Question

In a single-slit diffraction experiment, light of wavelength $\lambda$ illuminates the slit of width ‘a’. The diffraction pattern is observed on a screen kept at a distance D from the slits. Depict variation of intensity in the fringe pattern with the angular position of the fringes.

Hint:

Unlike Young's double-slit interference (where all bright fringes have equal intensity $4I_0$), single-slit diffraction fringes have unequal widths and rapidly diminishing intensities.

Answer 1

Concept:
• Single-slit diffraction produces a central bright maximum flanked by secondary minima and secondary maxima of rapidly decreasing intensity.

• Minima occur at angular positions $\sin \theta = \pm \frac{n \lambda}{a} \approx \pm \frac{n \lambda}{a}$ for integer $n = 1, 2, 3, \dots$.

• Secondary maxima occur approximately at $\sin \theta \approx \pm \left(n + \frac{1}{2}\right) \frac{\lambda}{a}$.

Step 1: Key features of diffraction intensity curve
1. Central Maximum: Positioned at $\theta = 0$ with maximum intensity $I_0$. Its angular width is $2\theta_0 = \frac{2\lambda}{a}$.
2. First Minima: Located at $\theta = \pm \frac{\lambda}{a}$ with zero intensity ($I = 0$).
3. Secondary Maxima: Located near $\theta = \pm \frac{3\lambda}{2a}, \pm \frac{5\lambda}{2a}, \dots$
Intensity drops rapidly: $I_1 \approx \frac{I_0}{22}$ (about $4.5\%$ of $I_0$), $I_2 \approx \frac{I_0}{61}$ (about $1.6\%$ of $I_0$).

Step 2: Diagram depiction
The graph plots Intensity $I$ along the vertical axis against angular position $\theta$ (or path parameter $\beta = \frac{\pi a \sin \theta}{\lambda}$) along the horizontal axis.

Step 3: Conclusion
The intensity pattern consists of a central peak of maximum height $I_0$ flanked symmetrically by secondary peaks whose heights decrease rapidly on either side.