Question

In a single-slit diffraction pattern, the angle of diffraction for the second minimum is $60^\circ$. Find the width of the slit in terms of $\lambda$.

Hint:

Single-slit {minima}: $a\sin\theta=m\lambda\ (m=1,2,\ldots)$; {central maximum} at $m=0$. Always verify that the stated order refers to the {minima}, not side maxima (which occur near $a\sin\theta \approx (m+\tfrac12)\lambda$).

Answer 1

Step 1: Fraunhofer single-slit intensity and minima condition.
Treat the slit of width $a$ as a line of secondary sources (Huygens principle). The complex amplitude at an angle $\theta$ is the phasor sum across the slit and equals \[ E(\theta) \propto \frac{\sin \beta}{\beta}, \qquad \beta \equiv \frac{\pi a}{\lambda}\sin\theta. \] Hence the intensity \[ I(\theta)=I_0\left(\frac{\sin\beta}{\beta}\right)^2. \] Minima occur when $\sin\beta=0$ with $\beta\neq 0$, i.e., \[ \beta = m\pi (m=1,2,3,\ldots)\ ⇒\ a\sin\theta = m\lambda. \] Step 2: Use the given order and angle.
“Second minimum” $⇒ m=2$, and $\theta=60^\circ$: \[ a\sin 60^\circ = 2\lambda \ ⇒\ a\left(\frac{\sqrt{3}}{2}\right)=2\lambda. \] Step 3: Solve for $a$.
\[ a = \frac{2\lambda}{(\sqrt{3}/2)}=\frac{4\lambda}{\sqrt{3}}. \] Final Answer: The slit width is \(\displaystyle a=\frac{4\lambda}{\sqrt{3}}.\)