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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Physics
Chapter
Wave Optics

BITSAT 2012 · Physics · Wave Optics

In a Young's double slit experiment the angular width of a fringe...

In a Young's double slit experiment the angular width of a fringe formed on a distant screen is 1∘1^{{}^{\circ}}. The wavelength fo the light used is 6280A˚6280 \AA. What is the distance between the two coherent sources?

Options

  1. A0.036mm0.036 \mathrm{ mm}
  2. B0.12mm0.12 \mathrm{ mm}
  3. C6mm6 \mathrm{ mm}
  4. D4mm4 \mathrm{ mm}

Correct answer

A. 0.036mm0.036 \mathrm{ mm}

Step-by-step solution

The angular fringe width is given by α=λd\alpha=\frac{\lambda}{\mathrm{d}} where λ\lambda is wavelength and disthedistancebetweentwocoherentsources.Thus\mathrm{d} is the distance between two coherent sources. Thusd=λα Given, λ=6280A˚,α=1∘=π180radian\begin{array}{l} \mathrm{d}=\frac{\lambda}{\alpha} \\ \text{ Given, } \lambda=6280 \AA, \alpha=1^{{}^{\circ}}=\frac{\pi}{180} \mathrm{radian} \end{array}Thusd=6280×10−103.14×180=3.6×10−5m=0.036mmThus \mathrm{d}=\frac{6280 \times 10^{-10}}{3.14} \times180=3.6 \times10^{-5} \mathrm{ m}=0.036 \mathrm{ mm}
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