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JEE Main 20172 April
Exam
JEE Main
Year
2017
Date
2 April
Subject
Physics
Chapter
Waves and Sound

JEE Main 2017 · 2 April · Physics · Waves and Sound

An observer is moving with half the speed of light towards a...

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10GHz10GHz. What is the frequency of the microwave measured by the observer? (speed of light =3×108ms−1= 3 \times {10}^{8}m{s}^{- 1})

Options

  1. A15.3GHz15 . 3GHz
  2. B10.1GHz10 . 1GHz
  3. C12.1GHz12 . 1GHz
  4. D17.3GHz17 . 3GHz

Correct answer

D. 17.3GHz17 . 3GHz

Step-by-step solution

\dfrac{c}{{f}^{′}} = \dfrac{c}{f}\sqrt{\dfrac{1 - \beta}{1 + \beta}}, \left[β=vcβ=12\begin{array}{ll}\beta = \dfrac{v}{c} \\ \beta = \dfrac{1}{2}\end{array}\right]

⇒1f′=1f1−121+12=131f⇒ \dfrac{1}{{f}^{′}} = \dfrac{1}{f}\sqrt{\dfrac{1 - \dfrac{1}{2}}{1 + \dfrac{1}{2}}} = \sqrt{\dfrac{1}{3}}\dfrac{1}{f}

⇒f′=3f⇒ {f}^{′} = \sqrt{3}f

⇒f′=1.73(10)⇒ {f}^{′} = 1.73\left(10\right)

⇒f′=17.3GHz⇒ {f}^{′} = 17.3GHz
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