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JEE Main 2021 (16 Mar, Shift 1) · YDSE: Fringe Width & Fringe Positions

YDSE: Fringe Width & Fringe Positions question from JEE Main 2021 (16 Mar, Shift 1)

Q45JEE Main 2021 · 16 MAR · SHIFT 1Easy

YDSE: Fringe Width & Fringe Positions

A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 600

  1. λ=βdD=6×10−3×1×10−310\lambda = \frac{\beta d}{D} = \frac{6\times10^{-3}\times 1\times10^{-3}}{10}
  2. λ=6×10−7 m=600\lambda = 6\times10^{-7}\text{ m} = 600 nm, so x=600x = 600.
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