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MHT CET 2024 (3 May, Shift 2) · Diffraction at a Single Slit

Diffraction at a Single Slit question from MHT CET 2024 (3 May, Shift 2)

Q32MHT CET 2024 · 3 MAY · SHIFT 2Medium

Diffraction at a Single Slit

A screen is placed at 50 cm from a single slit, which is illuminated with light of wavelength 600 nm. If separation between the 11st and 33rd minima in the diffraction pattern is 3 mm then slit width is

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Answer: (a)

  1. Minima on the screen: yn=nDλay_n = \dfrac{nD\lambda}{a}, so y3−y1=2Dλay_3 - y_1 = \dfrac{2D\lambda}{a}.
  2. a=2Dλ3×10−3=2×0.5×600×10−93×10−3=2×10−4 ma = \dfrac{2D\lambda}{3\times10^{-3}} = \dfrac{2 \times 0.5 \times 600\times10^{-9}}{3\times10^{-3}} = 2\times10^{-4}\text{ m}
  3. So the answer is (a) 0.2 mm0.2\text{ mm}.
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