Q143 / 160 · AP EAMCET 2024 (20 May Shift 1) · Physics

In Young's double slit experiment, two slits are placed 2 mm from each other. Interference pattern is observed on a screen placed 2 m from the plane of the slits. Then the fringe width for a light of wavelength 400 nm is

  • A
    $0.4 \times 10^{-6} \mathrm{~m}$
  • B
    $4 \times 10^{-6} \mathrm{~m}$
  • Correct answer
    C
    $0.4 \times 10^{-3} \mathrm{~m}$
  • D
    $400 m$

Solution

In YDSE, $\mathrm{d}=2 \mathrm{~mm}=2 \times 10^{-5} \mathrm{~m}, \mathrm{D}=2 \mathrm{~m}, \lambda=400 \mathrm{~nm}$ $\therefore$ Fringe width, $\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{400 \times 10^{-9} \times 2}{2 \times 10^{-3}}$ $=0.4 \times 10^{-3} \mathrm{~m}$

Hence, the correct answer is option: 'C'.

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