In Young's double-slit experiment, the distance between two slits is d = 1 mm. Two light…
Physics · JEE Main · NTA Exams — Optics
In Young's double-slit experiment, the distance between two slits is d = 1 mm. Two light waves of wavelength \(\lambda_{1}=500 \mathrm{~nm}\) and \(\lambda_{2}=700 \mathrm{~nm}\), respectively are incident on the slits, the least distance from the central maxima where the bright fringes due to both the wavelengths coincide for the first time is [given, distance between the slits and the screen is 0.8 m]
- \(2.5 \mathrm{~mm}\)
- \(4.9 \mathrm{~mm}\)
- \(3.5 \mathrm{~mm}\)
- \(2.8 \mathrm{~mm}\)
Answer
(D) 2.8 ~mm
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