JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2023
- Medium
- Numerical
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 \(.............mm\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(48\)
Step-by-step Solution
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 found.
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