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JEE Main 201212 MayOnline
Exam
JEE Main
Year
2012
Date
12 May
Mode
Online
Subject
Physics
Chapter
Wave Optics

JEE Main 2012 · 12 May · Online · Physics · Wave Optics

The first diffraction minimum due to the single slit diffraction is...

The first diffraction minimum due to the single slit diffraction is seen at θ=30∘\theta=30^{\circ} for a light of wavelength 5000A˚5000 Å falling perpendicularly on the slit. The width of the slit is

Options

  1. A2.5×10−5 cm2.5 \times 10^{-5} \mathrm{~cm}
  2. B1.25×10−5 cm1.25 \times 10^{-5} \mathrm{~cm}
  3. C10×10−5 cm10 \times 10^{-5} \mathrm{~cm}
  4. D5×10−5 cm5 \times 10^{-5} \mathrm{~cm}

Correct answer

C. 10×10−5 cm10 \times 10^{-5} \mathrm{~cm}

Step-by-step solution

For first minimum, dsin⁡θ=λd \sin \theta=\lambda⇒\quadd=λsin⁡θ=5000×10−8 cmsin⁡30∘=5000×10−8 cm1/2=10×10−5 cm\begin{aligned} \Rightarrow \quadd & =\frac{\lambda}{\sin \theta}=\frac{5000 \times 10^{-8} \mathrm{~cm}}{\sin 30^{{}^{\circ}}}\\ & =\frac{5000 \times 10^{-8} \mathrm{~cm}}{1 / 2}=10 \times10^{-5} \mathrm{~cm} \end{aligned}$$
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