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JEE Main 2020 (7 Jan, Shift 2) · YDSE: Fringe Width & Fringe Positions

YDSE: Fringe Width & Fringe Positions question from JEE Main 2020 (7 Jan, Shift 2)

Q66JEE Main 2020 · 7 JAN · SHIFT 2Medium

YDSE: Fringe Width & Fringe Positions

In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is:

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Answer: (b)

  1. β=λDd=589×10−9×1.50.15×10−3\beta = \frac{\lambda D}{d} = \frac{589\times10^{-9} \times 1.5}{0.15\times10^{-3}}
  2. β=5.89×10−3 m\beta = 5.89\times10^{-3}\text{ m}
  3. So the answer is (b) 5.95.9 mm.
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