Question:
A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 \(\mathrm{\AA}\) and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :
A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 \(\mathrm{\AA}\) and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :
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Dark bands of a single slit satisfy \(a\sin\theta = n\lambda\), so \(y_n = n\lambda D/a\) for small angles.
Updated On: Nov 11, 2026
- 3 mm
- 9 mm
- 4.5 mm
- 1.5 mm
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The Correct Option is B
Solution and Explanation
Condition for the n-th minimum: \(a\sin\theta = n\lambda\), and for small angles \(\sin\theta \approx y/D\), giving \(y_n = \frac{n\lambda D}{a}\).
\(\lambda = 6000\,\mathrm{\AA} = 6\times10^{-7}\) m, \(D = 0.5\) m, \(a = 0.1\) mm \(= 10^{-4}\) m, \(n = 3\).
\(y_3 = \frac{3\times 6\times10^{-7}\times 0.5}{10^{-4}} = 9\times10^{-3}\) m = 9 mm.
Answer: (2) 9 mm
\(\lambda = 6000\,\mathrm{\AA} = 6\times10^{-7}\) m, \(D = 0.5\) m, \(a = 0.1\) mm \(= 10^{-4}\) m, \(n = 3\).
\(y_3 = \frac{3\times 6\times10^{-7}\times 0.5}{10^{-4}} = 9\times10^{-3}\) m = 9 mm.
Answer: (2) 9 mm
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