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MHT CET 2025 (19 Apr, Shift 2) · Diffraction at a Single Slit

Diffraction at a Single Slit question from MHT CET 2025 (19 Apr, Shift 2)

Q15MHT CET 2025 · 19 APR · SHIFT 2Easy

Diffraction at a Single Slit

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If the distance between the first minimum on either side of the central maximum is 1 cm, the wavelength of light used is

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Answer: (c)

  1. Distance between first minima on either side (central maximum width): W=2DλaW = \dfrac{2D\lambda}{a}.
  2. λ=Wa2D=10−2×0.2×10−32×2=5×10−7 m\lambda = \dfrac{Wa}{2D} = \dfrac{10^{-2} \times 0.2\times10^{-3}}{2 \times 2} = 5\times10^{-7}\text{ m}
  3. So the answer is (c) 5000 A˚5000\text{ \AA}.
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