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MHT CET 2024 (9 May, Shift 1) · Diffraction at a Single Slit

Diffraction at a Single Slit question from MHT CET 2024 (9 May, Shift 1)

Q37MHT CET 2024 · 9 MAY · SHIFT 1Medium

Diffraction at a Single Slit

Two wavelength 590 nm and 596 nm of sodium light are used one after other, to study the diffraction taking place at a single slit of aperture 2.4 mm. The distance between the slit and screen is 2 m. The separation between the positions of first secondary maximum of the diffraction pattern obtained in the two cases is

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Answer: (a)

  1. First secondary maximum: y=3λD2ay = \dfrac{3\lambda D}{2a}, so Δy=3D Δλ2a\Delta y = \dfrac{3D\,\Delta\lambda}{2a}
  2. Δλ=596−590=6\Delta\lambda = 596 - 590 = 6 nm
  3. Δy=3×2×6×10−92×2.4×10−3=7.5×10−6\Delta y = \dfrac{3 \times 2 \times 6 \times 10^{-9}}{2 \times 2.4 \times 10^{-3}} = 7.5 \times 10^{-6} m
  4. So the answer is (a).
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