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 slit.
(i) Depict variation of intensity in the fringe pattern with the angular position of the fringes.
(ii) How is the linear width of central maximum affected when separation between the slit and the screen is decreased ?

Hint:

For single-slit diffraction, \[ W=\frac{2D\lambda}{a} \] Increase \(D\) or \(\lambda\) to obtain a wider central maximum.

Answer 1

Concept: Single-slit diffraction occurs due to interference among secondary wavelets emerging from different parts of the slit. The central maximum is the brightest and widest region of the diffraction pattern.

Step 1: Intensity distribution in diffraction pattern. The intensity distribution consists of:
• A broad and bright central maximum.
• Secondary maxima of much smaller intensity.
• Intensity decreases rapidly away from the centre. The schematic graph is: \[ \text{Intensity} \] \[ \includegraphics[width=7cm]{diffraction_intensity_placeholder} \] (Students should draw a broad central peak with smaller side maxima on both sides.)

Step 2: Width of central maximum. For single slit diffraction, \[ W=\frac{2D\lambda}{a} \] where \[ W \] is the linear width of the central maximum.

Step 3: Effect of decreasing screen distance. Since \[ W\propto D \] the width is directly proportional to the screen distance. If \[ D \] decreases, \[ W \] also decreases. Hence the central maximum becomes narrower. Final Answer: \[ W=\frac{2D\lambda}{a} \] Since \(W\propto D\), decreasing the slit-screen distance decreases the linear width of the central maximum.