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An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 108 ms–1)
- a)17.3 GHz
- b)15.3 GHz
- c)10.1 GHz
- d)12.1 GHz
Correct answer is option 'A'. Can you explain this answer?
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An observer is moving with half the speed of light towards a stationar...
Doppler effect in light (speed of observer is not very small compare to speed of light)
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An observer is moving with half the speed of light towards a stationar...
Given:
- Speed of light (c) = 3 x 10^8 m/s
- Frequency of the stationary microwave source (f) = 10 GHz (10 x 10^9 Hz)
- The observer is moving with half the speed of light towards the source.
To find:
The frequency of the microwave measured by the observer.
When an observer is moving towards a stationary source, the frequency of the waves observed by the observer will be higher than the frequency emitted by the source. This phenomenon is known as the Doppler effect.
The formula to calculate the observed frequency (f') in terms of the emitted frequency (f) and the velocity of the observer (v) is:
f' = f * (c + v) / (c - v)
Where:
- f' is the observed frequency
- f is the emitted frequency
- c is the speed of light
- v is the velocity of the observer
Substituting the given values into the formula:
f' = 10 x 10^9 Hz * (3 x 10^8 m/s + (0.5 * 3 x 10^8 m/s)) / (3 x 10^8 m/s - (0.5 * 3 x 10^8 m/s))
Simplifying the equation:
f' = 10 x 10^9 Hz * (1 + 0.5) / (1 - 0.5)
f' = 10 x 10^9 Hz * (1.5) / (0.5)
f' = 10 x 10^9 Hz * 3
f' = 30 x 10^9 Hz
Converting the frequency to GHz:
f' = 30 GHz
Answer:
The frequency of the microwave measured by the observer is 30 GHz.
- Speed of light (c) = 3 x 10^8 m/s
- Frequency of the stationary microwave source (f) = 10 GHz (10 x 10^9 Hz)
- The observer is moving with half the speed of light towards the source.
To find:
The frequency of the microwave measured by the observer.
Explanation:
When an observer is moving towards a stationary source, the frequency of the waves observed by the observer will be higher than the frequency emitted by the source. This phenomenon is known as the Doppler effect.
The formula to calculate the observed frequency (f') in terms of the emitted frequency (f) and the velocity of the observer (v) is:
f' = f * (c + v) / (c - v)
Where:
- f' is the observed frequency
- f is the emitted frequency
- c is the speed of light
- v is the velocity of the observer
Substituting the given values into the formula:
f' = 10 x 10^9 Hz * (3 x 10^8 m/s + (0.5 * 3 x 10^8 m/s)) / (3 x 10^8 m/s - (0.5 * 3 x 10^8 m/s))
Simplifying the equation:
f' = 10 x 10^9 Hz * (1 + 0.5) / (1 - 0.5)
f' = 10 x 10^9 Hz * (1.5) / (0.5)
f' = 10 x 10^9 Hz * 3
f' = 30 x 10^9 Hz
Converting the frequency to GHz:
f' = 30 GHz
Answer:
The frequency of the microwave measured by the observer is 30 GHz.
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An observer is moving with half the speed of light towards a stationar...
Doppler effect in light (speed of observer is not very small compare to speed of light)
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An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 108 ms–1)a)17.3 GHzb)15.3 GHzc)10.1 GHzd)12.1 GHzCorrect answer is option 'A'. Can you explain this answer?
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