A screen has two slits, each of width $w$, with their centres at a distance $2w$ apart. It is illuminated by a monochromatic plane wave travelling along the $x$-axis.
The intensity of the interference pattern, measured on a distant screen, at an angle $\theta = n\lambda/w$ to the $x$-axis is
The given problem involves the interference pattern produced by two closely spaced slits of equal width \(w\), with their centers separated by a distance \(2w\). This configuration is illuminated by a monochromatic plane wave along the \(x\)-axis. We are to find the condition under which the intensity of the interference pattern at an angle \(\theta = n\lambda/w\) is zero.
The intensity of the interference pattern is determined by the principle of superposition of the light waves from the two slits, resulting in constructive and destructive interference.
- The path difference (\(\Delta x\)) between the light from the two slits at angle \(\theta\) is given by: \(\Delta x = 2w \sin \theta\).
- The condition for destructive interference (zero intensity) is when the path difference is an odd multiple of half-wavelengths: \(\Delta x = (m + \frac{1}{2}) \lambda\), where \(m\) is an integer.
- Substituting the path difference formula: \(2w \sin \theta = (m + \frac{1}{2}) \lambda\).
- Given that \(\theta = n\lambda/w\), we substitute: \(\sin \theta \approx \theta = n\lambda/w\).
- Recalculate: \(2w \cdot \frac{n\lambda}{w} = (m + \frac{1}{2})\lambda\) \(2n\lambda = (m + \frac{1}{2})\lambda\) \(2n = m + \frac{1}{2}\).
- From this equation, for integer \(n\), \((m + \frac{1}{2})\) is never an integer unless \(n\) is a half-integer like \(1/2, 3/2\), etc., i.e., odd integers cancel with half.
- Thus, zero intensity occurs for integer multiples of \(n\): \(n = 1, 2, 3, \ldots\).
Therefore, the correct answer is that the intensity is zero for \(n = 1, 2, 3 \ldots\).
Correct Option: Zero for \(n = 1, 2, 3 \ldots\)
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