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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Physics
Chapter
Wave Optics

JIPMER 2019 · Physics · Wave Optics

In YDSE angular fringe width is 0.1^ wavelength of light used is...

In YDSE angular fringe width is 0.1′0.1^{\prime} wavelength of light used is 6000A˚6000 Å. Find the distance between two slits.

Options

  1. A2μm2 \mu \mathrm{m}
  2. B3μm3 \mu \mathrm{m}
  3. C5μm5 \mu \mathrm{m}
  4. D6μm6 \mu \mathrm{m}

Correct answer

D. 6μm6 \mu \mathrm{m}

Step-by-step solution

In YDSE, angular fringe width θ=0.1\theta=0.1 and wavelength of light, λ=6000A˚=6×10−7 m\lambda=6000 Å=6 \times 10^{-7} \mathrm{~m} If ' dd ' be the distance between slits then, θ=λd\theta=\frac{\lambda}{d} ⇒d=λθ=6×10−70.1=6×10−6=6μm\Rightarrow \quad \mathrm{d}=\frac{\lambda}{\theta}=\frac{6 \times 10^{-7}}{0.1}=6 \times 10^{-6}=6 \mu \mathrm{m}
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