The maximum intensity in Young’s double slit experiment is I0. Distance between the slits…
Physics · JEE Advanced · NTA Exams — Optics
The maximum intensity in Young’s double slit experiment is I0. Distance between the slits is d = 5λ, where λ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = 10d
- \(\frac{I_{0}}{2}\)
- \(\frac{3}{4} I_{0}\)
- \(I_{0}\)
- \(\frac{I_{0}}{4}\)
Answer
(A) I_ 0 2
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