Question

Write formula for the position of minima in the diffraction pattern of a single slit.

Hint:

For a single slit, a dark fringe forms when the extreme-ray path difference across the slit width is a whole number of wavelengths.

Answer 1

Step 1: Concept. In single-slit Fraunhofer diffraction, a slit of width \(a\) is illuminated by monochromatic light of wavelength \(\lambda\). To locate a dark fringe (minimum) at an angle \(\theta\) from the central axis, the slit is imagined to be divided into pairs of parallel strips. A minimum occurs when the path difference between the extreme rays of the slit equals a whole number of wavelengths, so that the wavelets cancel in pairs.

Step 2: Condition for minima. The extreme-ray path difference is \(a\sin\theta\). For total destructive interference,
\[ a\sin\theta = n\lambda \]
Step 3: State the formula. Hence the position (angular direction) of the \(n^{th}\) minimum is
\[\boxed{\,a\sin\theta = n\lambda,\quad n = 1,\,2,\,3,\dots\,}\]
where \(a\) = slit width, \(\theta\) = angle of the minimum from the central maximum, \(\lambda\) = wavelength, and \(n\) is a non-zero integer. Note \(n=0\) is excluded because it gives the central bright maximum, not a minimum.