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JEE Main 2025 (4 Apr, Shift 2) · Diffraction at a Single Slit

Diffraction at a Single Slit question from JEE Main 2025 (4 Apr, Shift 2)

Q24JEE Main 2025 · 4 APR · SHIFT 2Medium

Diffraction at a Single Slit

In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the central maximum of the single slit diffraction pattern, then the width of each slit is x×10−3 cmx \times 10^{-3}\text{ cm}, where xx-value is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 15

  1. Central maximum width 2λDa\dfrac{2\lambda D}{a} holds 20 fringes of width λDd\dfrac{\lambda D}{d}.
  2. 2λDa=20 λDd⇒a=d10=0.15\dfrac{2\lambda D}{a} = 20\,\dfrac{\lambda D}{d} \Rightarrow a = \dfrac{d}{10} = 0.15 mm
  3. a=0.015a = 0.015 cm =15×10−3= 15 \times 10^{-3} cm, so x=15x = 15.
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