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=3108 ms1)
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Hey dear,
● Answer- 17.3 GHz
● Explanation-
# Given-
v = c/2
f = 10 GHz
# Solution-
According to Doppler effect in light, let's calculate relative frequency-
f' = f√[(1+v/c)(1-v/c)]
f' = 10√[(1+c/2c)(1-c/2c)]
f' = 10√[(3/2)(1/2)]
f' = 10√3
f' = 10 × 1.73
f' = 17.3 GHz.
Frequency of microwave measured by observer is 17.3 GHz.
Hope this is useful...
● Answer- 17.3 GHz
● Explanation-
# Given-
v = c/2
f = 10 GHz
# Solution-
According to Doppler effect in light, let's calculate relative frequency-
f' = f√[(1+v/c)(1-v/c)]
f' = 10√[(1+c/2c)(1-c/2c)]
f' = 10√[(3/2)(1/2)]
f' = 10√3
f' = 10 × 1.73
f' = 17.3 GHz.
Frequency of microwave measured by observer is 17.3 GHz.
Hope this is useful...
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