Question:
In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is \(6600\,\mathrm{\AA}\), then wavelength of first maximum will be:
In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is \(6600\,\mathrm{\AA}\), then wavelength of first maximum will be:
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First minimum: a sin(theta) = lambda. First secondary maximum: a sin(theta) = 3 lambda'/2. Equate the two at the same angle.
Updated On: Nov 11, 2026
- \(3300\,\mathrm{\AA}\)
- \(4400\,\mathrm{\AA}\)
- \(5500\,\mathrm{\AA}\)
- \(6600\,\mathrm{\AA}\)
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The Correct Option is B
Solution and Explanation
At the given angle, the first minimum of red light satisfies \(a\sin\theta=\lambda_r=6600\,\mathrm{\AA}\).
The first (secondary) maximum of another wavelength \(\lambda'\) lies at \(a\sin\theta=\frac{3\lambda'}{2}\).
Equating: \(\frac{3\lambda'}{2}=6600\Rightarrow\lambda'=\frac{2\times6600}{3}=4400\,\mathrm{\AA}\).
Answer: (2) \(4400\,\mathrm{\AA}\)
The first (secondary) maximum of another wavelength \(\lambda'\) lies at \(a\sin\theta=\frac{3\lambda'}{2}\).
Equating: \(\frac{3\lambda'}{2}=6600\Rightarrow\lambda'=\frac{2\times6600}{3}=4400\,\mathrm{\AA}\).
Answer: (2) \(4400\,\mathrm{\AA}\)
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