Q120 / 200 · NEET 2024 (Re-NEET) · Physics

Two slits in Young's double slit experiment are 1.5 mm apart and the screen is placed at a distance of 1 m from the slits. If the wavelength of light used is $600 \times 10^{-9} \mathrm{~m}$ then the fringe separation is

  • A
    $4 \times 10^{-5} \mathrm{~m}$
  • B
    $9 \times 10^{-8} \mathrm{~m}$
  • C
    $4 \times 10^{-7} \mathrm{~m}$
  • Correct answer
    D
    $4 \times 10^{-4} \mathrm{~m}$

Solution

Fringe width $=$ Fringe separation $(\beta)=\frac{\lambda D}{d}$ $\Rightarrow \quad \beta=\frac{600 \times 10^{-9} \times 1}{1.5 \times 10^{-3}}=\frac{6 \times 10^{-7}}{1.5 \times 10^{-3}}=4 \times 10^{-4} \mathrm{~m}$

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

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