}{6}\end{aligned}$ Here $\lambda=628 \mathrm{~nm}$ After solving $\mathrm{A}=6.07 \mu \mathrm{~m}$ Approximate Method : $\begin{aligned} & 	heta=	heta_1+	heta_2 \ & \Rightarrow \frac{\pi}{6}=\frac{2 \lambda}{a}+\frac{3 \lambda}{a} \ & \Rightarrow \frac{\pi}{6}=\frac{5}{a}(628 \mathrm{~nm}) \ & \Rightarrow a=6 \mu \mathrm{~m}\end{aligned}$

If the measured angular separation between the second minimum to the left of the central maximum and the…

Solution