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MHT CET 2024 (15 May, Shift 2) · YDSE: Fringe Width & Fringe Positions

YDSE: Fringe Width & Fringe Positions question from MHT CET 2024 (15 May, Shift 2)

Q26MHT CET 2024 · 15 MAY · SHIFT 2Medium

YDSE: Fringe Width & Fringe Positions

Sodium light (λ=6×10−7 m)(\lambda = 6 \times 10^{-7}\text{ m}) is used to produce interference pattern. The observed fringe width is 0.12 mm. The angle between the two wave trains is

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Answer: (b)

  1. Fringe width β=λDd=λθ\beta = \frac{\lambda D}{d} = \frac{\lambda}{\theta}, where θ=dD\theta = \frac{d}{D} is the angle between the wave trains.
  2. θ=λβ=6×10−70.12×10−3=5×10−3 rad\theta = \frac{\lambda}{\beta} = \frac{6\times10^{-7}}{0.12\times10^{-3}} = 5\times10^{-3}\text{ rad}
  3. So the answer is (b) 5×10−35\times10^{-3} rad.
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