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MHT CET 2025 (23 Apr, Shift 1) · YDSE: Fringe Width & Fringe Positions

YDSE: Fringe Width & Fringe Positions question from MHT CET 2025 (23 Apr, Shift 1)

Q11MHT CET 2025 · 23 APR · SHIFT 1Easy

YDSE: Fringe Width & Fringe Positions

In Young's double slit experiment the wavelength of light used is 6000 A˚6000\ \text{Å}, the screen is 40 cm from the slits and the fringe width is 0.012 cm, the distance between two slits is

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Answer: (d)

  1. β=λDd⇒d=λDβ\beta = \frac{\lambda D}{d} \Rightarrow d = \frac{\lambda D}{\beta}
  2. d=6×10−7×0.41.2×10−4=2×10−3 md = \frac{6\times10^{-7}\times 0.4}{1.2\times10^{-4}} = 2\times10^{-3}\text{ m}
  3. So d=0.2d = 0.2 cm, answer (d).
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