JEE Mains · Physics · STD 12 - 10. Wave optics
In Young's double slit experiment, monochromatic light of wavelength \(5000\) \(A\) is used. The slits are \(1.0 \mathrm{~mm}\) apart and screen is placed at \(1.0 \mathrm{~m}\) away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is _______ \(10^{-6} \mathrm{~m}\).
- A \(121\)
- B \(122\)
- C \(124\)
- D \(125\)
Answer & Solution
Correct Answer
(D) \(125\)
Step-by-step Solution
Detailed explanation
Let intensity of light on screen due to each slit is \(\mathrm{I}_0\) So internity at centre of screen is \(4 \mathrm{I}_0\) Intensity at distance y from centre- \(\mathrm{I}=\mathrm{I}_0+\mathrm{I}_0+2 \sqrt{\mathrm{I}_0 \mathrm{I}_0} \cos \phi\)…
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