367734
A galaxy is moving away from the Earth so that a spectral line at 600 nm is observed at 601 nm . Then the speed of the galaxy with respect to the Earth is
1 \({500 {~km} {~s}^{-1}}\)
2 \({50 {~km} {~s}^{-1}}\)
3 \({200 {~km} {~s}^{-1}}\)
4 \({20 {~km} {~s}^{-1}}\)
Explanation:
Speed of galaxy can be calculated as \({\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}}\) \({\Rightarrow \dfrac{1}{600}=\dfrac{v}{3 \times 10^{8}}}\) \(v = \frac{{3 \times {{10}^8}}}{{600}} = 500\;\,km/{s^{ - 1}}\)
KCET - 2024
PHXII10:WAVE OPTICS
367735
Light emitted from a distant stationary star is observed at frequency 5000 \(MHz\). If the star starts approaching us with speed \(9 \times {10^5}m{\rm{/}}s\) then the observed frequency will be
1 \(5025MHz\)
2 \(5015MHz\)
3 \(5005MHz\)
4 \(5000MHz\)
Explanation:
Given that \(v = 5000MHz\,\,\& \,\,u = 9 \times {10^5}m/s\) The apparent frequency is \(v' = v\left( {1 + \frac{u}{c}} \right)\) The \(\left( + \right)\) ve shows that the relative separation between observer and source is increasing \(v' = 5000MHz\left( {1 + \frac{{9 \times {{10}^5}}}{{3 \times {{10}^8}}}} \right) = 5015MHz\)
PHXII10:WAVE OPTICS
367736
Assertion : In everyday life the Doppler's effect is observed readily for sound waves than light waves. Reason : Velocity of light is greater than that of sound.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
For both \(v=n \lambda\). For light, we denote \(v=c, \mathrm{c}\) is very high. The Doppler effect is more readily observed in sound waves due to their larger wavelengths, while the same is not the case with light waves because they have much shorter wavelengths in everyday life. So correct option is (2).
PHXII10:WAVE OPTICS
367737
It is believed that the universe is expanding and hence the distant stars are receding from us. Light from such a star will show
1 Shift in frequency towards shorter wavelengths
2 Shift in frequency towards longer wavelengths
3 A shift in frequency sometimes towards longer and sometimes towards shorter wavelengths
4 No shift in frequency but a decrease in intensity
Explanation:
Conceptual Question
PHXII10:WAVE OPTICS
367738
The \(6563\mathop A\limits^ \circ {H_ \propto }\) sign line emitted by hydrogen in a star is found to be red-shifted by \(15\mathop A\limits^ \circ \). Estimate the speed with the star receding from the earth.
1 \(8.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
2 \(7.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
3 \(6.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) approaching the earth
4 \(6.86 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
Explanation:
Given, wavelength of \({H_\alpha },\) \(\lambda = 6563\mathop A\limits^o = 6563 \times {10^{ - 10}}{\mkern 1mu} m\) Red-shift \(\Delta \lambda = 15\mathop A\limits^ \circ \) As the wavelength is found to be red-shifted, hence star is receding away from earth \(\Delta \lambda = v\frac{\lambda }{c} \Rightarrow v = \frac{{\Delta \lambda \cdot c}}{\lambda } = - \frac{{15 \times 3 \times {{10}^8}}}{{6563}}\) \(v = 6.86 \times {10^5}m/s\)
367734
A galaxy is moving away from the Earth so that a spectral line at 600 nm is observed at 601 nm . Then the speed of the galaxy with respect to the Earth is
1 \({500 {~km} {~s}^{-1}}\)
2 \({50 {~km} {~s}^{-1}}\)
3 \({200 {~km} {~s}^{-1}}\)
4 \({20 {~km} {~s}^{-1}}\)
Explanation:
Speed of galaxy can be calculated as \({\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}}\) \({\Rightarrow \dfrac{1}{600}=\dfrac{v}{3 \times 10^{8}}}\) \(v = \frac{{3 \times {{10}^8}}}{{600}} = 500\;\,km/{s^{ - 1}}\)
KCET - 2024
PHXII10:WAVE OPTICS
367735
Light emitted from a distant stationary star is observed at frequency 5000 \(MHz\). If the star starts approaching us with speed \(9 \times {10^5}m{\rm{/}}s\) then the observed frequency will be
1 \(5025MHz\)
2 \(5015MHz\)
3 \(5005MHz\)
4 \(5000MHz\)
Explanation:
Given that \(v = 5000MHz\,\,\& \,\,u = 9 \times {10^5}m/s\) The apparent frequency is \(v' = v\left( {1 + \frac{u}{c}} \right)\) The \(\left( + \right)\) ve shows that the relative separation between observer and source is increasing \(v' = 5000MHz\left( {1 + \frac{{9 \times {{10}^5}}}{{3 \times {{10}^8}}}} \right) = 5015MHz\)
PHXII10:WAVE OPTICS
367736
Assertion : In everyday life the Doppler's effect is observed readily for sound waves than light waves. Reason : Velocity of light is greater than that of sound.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
For both \(v=n \lambda\). For light, we denote \(v=c, \mathrm{c}\) is very high. The Doppler effect is more readily observed in sound waves due to their larger wavelengths, while the same is not the case with light waves because they have much shorter wavelengths in everyday life. So correct option is (2).
PHXII10:WAVE OPTICS
367737
It is believed that the universe is expanding and hence the distant stars are receding from us. Light from such a star will show
1 Shift in frequency towards shorter wavelengths
2 Shift in frequency towards longer wavelengths
3 A shift in frequency sometimes towards longer and sometimes towards shorter wavelengths
4 No shift in frequency but a decrease in intensity
Explanation:
Conceptual Question
PHXII10:WAVE OPTICS
367738
The \(6563\mathop A\limits^ \circ {H_ \propto }\) sign line emitted by hydrogen in a star is found to be red-shifted by \(15\mathop A\limits^ \circ \). Estimate the speed with the star receding from the earth.
1 \(8.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
2 \(7.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
3 \(6.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) approaching the earth
4 \(6.86 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
Explanation:
Given, wavelength of \({H_\alpha },\) \(\lambda = 6563\mathop A\limits^o = 6563 \times {10^{ - 10}}{\mkern 1mu} m\) Red-shift \(\Delta \lambda = 15\mathop A\limits^ \circ \) As the wavelength is found to be red-shifted, hence star is receding away from earth \(\Delta \lambda = v\frac{\lambda }{c} \Rightarrow v = \frac{{\Delta \lambda \cdot c}}{\lambda } = - \frac{{15 \times 3 \times {{10}^8}}}{{6563}}\) \(v = 6.86 \times {10^5}m/s\)
367734
A galaxy is moving away from the Earth so that a spectral line at 600 nm is observed at 601 nm . Then the speed of the galaxy with respect to the Earth is
1 \({500 {~km} {~s}^{-1}}\)
2 \({50 {~km} {~s}^{-1}}\)
3 \({200 {~km} {~s}^{-1}}\)
4 \({20 {~km} {~s}^{-1}}\)
Explanation:
Speed of galaxy can be calculated as \({\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}}\) \({\Rightarrow \dfrac{1}{600}=\dfrac{v}{3 \times 10^{8}}}\) \(v = \frac{{3 \times {{10}^8}}}{{600}} = 500\;\,km/{s^{ - 1}}\)
KCET - 2024
PHXII10:WAVE OPTICS
367735
Light emitted from a distant stationary star is observed at frequency 5000 \(MHz\). If the star starts approaching us with speed \(9 \times {10^5}m{\rm{/}}s\) then the observed frequency will be
1 \(5025MHz\)
2 \(5015MHz\)
3 \(5005MHz\)
4 \(5000MHz\)
Explanation:
Given that \(v = 5000MHz\,\,\& \,\,u = 9 \times {10^5}m/s\) The apparent frequency is \(v' = v\left( {1 + \frac{u}{c}} \right)\) The \(\left( + \right)\) ve shows that the relative separation between observer and source is increasing \(v' = 5000MHz\left( {1 + \frac{{9 \times {{10}^5}}}{{3 \times {{10}^8}}}} \right) = 5015MHz\)
PHXII10:WAVE OPTICS
367736
Assertion : In everyday life the Doppler's effect is observed readily for sound waves than light waves. Reason : Velocity of light is greater than that of sound.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
For both \(v=n \lambda\). For light, we denote \(v=c, \mathrm{c}\) is very high. The Doppler effect is more readily observed in sound waves due to their larger wavelengths, while the same is not the case with light waves because they have much shorter wavelengths in everyday life. So correct option is (2).
PHXII10:WAVE OPTICS
367737
It is believed that the universe is expanding and hence the distant stars are receding from us. Light from such a star will show
1 Shift in frequency towards shorter wavelengths
2 Shift in frequency towards longer wavelengths
3 A shift in frequency sometimes towards longer and sometimes towards shorter wavelengths
4 No shift in frequency but a decrease in intensity
Explanation:
Conceptual Question
PHXII10:WAVE OPTICS
367738
The \(6563\mathop A\limits^ \circ {H_ \propto }\) sign line emitted by hydrogen in a star is found to be red-shifted by \(15\mathop A\limits^ \circ \). Estimate the speed with the star receding from the earth.
1 \(8.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
2 \(7.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
3 \(6.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) approaching the earth
4 \(6.86 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
Explanation:
Given, wavelength of \({H_\alpha },\) \(\lambda = 6563\mathop A\limits^o = 6563 \times {10^{ - 10}}{\mkern 1mu} m\) Red-shift \(\Delta \lambda = 15\mathop A\limits^ \circ \) As the wavelength is found to be red-shifted, hence star is receding away from earth \(\Delta \lambda = v\frac{\lambda }{c} \Rightarrow v = \frac{{\Delta \lambda \cdot c}}{\lambda } = - \frac{{15 \times 3 \times {{10}^8}}}{{6563}}\) \(v = 6.86 \times {10^5}m/s\)
367734
A galaxy is moving away from the Earth so that a spectral line at 600 nm is observed at 601 nm . Then the speed of the galaxy with respect to the Earth is
1 \({500 {~km} {~s}^{-1}}\)
2 \({50 {~km} {~s}^{-1}}\)
3 \({200 {~km} {~s}^{-1}}\)
4 \({20 {~km} {~s}^{-1}}\)
Explanation:
Speed of galaxy can be calculated as \({\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}}\) \({\Rightarrow \dfrac{1}{600}=\dfrac{v}{3 \times 10^{8}}}\) \(v = \frac{{3 \times {{10}^8}}}{{600}} = 500\;\,km/{s^{ - 1}}\)
KCET - 2024
PHXII10:WAVE OPTICS
367735
Light emitted from a distant stationary star is observed at frequency 5000 \(MHz\). If the star starts approaching us with speed \(9 \times {10^5}m{\rm{/}}s\) then the observed frequency will be
1 \(5025MHz\)
2 \(5015MHz\)
3 \(5005MHz\)
4 \(5000MHz\)
Explanation:
Given that \(v = 5000MHz\,\,\& \,\,u = 9 \times {10^5}m/s\) The apparent frequency is \(v' = v\left( {1 + \frac{u}{c}} \right)\) The \(\left( + \right)\) ve shows that the relative separation between observer and source is increasing \(v' = 5000MHz\left( {1 + \frac{{9 \times {{10}^5}}}{{3 \times {{10}^8}}}} \right) = 5015MHz\)
PHXII10:WAVE OPTICS
367736
Assertion : In everyday life the Doppler's effect is observed readily for sound waves than light waves. Reason : Velocity of light is greater than that of sound.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
For both \(v=n \lambda\). For light, we denote \(v=c, \mathrm{c}\) is very high. The Doppler effect is more readily observed in sound waves due to their larger wavelengths, while the same is not the case with light waves because they have much shorter wavelengths in everyday life. So correct option is (2).
PHXII10:WAVE OPTICS
367737
It is believed that the universe is expanding and hence the distant stars are receding from us. Light from such a star will show
1 Shift in frequency towards shorter wavelengths
2 Shift in frequency towards longer wavelengths
3 A shift in frequency sometimes towards longer and sometimes towards shorter wavelengths
4 No shift in frequency but a decrease in intensity
Explanation:
Conceptual Question
PHXII10:WAVE OPTICS
367738
The \(6563\mathop A\limits^ \circ {H_ \propto }\) sign line emitted by hydrogen in a star is found to be red-shifted by \(15\mathop A\limits^ \circ \). Estimate the speed with the star receding from the earth.
1 \(8.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
2 \(7.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
3 \(6.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) approaching the earth
4 \(6.86 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
Explanation:
Given, wavelength of \({H_\alpha },\) \(\lambda = 6563\mathop A\limits^o = 6563 \times {10^{ - 10}}{\mkern 1mu} m\) Red-shift \(\Delta \lambda = 15\mathop A\limits^ \circ \) As the wavelength is found to be red-shifted, hence star is receding away from earth \(\Delta \lambda = v\frac{\lambda }{c} \Rightarrow v = \frac{{\Delta \lambda \cdot c}}{\lambda } = - \frac{{15 \times 3 \times {{10}^8}}}{{6563}}\) \(v = 6.86 \times {10^5}m/s\)
367734
A galaxy is moving away from the Earth so that a spectral line at 600 nm is observed at 601 nm . Then the speed of the galaxy with respect to the Earth is
1 \({500 {~km} {~s}^{-1}}\)
2 \({50 {~km} {~s}^{-1}}\)
3 \({200 {~km} {~s}^{-1}}\)
4 \({20 {~km} {~s}^{-1}}\)
Explanation:
Speed of galaxy can be calculated as \({\dfrac{\Delta \lambda}{\lambda}=\dfrac{v}{c}}\) \({\Rightarrow \dfrac{1}{600}=\dfrac{v}{3 \times 10^{8}}}\) \(v = \frac{{3 \times {{10}^8}}}{{600}} = 500\;\,km/{s^{ - 1}}\)
KCET - 2024
PHXII10:WAVE OPTICS
367735
Light emitted from a distant stationary star is observed at frequency 5000 \(MHz\). If the star starts approaching us with speed \(9 \times {10^5}m{\rm{/}}s\) then the observed frequency will be
1 \(5025MHz\)
2 \(5015MHz\)
3 \(5005MHz\)
4 \(5000MHz\)
Explanation:
Given that \(v = 5000MHz\,\,\& \,\,u = 9 \times {10^5}m/s\) The apparent frequency is \(v' = v\left( {1 + \frac{u}{c}} \right)\) The \(\left( + \right)\) ve shows that the relative separation between observer and source is increasing \(v' = 5000MHz\left( {1 + \frac{{9 \times {{10}^5}}}{{3 \times {{10}^8}}}} \right) = 5015MHz\)
PHXII10:WAVE OPTICS
367736
Assertion : In everyday life the Doppler's effect is observed readily for sound waves than light waves. Reason : Velocity of light is greater than that of sound.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
For both \(v=n \lambda\). For light, we denote \(v=c, \mathrm{c}\) is very high. The Doppler effect is more readily observed in sound waves due to their larger wavelengths, while the same is not the case with light waves because they have much shorter wavelengths in everyday life. So correct option is (2).
PHXII10:WAVE OPTICS
367737
It is believed that the universe is expanding and hence the distant stars are receding from us. Light from such a star will show
1 Shift in frequency towards shorter wavelengths
2 Shift in frequency towards longer wavelengths
3 A shift in frequency sometimes towards longer and sometimes towards shorter wavelengths
4 No shift in frequency but a decrease in intensity
Explanation:
Conceptual Question
PHXII10:WAVE OPTICS
367738
The \(6563\mathop A\limits^ \circ {H_ \propto }\) sign line emitted by hydrogen in a star is found to be red-shifted by \(15\mathop A\limits^ \circ \). Estimate the speed with the star receding from the earth.
1 \(8.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
2 \(7.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
3 \(6.9 \times {10^5}{\rm{ }}m{\rm{/}}s\) approaching the earth
4 \(6.86 \times {10^5}{\rm{ }}m{\rm{/}}s\) receding the earth
Explanation:
Given, wavelength of \({H_\alpha },\) \(\lambda = 6563\mathop A\limits^o = 6563 \times {10^{ - 10}}{\mkern 1mu} m\) Red-shift \(\Delta \lambda = 15\mathop A\limits^ \circ \) As the wavelength is found to be red-shifted, hence star is receding away from earth \(\Delta \lambda = v\frac{\lambda }{c} \Rightarrow v = \frac{{\Delta \lambda \cdot c}}{\lambda } = - \frac{{15 \times 3 \times {{10}^8}}}{{6563}}\) \(v = 6.86 \times {10^5}m/s\)
