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JEE Mains · Physics · STD 12 - 10. Wave optics

Two light waves of wavelengths \(800\) and \(600\,nm\) are used in Young's double slit experiment to obtain interference fringes on a screen placed \(7\,m\) away from plane of slits. If the two slits are separated by \(0.35\,mm\), then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be \(.............mm\).

  1. A \(45\)
  2. B \(46\)
  3. C \(48\)
  4. D \(47\)
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Answer & Solution

Correct Answer

(C) \(48\)

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Detailed explanation

Correct answer is \(\mathbf{4 8}\) \(\omega_1=\frac{\lambda_1 D}{d} \& \omega_2=\frac{\lambda_2 D}{d} \) \(\omega_1=16 \mathrm{~mm} \& \omega_2=12 \mathrm{~mm} \) \(\text { so LCM }\left(\omega_1, \omega_2\right)=48 \mathrm{~mm}\) so at 48 mm distance both bright fringes will be…
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