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JIPMER 2014
Exam
JIPMER
Year
2014
Subject
Physics
Chapter
Wave Optics

JIPMER 2014 · Physics · Wave Optics

If prism angle α=1^ , μ=1.54 , distance between screen and prism...

If prism angle α=1∘,μ=1.54\alpha=1^{\circ}, \quad \mu=1.54, distance between screen and prism (D)=0.7 m(D)=0.7 \mathrm{~m}, distance between prism and source a=0.3 ma=0.3 \mathrm{~m}, λ=180πnm\lambda=180 \pi \mathrm{nm}, then in Fresnal biprism find the value of β\beta (fringe width).

Options

  1. A10−4 m10^{-4} \mathrm{~m}
  2. B10−3 mm10^{-3} \mathrm{~mm}
  3. C10−4×πm10^{-4} \times \pi \mathrm{m}
  4. Dπ×10−3 m\pi \times 10^{-3} \mathrm{~m}

Correct answer

B. 10−3 mm10^{-3} \mathrm{~mm}

Step-by-step solution

a=0.3α=1∘,μ=1.54,a=0.3 \quad \alpha=1^{\circ}, \quad \mu=1.54, \quad we know that d=2a(μ−1)αd=2 a(\mu-1) \alpha =2×0.3(1.54−1)×1=2×0.3×0.54=0.324\begin{aligned} & =2 \times 0.3(1.54-1) \times 1 \\ & =2 \times 0.3 \times 0.54=0.324\end{aligned} Fringe width F=DλdF=\frac{D \lambda}{d} =0.7×180π×10−90.324=\frac{0.7 \times 180 \pi \times 10^{-9}}{0.324} =7×180π×10−6324=10−3 mm=\frac{7 \times 180 \pi \times 10^{-6}}{324}=10^{-3} \mathrm{~mm}
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