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NEET 2024
Exam
NEET
Year
2024
Subject
Physics
Chapter
Wave Optics

NEET 2024 · Physics · Wave Optics

Two slits in Young's double slit experiment are 1.5 mm apart and...

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 600×10−9 m600 \times 10^{-9} \mathrm{~m} then the fringe separation is

Options

  1. A4×10−5 m4 \times 10^{-5} \mathrm{~m}
  2. B9×10−8 m9 \times 10^{-8} \mathrm{~m}
  3. C4×10−7 m4 \times 10^{-7} \mathrm{~m}
  4. D4×10−4 m4 \times 10^{-4} \mathrm{~m}

Correct answer

D. 4×10−4 m4 \times 10^{-4} \mathrm{~m}

Step-by-step solution

Fringe width = Fringe separation (β)=λDd(\beta)=\frac{\lambda D}{d} ⇒β=600×10−9×11.5×10−3=6×10−71.5×10−3=4×10−4 m\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}
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