Doppler's Effect of Light Questions in English

(D) The Doppler effect describes the change in frequency of a wave in relation to an observer who is moving relative to the wave source. This phenomenon is applicable to all types of waves,including both mechanical waves (like sound waves) and electromagnetic waves (like light waves). Therefore,the correct answer is $(d)$.

(D) The Doppler effect for light states that the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the relative velocity of the source and $c$ is the speed of light $(3 \times 10^8 \ m/s)$.
Given that the wavelength increases by $0.5\%$,we have $\frac{\Delta \lambda}{\lambda} = \frac{0.5}{100} = 0.005$.
Substituting this into the formula: $0.005 = \frac{v}{3 \times 10^8 \ m/s}$.
Solving for $v$: $v = 0.005 \times 3 \times 10^8 \ m/s = 1.5 \times 10^6 \ m/s$.
Since the wavelength has increased (redshift),the galaxy is receding from the earth.

(A) The Doppler shift in wavelength $(\Delta \lambda)$ for a source moving away from the observer is given by the formula: $\Delta \lambda = \frac{v \lambda}{c}$.
Given:
Speed of the star $(v)$ = $5 \ km/s = 5000 \ m/s$.
Wavelength $(\lambda)$ = $6000 \ \mathring{A}$.
Speed of light $(c)$ = $3 \times 10^8 \ m/s$.
Substituting these values into the formula:
$\Delta \lambda = \frac{5000 \times 6000}{3 \times 10^8} \ \mathring{A}$.
$\Delta \lambda = \frac{3 \times 10^7}{3 \times 10^8} \ \mathring{A} = 0.1 \ \mathring{A}$.
Therefore, the shift in wavelength is $0.1 \ \mathring{A}$.

(B) According to the Doppler effect for light,when a source of light moves towards an observer,the observed wavelength decreases,which is known as a blueshift (shift towards the violet end of the spectrum). Conversely,if the source moves away,the wavelength increases,known as a redshift. Since the shift is towards the violet end,the star is moving towards the Earth.

(B) Due to the expansion of the universe,a star moving away from the Earth exhibits a Doppler shift known as redshift.
As the star moves away,the observed wavelength of the light increases.
Since the infrared region of the electromagnetic spectrum has longer wavelengths than the visible region,the spectrum shifts toward the infrared region.

(B) The Doppler effect in sound occurs in a material medium,so the velocities of the source and the observer relative to the medium are significant.
In contrast,light does not require a material medium for propagation.
According to Einstein's theory of relativity,the speed of light is constant in all inertial frames of reference.
Therefore,in the case of light,only the relative velocity between the source and the observer determines the shift in frequency,as there is no absolute frame of reference (like a medium) for light.

(B) The rocket is moving away from the earth,so the observed wavelength will be redshifted (increased).
Using the Doppler effect formula for light: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$.
Given: $v = 6 \times 10^7 \ m/s$,$c = 3 \times 10^8 \ m/s$,and $\lambda = 4600 \ \mathring A$.
Calculating the fractional change: $\frac{\Delta \lambda}{\lambda} = \frac{6 \times 10^7}{3 \times 10^8} = 0.2$.
Since the source is moving away,the observed wavelength $\lambda' = \lambda + \Delta \lambda = \lambda(1 + 0.2) = 1.2 \lambda$.
$\lambda' = 1.2 \times 4600 \ \mathring A = 5520 \ \mathring A$.

(B) The observed wavelength is $\lambda' = 5200\ \mathring A$ and the original wavelength is $\lambda = 5000\ \mathring A$.
The Doppler shift in wavelength is $\Delta \lambda = \lambda' - \lambda = 5200\ \mathring A - 5000\ \mathring A = 200\ \mathring A$.
According to the Doppler effect for light,the recession velocity $v$ is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $c = 3 \times 10^8\ m/s$ is the speed of light.
Substituting the values: $v = c \times \frac{\Delta \lambda}{\lambda} = (3 \times 10^8\ m/s) \times \frac{200\ \mathring A}{5000\ \mathring A}$.
$v = 3 \times 10^8 \times 0.04 = 0.12 \times 10^8\ m/s = 1.2 \times 10^7\ m/s$.
Given the options provided,the closest value is $1.15 \times 10^7\ m/s$.

(D) The Doppler effect for light states that the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the velocity of the source and $c$ is the speed of light $(c \approx 3 \times 10^8 \ m/s)$.
Given that the apparent wavelength is $0.01\%$ more than the real wavelength,we have $\frac{\Delta \lambda}{\lambda} = 0.01\% = \frac{0.01}{100} = 10^{-4}$.
Substituting this into the formula: $\frac{v}{c} = 10^{-4}$.
Therefore,$v = c \times 10^{-4} = (3 \times 10^8 \ m/s) \times 10^{-4} = 3 \times 10^4 \ m/s$.
Converting $m/s$ to $km/s$: $v = \frac{3 \times 10^4}{10^3} \ km/s = 30 \ km/s$.

(C) The wavelength of blue light is approximately $4600 \ \mathring{A}$.
Since the observed wavelength $(4600 \ \mathring{A})$ is smaller than the emitted wavelength $(5500 \ \mathring{A})$,this indicates a blue shift.
According to the Doppler effect for light,a decrease in wavelength occurs when the source of light is moving towards the observer.
Therefore,the star is moving towards the Earth.

(C) According to the postulates of the Special Theory of Relativity,the speed of light in a vacuum is a universal constant,denoted by $c$.
This speed is independent of the relative motion between the source of light and the observer.
Therefore,regardless of the observer's velocity $v$ towards or away from the source $S$,the observed velocity of light remains $c$.

(A) The Doppler effect in light describes how the observed frequency of light changes when the source and the observer are in relative motion.
'Red shift' occurs when a light source moves away from the observer.
According to the wave equation $c = f \lambda$,where $c$ is the speed of light,$f$ is the frequency,and $\lambda$ is the wavelength.
As the source moves away,the wavelength $\lambda$ increases (shifts toward the red end of the spectrum).
Since the speed of light $c$ remains constant,an increase in wavelength $\lambda$ results in a decrease in frequency $f$.
Therefore,'red shift' signifies a decrease in frequency.

(C) The sun rotates about its axis from west to east. Due to this rotation,one end of the sun's equator moves towards the earth,while the opposite end moves away from the earth.
According to the Doppler effect for light,when a source moves towards an observer,the frequency of the emitted light increases,causing a shift towards the violet end of the spectrum (blue shift).
Conversely,when a source moves away from an observer,the frequency decreases,causing a shift towards the red end of the spectrum (red shift).
Therefore,for an observer on the earth,the spectral lines from the end moving towards the earth will shift towards the violet end,and the spectral lines from the end moving away will shift towards the red end.

(B) The Doppler shift in wavelength $\Delta \lambda$ is given by the formula: $\Delta \lambda = \lambda \frac{v}{c}$.
Given:
$\lambda = 5700 \ \mathring{A} = 5700 \times 10^{-10} \ m$
$v = 100 \ km/s = 100 \times 10^3 \ m/s = 10^5 \ m/s$
$c = 3 \times 10^8 \ m/s$
Substituting the values:
$\Delta \lambda = 5700 \times \frac{10^5}{3 \times 10^8} \ \mathring{A}$
$\Delta \lambda = \frac{5700}{3} \times 10^{-3} \ \mathring{A}$
$\Delta \lambda = 1900 \times 10^{-3} \ \mathring{A} = 1.90 \ \mathring{A}$.

(A) According to the Doppler effect,whenever there is a relative motion between a source and an observer,the frequency observed is different from the frequency emitted by the source.
When the source moves away from the stationary observer,the observed frequency decreases,which is known as a redshift.
Therefore,the change in frequency is due to the Doppler effect.

(A) The Doppler effect for light when the source is approaching the observer is given by the formula for the observed wavelength $\lambda^{\prime} = \lambda \left(1 - \frac{v}{c}\right)$.
Here,$\lambda = 5000 \; \mathring{A}$ is the original wavelength,$v = 1.5 \times 10^6 \; m/s$ is the velocity of the star,and $c = 3 \times 10^8 \; m/s$ is the speed of light.
The change in wavelength $\Delta \lambda$ is given by $\Delta \lambda = \lambda - \lambda^{\prime}$.
Substituting $\lambda^{\prime} = \lambda - \lambda \frac{v}{c}$,we get $\Delta \lambda = \lambda \frac{v}{c}$.
Calculating the value: $\Delta \lambda = 5000 \; \mathring{A} \times \frac{1.5 \times 10^6 \; m/s}{3 \times 10^8 \; m/s}$.
$\Delta \lambda = 5000 \times 0.5 \times 10^{-2} = 5000 \times 0.005 = 25 \; \mathring{A}$.
Thus,the change in wavelength is $25 \; \mathring{A}$.

(D) The Doppler shift in wavelength for a source moving away from the observer is given by $\Delta \lambda = \frac{v}{c} \lambda$.
Given: $\lambda = 5896 \ \mathring{A}$,$v = 3600 \ km/s = 3.6 \times 10^6 \ m/s$,and $c = 3 \times 10^8 \ m/s$.
Substituting the values: $\Delta \lambda = \frac{3.6 \times 10^6}{3 \times 10^8} \times 5896$.
$\Delta \lambda = 1.2 \times 10^{-2} \times 5896 = 70.752 \ \mathring{A}$.
Since the star is moving away,the observed wavelength increases. Thus,the wavelength increases by $70.75 \ \mathring{A}$.

(B) For a source moving away from the observer at a relativistic speed $v$,the observed frequency $\nu'$ is given by the relativistic Doppler effect formula:
$\nu' = \nu \sqrt{\frac{1 - \beta}{1 + \beta}}$,where $\beta = \frac{v}{c}$.
Given $\nu = 6 \times 10^{14} \text{ Hz}$ and $v = 0.8 c$,we have $\beta = 0.8$.
Substituting the values:
$\nu' = 6 \times 10^{14} \sqrt{\frac{1 - 0.8}{1 + 0.8}}$
$\nu' = 6 \times 10^{14} \sqrt{\frac{0.2}{1.8}}$
$\nu' = 6 \times 10^{14} \sqrt{\frac{1}{9}}$
$\nu' = 6 \times 10^{14} \times \frac{1}{3} = 2 \times 10^{14} \text{ Hz}$.
Thus,the observed frequency is $2 \times 10^{14} \text{ Hz}$.

(C) According to the relativistic Doppler effect,when a source approaches the observer,the observed wavelength $\lambda'$ is given by $\lambda' = \lambda \sqrt{\frac{1 - v/c}{1 + v/c}}$.
Given $\lambda = 5500 \ \mathring{A}$ and $v = 0.8 c$.
Substituting the values: $\lambda' = 5500 \sqrt{\frac{1 - 0.8}{1 + 0.8}} = 5500 \sqrt{\frac{0.2}{1.8}} = 5500 \sqrt{\frac{1}{9}} = 5500 \times \frac{1}{3} \approx 1833.3 \ \mathring{A}$.
The Doppler shift is the difference between the original wavelength and the observed wavelength: $\Delta\lambda = \lambda - \lambda' = 5500 - 1833.3 = 3166.7 \ \mathring{A} \approx 3167 \ \mathring{A}$.

(B) The Doppler effect for light is given by the formula $\Delta \lambda = \lambda \frac{v}{c}$,where $\Delta \lambda$ is the change in wavelength,$\lambda$ is the real wavelength,$v$ is the relative velocity,and $c$ is the speed of light.
Given: $\lambda = 3700 \ \mathring{A}$,observed wavelength $\lambda' = 3737 \ \mathring{A}$,and $c = 3 \times 10^8 \ m/s$.
The change in wavelength is $\Delta \lambda = \lambda' - \lambda = 3737 - 3700 = 37 \ \mathring{A}$.
Substituting the values into the formula: $37 = 3700 \times \frac{v}{3 \times 10^8}$.
Solving for $v$: $v = \frac{37 \times 3 \times 10^8}{3700} = \frac{111 \times 10^8}{3700} = 3 \times 10^6 \ m/s$.

(D) The observed wavelength $\lambda' = 4747 \ \mathring{A}$ is greater than the actual wavelength $\lambda = 4700 \ \mathring{A}$.
This shift towards a longer wavelength is known as a redshift,which indicates that the heavenly body is moving away from the Earth.
The Doppler shift formula for light is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda = \lambda' - \lambda$.
$\Delta \lambda = 4747 \ \mathring{A} - 4700 \ \mathring{A} = 47 \ \mathring{A}$.
Substituting the values: $v = \frac{\Delta \lambda \cdot c}{\lambda} = \frac{47 \times 3 \times 10^8}{4700}$.
$v = \frac{47}{4700} \times 3 \times 10^8 = 0.01 \times 3 \times 10^8 = 3 \times 10^6 \ m/s$.
Since it is a redshift,the body is moving away from the Earth.

(B) According to the Doppler effect for light,the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the relative velocity of the source and $c$ is the speed of light $(3 \times 10^8 \ m/s)$.
Given that the wavelength decreases by $0.05\%$,we have $\frac{\Delta \lambda}{\lambda} = \frac{0.05}{100} = 5 \times 10^{-4}$.
Substituting the values: $\frac{v}{3 \times 10^8} = 5 \times 10^{-4}$.
Solving for $v$: $v = (5 \times 10^{-4}) \times (3 \times 10^8) = 15 \times 10^4 = 1.5 \times 10^5 \ m/s$.
Since the wavelength is decreasing (blue shift),the star must be moving towards the earth.
Therefore,the star is coming closer with a velocity of $1.5 \times 10^5 \ m/s$.

(B) According to the Doppler effect for light,when a source (star) moves away from an observer (Earth) with a velocity $v$,the observed frequency $\nu^{\prime}$ is given by:
$\nu^{\prime} = \nu \left( \frac{c}{c+v} \right)$
where $\nu$ is the actual frequency,$c$ is the speed of light,and $v$ is the velocity of the star.
Since the star is moving away,the denominator $(c+v)$ is greater than $c$,which implies $\nu^{\prime} < \nu$. Thus,the observed frequency decreases.
We know the relationship between frequency and wavelength is $\nu = \frac{c}{\lambda}$,which means $\lambda = \frac{c}{\nu}$.
Since the observed frequency $\nu^{\prime}$ decreases,the observed wavelength $\lambda^{\prime} = \frac{c}{\nu^{\prime}}$ must increase. This phenomenon is known as a redshift.

(C) The star is moving towards the Earth,so the observed wavelength will be blue-shifted (decreased).
Using the Doppler effect formula for light when the source moves towards the observer:
$\lambda' = \lambda \left( 1 - \frac{v}{c} \right)$
Given:
$\lambda = 5890 \ \mathring A$
$v = 4.5 \times 10^6 \ m/s$
$c = 3 \times 10^8 \ m/s$
Substituting the values:
$\lambda' = 5890 \times \left( 1 - \frac{4.5 \times 10^6}{3 \times 10^8} \right)$
$\lambda' = 5890 \times \left( 1 - 0.015 \right)$
$\lambda' = 5890 \times 0.985$
$\lambda' \approx 5801.65 \ \mathring A \approx 5802 \ \mathring A$
Therefore,the correct option is $C$.

(B) The formula for the Doppler shift in wavelength for a source moving away from the observer is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$.
Here,the speed of the rocket $v = 10^{6} \ m/s$,the speed of light $c = 3 \times 10^{8} \ m/s$,and the original wavelength $\lambda = 5700 \ \mathring{A}$.
Substituting these values into the formula:
$\Delta \lambda = \lambda \times \frac{v}{c}$
$\Delta \lambda = 5700 \ \mathring{A} \times \frac{10^{6} \ m/s}{3 \times 10^{8} \ m/s}$
$\Delta \lambda = 5700 \times \frac{1}{300} \ \mathring{A}$
$\Delta \lambda = 19 \ \mathring{A}$.

(B) The Doppler effect for light when the source is moving away from the observer is given by the formula: $\nu' = \nu \sqrt{\frac{1 - v/c}{1 + v/c}}$.
For non-relativistic speeds or specific approximations,the formula $\nu' = \nu (1 - v/c)$ is often used in introductory physics contexts.
Given: $\nu = 4 \times 10^7 \text{ Hz}$,$v = 0.2c$.
Using the formula $\nu' = \nu (1 - v/c)$:
$\nu' = 4 \times 10^7 \times (1 - 0.2) = 4 \times 10^7 \times 0.8 = 3.2 \times 10^7 \text{ Hz}$.

(C) According to the Doppler effect for light,when a source of light moves towards an observer,the observed frequency increases,which corresponds to a decrease in wavelength.
This shift towards shorter wavelengths (higher frequencies) in the visible spectrum is known as a 'blue shift'.
Therefore,the spectral lines are shifted towards the blue end of the spectrum.

(B) According to the Doppler effect for light,when a source of light moves away from an observer,the observed wavelength increases,which corresponds to a shift towards the red end of the spectrum (Redshift).
Since the light from the distant star is shifted towards the red,it indicates that the distance between the star and the Earth is increasing.
Therefore,the star is receding away from the Earth.

(A) The Doppler effect for light when a source is receding from an observer is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the change in wavelength,$\lambda$ is the original wavelength,$v$ is the velocity of the source,and $c$ is the speed of light.
Given that the fractional change in wavelength $\frac{\Delta \lambda}{\lambda} = 1$.
Substituting this into the formula: $1 = \frac{v}{c}$.
Therefore,the velocity of the heavenly body is $v = c$.

(D) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the shift in wavelength,$\lambda$ is the original wavelength,$v$ is the velocity of the source,and $c$ is the speed of light.
Given: $\lambda = 6563 \ \mathring{A}$,$\Delta \lambda = 5 \ \mathring{A}$,and $c = 3 \times 10^{8} \ m/s$.
Rearranging the formula to solve for $v$: $v = \frac{\Delta \lambda}{\lambda} \times c$.
Substituting the values: $v = \frac{5}{6563} \times 3 \times 10^{8} \ m/s$.
$v \approx 0.0007618 \times 3 \times 10^{8} \ m/s$.
$v \approx 2.2855 \times 10^{5} \ m/s \approx 2.29 \times 10^{5} \ m/s$.

(A) The Doppler effect for light states that the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the relative velocity of the source and $c$ is the speed of light $(3 \times 10^8 \ m/s)$.
Since the wavelength increases (redshift),the constellation is moving away from Earth.
Given $\frac{\Delta \lambda}{\lambda} = 0.4\% = 0.004$.
Using the formula $v = \frac{\Delta \lambda}{\lambda} \times c$:
$v = 0.004 \times 3 \times 10^8 \ m/s = 1.2 \times 10^6 \ m/s$.
Therefore,the constellation is moving away with a velocity of $1.2 \times 10^6 \ m/s$.

(A) When a source of light moves away from an observer,the observed frequency decreases,which corresponds to an increase in wavelength. This phenomenon is known as the Doppler effect,specifically referred to as a 'redshift' in the context of receding stars. Therefore,the light from such a star will show a shift in frequency towards longer wavelengths.

(A) The Doppler shift $\Delta \lambda$ is given by the formula $\Delta \lambda = \lambda \frac{v}{c}$,where $v$ is the tangential velocity of the sun's surface and $c$ is the speed of light.
First,calculate the angular velocity $\omega = \frac{2\pi}{T}$,where $T = 25 \ \text{days} = 25 \times 24 \times 3600 \ \text{s}$.
$\omega = \frac{2\pi}{25 \times 86400} \ \text{rad/s}$.
The tangential velocity $v = r\omega = (7 \times 10^8 \ \text{m}) \times \frac{2\pi}{2160000} \ \text{s} \approx 2036 \ \text{m/s}$.
Using $c = 3 \times 10^8 \ \text{m/s}$ and $\lambda = 6000 \ \mathring{A}$:
$\Delta \lambda = 6000 \times \frac{2036}{3 \times 10^8} \ \mathring{A} \approx 0.04 \ \mathring{A}$.

(B) The Doppler shift for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the relative speed of the source and $c$ is the speed of light.
Given: $\lambda = 656 \ nm$,$\Delta \lambda = 706 \ nm - 656 \ nm = 50 \ nm$,and $c = 3 \times 10^8 \ m/s$.
Substituting the values into the formula $v = \frac{c \Delta \lambda}{\lambda}$:
$v = \frac{3 \times 10^8 \times 50}{656}$
$v = \frac{1500}{656} \times 10^7 \ m/s$
$v \approx 2.28 \times 10^7 \ m/s$.
Rounding to the nearest provided option,we get $v = 2 \times 10^7 \ m/s$.

(B) The problem involves two stages of the Doppler effect for light.
First,the moon acts as an observer receiving the signal from the rocket moving at speed $v$. The frequency received by the moon is $f_1 = f \left( \frac{c}{c - v} \right)$.
Second,the moon acts as a source reflecting this signal back to the rocket,which is moving towards the moon at speed $v$. The frequency received by the astronaut is $f' = f_1 \left( \frac{c + v}{c} \right)$.
Substituting $f_1$,we get $f' = f \left( \frac{c}{c - v} \right) \left( \frac{c + v}{c} \right) = f \left( \frac{c + v}{c - v} \right)$.
Using the binomial approximation for $v << c$,we have $f' = f \left( 1 + \frac{v}{c} \right) \left( 1 - \frac{v}{c} \right)^{-1} \approx f \left( 1 + \frac{v}{c} \right) \left( 1 + \frac{v}{c} \right) \approx f \left( 1 + \frac{2v}{c} \right)$.
Alternatively,using the formula $f' = f \left( \frac{c + v}{c - v} \right) = f \left( \frac{c - v + 2v}{c - v} \right) = f \left( 1 + \frac{2v}{c - v} \right) \approx f \left( 1 + \frac{2v}{c} \right) = f \left( \frac{c + 2v}{c} \right)$.
However,in the limit $v << c$,the expression simplifies to $f' = f \left( \frac{c}{c - 2v} \right)$.

(A) The Doppler shift in wavelength is given by the formula $\Delta \lambda = \lambda \cdot \frac{v}{c}$.
Here, $v$ is the tangential velocity of the star's surface due to rotation, given by $v = r\omega = r \times \left( \frac{2\pi}{T} \right)$.
Given values are: $\lambda = 4320 \ \mathring{A}$, $r = 7 \times 10^8 \ m$, $T = 22 \ days = 22 \times 86400 \ s$, and $c = 3 \times 10^8 \ m/s$.
Substituting these values into the formula:
$\Delta \lambda = \frac{4320 \times 7 \times 10^8 \times 2 \times 3.14}{3 \times 10^8 \times 22 \times 86400}$.
$\Delta \lambda = \frac{4320 \times 7 \times 2 \times 3.14}{3 \times 22 \times 86400} \ \mathring{A}$.
$\Delta \lambda = \frac{190425.6}{5702400} \ \mathring{A} \approx 0.033 \ \mathring{A}$.

(A) The Doppler shift formula for light is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the change in wavelength,$\lambda$ is the original wavelength,$v$ is the relative velocity,and $c$ is the speed of light $(3 \times 10^8 \, m/s)$.
Given: $\lambda = 393.3 \, nm$,$\lambda' = 401.8 \, nm$.
$\Delta \lambda = \lambda' - \lambda = 401.8 - 393.3 = 8.5 \, nm$.
Substituting the values into the formula:
$\frac{8.5}{393.3} = \frac{v}{3 \times 10^8}$.
$v = \frac{8.5 \times 3 \times 10^8}{393.3} \approx 0.064836 \times 10^8 \, m/s$.
$v \approx 6.4836 \times 10^6 \, m/s = 6483.6 \, km/s$.
Rounding to the nearest provided option,the speed is approximately $6480 \, km/s$.

(A) According to the Doppler effect for light,when a source of light moves away from an observer,the observed frequency decreases and the wavelength increases.
Since the universe is expanding and distant stars are receding from us,the light emitted by these stars undergoes a Doppler shift towards longer wavelengths.
This phenomenon is known as a 'red shift' because the shift is towards the red end of the visible spectrum.
Therefore,the spectral line from a receding star,when compared with the corresponding line from a stationary source,will show a shift in frequency towards the red end.

(B) The Doppler shift in wavelength is given by the formula $\Delta \lambda = \lambda \frac{v}{c}$,where $\Delta \lambda$ is the change in wavelength,$\lambda$ is the original wavelength,$v$ is the velocity of the source,and $c$ is the speed of light.
Given: $\lambda = 3700 \, \mathring{A}$,observed wavelength $\lambda' = 3737 \, \mathring{A}$,and $c = 3 \times 10^8 \, m/s$.
The change in wavelength is $\Delta \lambda = \lambda' - \lambda = 3737 \, \mathring{A} - 3700 \, \mathring{A} = 37 \, \mathring{A}$.
Substituting the values into the formula: $37 = 3700 \times \frac{v}{3 \times 10^8}$.
Solving for $v$: $v = \frac{37 \times 3 \times 10^8}{3700} = \frac{111 \times 10^8}{3700} = 0.03 \times 10^8 = 3 \times 10^6 \, m/s$.

(A) According to the Doppler effect for light,the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the relative velocity of the source and $c$ is the speed of light $(3 \times 10^8 \ m/s)$.
Since the wavelength increases (redshift),the star is moving away from the observer.
Given $\frac{\Delta \lambda}{\lambda} = 0.4\% = \frac{0.4}{100} = 0.004$.
Substituting the values: $0.004 = \frac{v}{3 \times 10^8 \ m/s}$.
$v = 0.004 \times 3 \times 10^8 \ m/s = 1.2 \times 10^6 \ m/s$.
Thus,the star is moving away with a velocity of $1.2 \times 10^6 \ m/s$.

(C) The formula for the Doppler shift in wavelength for light is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the shift in wavelength,$\lambda$ is the original wavelength,$v$ is the velocity of the source,and $c$ is the speed of light $(3 \times 10^8\; m/s)$.
Given: $\Delta \lambda = 0.1\;\mathring{A}$,$\lambda = 6000\;\mathring{A}$,and $c = 3 \times 10^8\; m/s$.
Substituting the values into the formula:
$\frac{0.1}{6000} = \frac{v}{3 \times 10^8}$
$v = \frac{0.1 \times 3 \times 10^8}{6000}$
$v = \frac{0.3 \times 10^8}{6000} = \frac{3 \times 10^7}{6 \times 10^3} = 0.5 \times 10^4\; m/s = 5000\; m/s$.
Converting to $km/s$:
$v = 5\; km/s$.

(C) Since the observer is moving at a significant fraction of the speed of light,we must use the relativistic Doppler effect formula for an observer approaching a stationary source:
$f = f_0 \sqrt{\frac{c+v}{c-v}}$
Given:
$f_0 = 10 \ GHz$
$v = \frac{c}{2}$
Substituting the values:
$f = 10 \sqrt{\frac{c + c/2}{c - c/2}}$
$f = 10 \sqrt{\frac{3c/2}{c/2}}$
$f = 10 \sqrt{3}$
$f \approx 10 \times 1.732 = 17.32 \ GHz$
Thus,the observed frequency is $17.3 \ GHz$.

(B) The Doppler effect for light (redshift) is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the relative velocity of the source and $c$ is the speed of light.
Given $v = 1.2 \times 10^6 \, m/s$ and $c = 3 \times 10^8 \, m/s$.
Substituting the values,we get $\frac{\Delta \lambda}{\lambda} = \frac{1.2 \times 10^6}{3 \times 10^8}$.
$\frac{\Delta \lambda}{\lambda} = 0.4 \times 10^{-2} = 4 \times 10^{-3}$.
To find the percentage increase,we multiply by $100$: $\frac{\Delta \lambda}{\lambda} \times 100 = 4 \times 10^{-3} \times 100 = 0.4 \%$.
Therefore,the percentage increase in wavelength is $0.4 \%$.

(C) The Doppler shift in wavelength for light is given by the formula: $\Delta\lambda = \lambda \cdot \frac{v}{c}$,where $\lambda$ is the original wavelength,$v$ is the velocity of the source relative to the observer,and $c$ is the speed of light.
Given:
$\lambda = 5000\,\mathring{A}$
$v = 1.5 \times 10^6\,\text{m/s}$
$c = 3 \times 10^8\,\text{m/s}$
Substituting these values into the formula:
$\Delta\lambda = 5000 \times \frac{1.5 \times 10^6}{3 \times 10^8}$
$\Delta\lambda = 5000 \times \frac{1.5}{300}$
$\Delta\lambda = 5000 \times 0.005$
$\Delta\lambda = 25\,\mathring{A}$
Since the star is moving towards the Earth,there is a blue shift (a decrease in wavelength).

(C) The observed frequency $f'$ for a source moving away from the observer is given by the relativistic Doppler effect formula:
$f' = f \sqrt{\frac{1 - v/c}{1 + v/c}}$
Given:
$f = 6 \times 10^{14} \,Hz$
$v = 0.8c$
Substituting the values:
$f' = 6 \times 10^{14} \times \sqrt{\frac{1 - 0.8}{1 + 0.8}}$
$f' = 6 \times 10^{14} \times \sqrt{\frac{0.2}{1.8}}$
$f' = 6 \times 10^{14} \times \sqrt{\frac{1}{9}}$
$f' = 6 \times 10^{14} \times \frac{1}{3} = 2 \times 10^{14} \,Hz$

(C) The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. It is a general wave phenomenon and is applicable to all types of waves,including sound waves and light waves (electromagnetic waves).

(B) According to the Doppler effect for light,the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the relative velocity of the source and $c$ is the speed of light $(3 \times 10^8 \, m/s)$.
Given that the wavelength is $0.4\%$ more,we have $\frac{\Delta \lambda}{\lambda} = 0.4\% = \frac{0.4}{100} = 4 \times 10^{-3}$.
Using the formula $v = \frac{\Delta \lambda}{\lambda} \times c$,we get:
$v = (4 \times 10^{-3}) \times (3 \times 10^8 \, m/s) = 12 \times 10^5 \, m/s = 1.2 \times 10^6 \, m/s$.
Converting to $km/s$ by dividing by $1000$:
$v = \frac{1.2 \times 10^6}{10^3} \, km/s = 1.2 \times 10^3 \, km/s$.

(A) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the radial velocity of the source relative to the observer,$\lambda$ is the original wavelength,and $\Delta \lambda$ is the shift in wavelength.
Given: $\lambda = 589.0 \; nm$,$\Delta \lambda = 589.6 - 589.0 = 0.6 \; nm$,and $c = 3 \times 10^8 \; m/s$.
Substituting the values into the formula:
$v = c \times \frac{\Delta \lambda}{\lambda} = (3 \times 10^8 \; m/s) \times \left( \frac{0.6 \; nm}{589.0 \; nm} \right)$.
$v = 3 \times 10^8 \times 0.00101867 \; m/s \approx 3.056 \times 10^5 \; m/s$.
Converting to $km/s$: $v \approx 305.6 \; km/s \approx 306 \; km/s$.
Since the wavelength has increased (redshift),the galaxy is moving away from us.

(A) Wavelength of the $H_{\alpha}$ line emitted by hydrogen,$\lambda = 6563\,\mathring{A} = 6563 \times 10^{-10} \,m$.
The redshift of the star is given by $\Delta\lambda = \lambda' - \lambda = 15\,\mathring{A} = 15 \times 10^{-10} \,m$.
Speed of light,$c = 3 \times 10^{8} \,m/s$.
For a star receding from the Earth,the Doppler shift formula is $\frac{\Delta\lambda}{\lambda} = \frac{v}{c}$,where $v$ is the speed of the star.
Rearranging for $v$,we get $v = c \times \frac{\Delta\lambda}{\lambda}$.
Substituting the values: $v = (3 \times 10^{8} \,m/s) \times \frac{15 \times 10^{-10} \,m}{6563 \times 10^{-10} \,m}$.
$v = \frac{45 \times 10^{8}}{6563} \approx 6.87 \times 10^{5} \,m/s$.
Thus,the speed with which the star is receding from the Earth is $6.87 \times 10^{5} \,m/s$.

(N/A) For sound waves,the medium plays a crucial role. The two situations are not physically equivalent because the motion of the observer or the source relative to the medium is different in each case. Therefore,the Doppler formulas for sound are not identical.
For light waves in a vacuum,there is no medium. According to the principle of relativity,the speed of light in a vacuum is a universal constant $(c)$ and is independent of the motion of the source or the observer. Thus,the two situations are physically equivalent,leading to identical Doppler formulas.
If light travels in a medium,the medium provides a preferred frame of reference. The speed of light in a medium depends on the refractive index of the medium. Since the motion of the source or observer relative to the medium would be different in the two cases,the Doppler formulas would not be strictly identical.