A star moving away from earth with a speed of 0.02c, while emitting light of frequency 6×1014Hz. What frequency will be observed on earth?
(Take c=3×108m/s)
A
0.24×1014Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.2×1014Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5.88×1014Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
0.12×1014Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C5.88×1014Hz The change in frequency is given by, Δff=vrc⇒Δf=f×vrc
∴Δf=6×1014×0.02cc=0.12×1014Hz
As the star is moving away from earth, the apparent frequency will be less than actual frequency, i.e. f′<f