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MHT CET 2025 (19 Apr, Shift 2) · YDSE: Film, Medium & Two Wavelengths

YDSE: Film, Medium & Two Wavelengths question from MHT CET 2025 (19 Apr, Shift 2)

Q30MHT CET 2025 · 19 APR · SHIFT 2Difficult

YDSE: Film, Medium & Two Wavelengths

In Young's double slit experiment let 'd' be the distance between two slits and 'D' be the distance between the slits and the screen. Using a monochromatic source of wavelength ' λ\lambda ', in an interference pattern, third minimum is observed exactly in front of one of the slits. If at the same point on the screen first minimum is to be obtained, the required change in the wavelength is [ d & Dd\ \&\ D are not changed].

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Answer: (c)

  1. Point opposite a slit: y=d2y = \frac{d}{2}, so path difference Δ=ydD=d22D\Delta = \frac{yd}{D} = \frac{d^2}{2D}.
  2. Third minimum: d22D=5λ2\frac{d^2}{2D} = \frac{5\lambda}{2}.
  3. First minimum with new λ′\lambda': d22D=λ′2⇒λ′=5λ\frac{d^2}{2D} = \frac{\lambda'}{2} \Rightarrow \lambda' = 5\lambda.
  4. Change =5λ−λ=4λ= 5\lambda - \lambda = 4\lambda. So the answer is (c) 4λ4\lambda.
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