$A$ star moves away from the Earth at a speed of $0.8\,c$ while emitting light of frequency $6 \times 10^{14} \,Hz$. What frequency will be observed on the Earth? ($c =$ speed of light)

For sound waves,the Doppler formula for frequency shift differs slightly between the two situations:
$(i)$ source at rest; observer moving,and
$(ii)$ source moving; observer at rest.
The exact Doppler formulas for the case of light waves in vacuum are,however,strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?

$A$ star producing light of wavelength $6000 \ \mathring{A}$ moves away from the Earth with a speed of $5 \ km/s$. Due to the Doppler effect, the shift in wavelength will be $...... \ \mathring{A}$ $(c = 3 \times 10^8 \ m/s)$.

$A$ heavenly body is receding from Earth such that the fractional change in wavelength $\lambda$ is $1$. What is its velocity?

The velocity of light emitted by a source $S$ observed by an observer $O$,who is at rest with respect to $S$,is $c$. If the observer moves towards $S$ with velocity $v$,the velocity of light as observed will be

In the context of the Doppler effect in light,the term 'red shift' signifies: