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  4. A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light, the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500 mathringA ?

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Q. A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light, the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength $ 6500\, \mathring{A}$ ?

ManipalManipal 2012Wave Optics

Solution:

$ x=\frac{(2n+1)\lambda D}{2a} $
For red light $ x=\frac{(4+1)D}{2a}\times 6500 $
For unknown wavelength of light,
$ x=\frac{(6+1)D}{2a}\times \lambda $
Accordingly, $ \therefore $ $ 5\times 6500=7\times \lambda $
$ \Rightarrow $ $ \lambda =\frac{5}{7}\times 6500 $
$ =\text{ }4642.8\mathring{A} $