Question

A parallel beam of monochromatic light falls normally on a single slit of width \( a \) and a diffraction pattern is observed on a screen placed at a distance \( D \) from the slit. Explain:
the formation of maxima and minima in the diffraction pattern, and
why the maxima go on becoming weaker and weaker with increasing order \( (n) \).

Hint:

Single-slit diffraction:

Minima: \( a\sin\theta = n\lambda \)
Central maximum brightest and widest
Intensity falls off for higher orders

Answer 1

Concept: This is Fraunhofer diffraction at a single slit. Every point of the slit acts as a source of secondary wavelets (Huygens’ principle). (I) Formation of maxima and minima: Minima (dark bands): Consider slit of width \( a \). If path difference between light from top and bottom of slit is: \[ a \sin \theta = n\lambda \quad (n = 1,2,3,\dots) \] then waves cancel pairwise by destructive interference → dark fringes. Thus, condition for minima: \[ a \sin \theta = n\lambda \] Maxima (bright bands): Between successive minima, waves interfere constructively to give bright regions. The central maximum occurs at \( \theta = 0 \) where all wavelets are in phase. Secondary maxima occur between minima due to partial constructive interference. (II) Why higher-order maxima become weaker:

As angle \( \theta \) increases, path differences across the slit increase.
Contributions from different parts of the slit increasingly cancel each other.
Only partial constructive interference occurs.
Energy spreads over a wider angular region.
Thus, intensity of higher-order maxima decreases. Additional explanation:

Central maximum is widest and brightest.
Intensity envelope decreases rapidly away from centre.
Energy conservation causes spreading of light.
Conclusion:

Minima occur due to complete destructive interference.
Higher-order maxima are weaker because interference becomes less constructive as angle increases.