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NEET 2013
Exam
NEET
Year
2013
Subject
Physics
Chapter
Wave Optics

NEET 2013 · Physics · Wave Optics

In Young's double slit experiment, the slits are 2 ~mm apart and...

In Young's double slit experiment, the slits are 2 mm2 \mathrm{~mm} apart and are illuminated by photons of two wavelengths λ1=12000A˚\lambda_1=12000 Å \quad and λ2=10000A˚\lambda_2=10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m2 \mathrm{~m} from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Options

  1. A8 mm8 \mathrm{~mm}
  2. B6 mm6 \mathrm{~mm}
  3. C4 mm4 \mathrm{~mm}
  4. D3 mm3 \mathrm{~mm}

Correct answer

B. 6 mm6 \mathrm{~mm}

Step-by-step solution

Given λ1=12000A˚\lambda_1=12000 Å and λ2=10000A˚\lambda_2=10000 Å, D=2 cmD=2 \mathrm{~cm} and d=2 mm=2×10−3 cmd=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~cm}. We have λ1λ2=h2h1=1200010000=65\frac{\lambda_1}{\lambda_2}=\frac{h_2}{h_1}=\frac{12000}{10000}=\frac{6}{5}as x=u1λ1Dd=5×12000×10−10×22×10−3=5×1.2×104×10−10×103=6 mmx=\frac{u_1 \lambda_1 D}{d}=\frac{5 \times 12000 \times 10^{-10} \times 2}{2 \times 10^{-3}}=5 \times 1.2 \times 10^4 \times 10^{-10} \times 10^3=6 \mathrm{~mm}
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