Doppler's Effect of Light Questions in English

(N/A) The Doppler effect for light is the change in the observed frequency (or wavelength) of light due to the relative motion between the source of light and the observer.
When the source moves away from the observer,the wavefronts have to travel a greater distance,leading to a longer time interval between the arrival of successive wavefronts. This results in a decrease in the observed frequency and an increase in wavelength,known as 'red shift'.
When the source moves towards the observer,the wavefronts arrive at shorter intervals,leading to an increase in observed frequency and a decrease in wavelength,known as 'blue shift'.
For light,the Doppler shift formula is given by:
$\Delta \nu = -\nu \frac{v_{radial}}{c}$
where $\Delta \nu$ is the change in frequency,$\nu$ is the original frequency,$v_{radial}$ is the relative velocity of the source along the line of sight,and $c$ is the speed of light.
The fractional change in frequency is:
$\frac{\Delta \nu}{\nu} = -\frac{v_{radial}}{c}$

(N/A) For light,the Doppler shift is defined by the change in frequency or wavelength due to the relative motion between the source and the observer.
If the source is moving with a velocity $v$ relative to the observer (where $v \ll c$),the Doppler shift in frequency $\Delta \nu$ is given by:
$\Delta \nu = \nu \left( \frac{v}{c} \right) \cos \theta$
where:
$\nu$ is the original frequency,
$c$ is the speed of light,
$v$ is the relative velocity,
$\theta$ is the angle between the direction of motion and the line of sight.
Alternatively,the shift in wavelength $\Delta \lambda$ is given by:
$\Delta \lambda = \lambda \left( \frac{v}{c} \right) \cos \theta$

(D) The Doppler shift in wavelength for a source moving away from the observer is given by the formula: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$.
Given:
Speed of the galaxy $v = 286 \, km/s = 286 \times 10^{3} \, m/s$.
Speed of light $c = 3 \times 10^{8} \, m/s$.
Wavelength $\lambda = 630 \, nm = 630 \times 10^{-9} \, m$.
Rearranging the formula to find the shift $\Delta \lambda$:
$\Delta \lambda = \frac{v}{c} \times \lambda$.
Substituting the values:
$\Delta \lambda = \frac{286 \times 10^{3}}{3 \times 10^{8}} \times 630 \times 10^{-9}$.
$\Delta \lambda = \frac{286}{3 \times 10^{5}} \times 630 \times 10^{-9}$.
$\Delta \lambda = \frac{286 \times 630}{3} \times 10^{-14} = 286 \times 210 \times 10^{-14} = 60060 \times 10^{-14} = 6.006 \times 10^{-10} \, m$.
Comparing this with $x \times 10^{-10} \, m$,we get $x \approx 6$.

(C) The Doppler effect for light 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 relative velocity,and $c$ is the speed of light.
Given: $\lambda = 5890 \ \mathring{A}$,observed wavelength $\lambda' = 5896 \ \mathring{A}$.
Change in wavelength $\Delta \lambda = \lambda' - \lambda = 5896 - 5890 = 6 \ \mathring{A}$.
Speed of light $c = 3 \times 10^8 \ \text{m/s} = 3 \times 10^5 \ \text{km/s}$.
Using the formula: $v = c \times \frac{\Delta \lambda}{\lambda}$.
$v = (3 \times 10^5 \ \text{km/s}) \times \frac{6 \ \mathring{A}}{5890 \ \mathring{A}}$.
$v = \frac{18 \times 10^5}{5890} \ \text{km/s} \approx 305.6 \ \text{km/s}$.
Rounding to the nearest integer,the speed is approximately $306 \ \text{km/s}$.

(C) Given:
$\lambda_{\text{emitted}} = 670 \; nm$
$\lambda_{\text{obs}} = 670.7 \; nm$
$c = 3 \times 10^{8} \; m/s$
The Doppler shift formula for light when $v << c$ is given by:
$\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$
Where $\Delta \lambda = \lambda_{\text{obs}} - \lambda_{\text{emitted}} = 670.7 - 670 = 0.7 \; nm$.
Substituting the values:
$\frac{0.7}{670} = \frac{v}{3 \times 10^{8}}$
$v = \frac{0.7 \times 3 \times 10^{8}}{670}$
$v = \frac{2.1 \times 10^{8}}{670} \approx 0.003134 \times 10^{8} \; m/s$
$v \approx 3.13 \times 10^{5} \; m/s$.

(C) red shift in the spectrum of light from a galaxy indicates that the observed wavelength of the light is longer than the emitted wavelength.
According to the Doppler effect for light,an increase in wavelength (a shift towards the red end of the spectrum) occurs when the source of light is moving away from the observer.
Therefore,a red shift implies that the galaxy is receding from the earth.

(B) The apparent wavelength $\lambda'$ for a source moving away from the observer is given by the Doppler effect formula: $\lambda' = \lambda \sqrt{\frac{c+v}{c-v}}$.
Given that $\lambda' = \lambda + 0.02\% \text{ of } \lambda = \lambda(1 + 0.0002) = 1.0002\lambda$.
Thus,$\sqrt{\frac{c+v}{c-v}} = 1.0002$.
Squaring both sides,$\frac{c+v}{c-v} = (1.0002)^2 \approx 1.0004$.
Using the approximation for $v \ll c$,$\frac{\Delta \lambda}{\lambda} \approx \frac{v}{c}$.
Here,$\frac{\Delta \lambda}{\lambda} = \frac{0.02}{100} = 0.0002$.
Therefore,$v = 0.0002 \times c$.
Given $c = 3 \times 10^5 \, km/s$,we have $v = 0.0002 \times 3 \times 10^5 = 60 \, km/s$.

(B) The Doppler effect for light is given by the formula $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $\Delta \lambda$ is the change in wavelength,$\lambda$ is the actual wavelength,$v$ is the velocity of the star,and $c$ is the speed of light.
Given that the apparent wavelength is $0.02 \%$ more than the actual wavelength,we have $\frac{\Delta \lambda}{\lambda} = 0.02 \% = \frac{0.02}{100} = 2 \times 10^{-4}$.
Substituting the values into the formula: $2 \times 10^{-4} = \frac{v}{3 \times 10^8 \ m/s}$.
Solving for $v$: $v = (2 \times 10^{-4}) \times (3 \times 10^8 \ m/s) = 6 \times 10^4 \ m/s$.
Converting to $km/s$: $v = \frac{6 \times 10^4}{10^3} \ km/s = 60 \ km/s$.

(C) According to the Doppler effect for light,when a source of light moves towards an observer,the observed frequency $f'$ increases compared to the source frequency $f$.
This is given by the formula $f' = f \sqrt{\frac{c+v}{c-v}}$,where $v$ is the velocity of the source towards the observer and $c$ is the speed of light.
Since the frequency increases,the wavelength $\lambda$ decreases $(\lambda = c/f)$.
$A$ decrease in wavelength corresponds to a shift towards the blue end of the visible spectrum,which is known as a 'blue shift'.
Therefore,the yellow light will appear to shift towards the blue color.

(A) Given,the original wavelength $\lambda = 600 \ nm = 600 \times 10^{-9} \ m$.
The observed wavelength $\lambda' = 601 \ nm = 601 \times 10^{-9} \ m$.
The change in wavelength is $\Delta \lambda = \lambda' - \lambda = (601 - 600) \times 10^{-9} \ m = 1 \times 10^{-9} \ m$.
According to the Doppler effect for light,the relation is $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the speed of the galaxy and $c$ is the speed of light $(c = 3 \times 10^8 \ m/s)$.
Therefore,$v = c \cdot \frac{\Delta \lambda}{\lambda} = (3 \times 10^8 \ m/s) \cdot \frac{1 \times 10^{-9} \ m}{600 \times 10^{-9} \ m}$.
$v = 3 \times 10^8 \cdot \frac{1}{600} = \frac{3 \times 10^8}{600} = 0.5 \times 10^6 \ m/s$.
$v = 500,000 \ m/s = 500 \ km/s$.

(A) The formula for the Doppler shift in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$, where $v$ is the velocity of the source, $c$ is the speed of light, $\Delta \lambda$ is the shift in wavelength, and $\lambda$ is the original wavelength.
Given: $\Delta \lambda = 0.1 \text{ Å}$, $\lambda = 6000 \text{ Å}$, and $c = 3 \times 10^8 \text{ m/s}$.
Rearranging the formula to solve for $v$: $v = \frac{\Delta \lambda}{\lambda} \times c$.
Substituting the values: $v = \frac{0.1}{6000} \times 3 \times 10^8 \text{ m/s}$.
$v = \frac{1}{60000} \times 3 \times 10^8 \text{ m/s} = \frac{3 \times 10^8}{6 \times 10^4} \text{ m/s} = 0.5 \times 10^4 \text{ m/s} = 5000 \text{ m/s}$.
Converting to $\text{km/s}$: $v = 5 \text{ km/s}$.

(B) 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 $(3 \times 10^8 \, \text{m} \, \text{s}^{-1})$.
Given: $\lambda = 6600 \, \text{Å}$, $\Delta \lambda = 22 \, \text{Å}$.
Since the light is red-shifted, the wavelength increases, which implies the star is receding away from the earth.
Using the formula: $v = c \times \frac{\Delta \lambda}{\lambda}$.
Substituting the values: $v = (3 \times 10^8 \, \text{m} \, \text{s}^{-1}) \times \frac{22 \, \text{Å}}{6600 \, \text{Å}}$.
$v = (3 \times 10^8) \times \frac{1}{300} = 10^6 \, \text{m} \, \text{s}^{-1} = 10 \times 10^5 \, \text{m} \, \text{s}^{-1}$.
Thus, the star is receding away from the earth with a speed of $10 \times 10^5 \, \text{m} \, \text{s}^{-1}$.