Question 186: Wave Optics

Question

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 _______ $\mu$m.

Accepted Answer

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

Wave Optics

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 _______ \mum.

Integer / Numeric Answer Type