283060
The phase difference between the following two waves \(y_2\) and \(y_1\) is \(\mathrm{y}_1=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) ; \mathrm{y}_2=\mathrm{b} \cos \left(\omega \mathrm{t}-\mathrm{kx}+\frac{\pi}{3}\right)\)
283065
What is the wavelength of light of frequency \(5 \times 10^{14} \mathrm{~Hz}\) in glass of refractive index 1.5? velocity of light of air \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
1 \(3600 \AA\)
2 \(4500 \AA\)
3 \(4000 \AA\)
4 \(5000 \AA\)
Explanation:
: Given, \(\mathrm{f}=5 \times 10^{14} \mathrm{~Hz}\), Refractive index, \(\mu=1.5\) Velocity of light in air (c) \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) Since, frequency does not change with the change in medium, \(\therefore\) Frequency in vacuum \(=\) frequency in medium \(\frac{\lambda_{\text {vacuum }}}{\mu}=\lambda_{\text {medium }}\) \(\lambda_{\text {medium }}=\frac{\mathrm{c} / \mathrm{f}}{\mu}=\frac{\frac{3 \times 10^8}{5 \times 10^{14}}}{1.5}\) \(=0.4 \times 10^{-6} \mathrm{~m}\) \(=4000 \times 10^{-10} \mathrm{~m}=4000 \AA\)
MHT-CET 2020
WAVE OPTICS
283070
A light wave has a frequency of \(4 \times \mathbf{1 0}^{14} \mathrm{~Hz}\) and a wavelength of \(5 \times 10^{-7} \mathrm{~m}\) in a medium. The refractive index of the medium is
283079
For a colour of light the wavelength for air is \(6000 \AA\) and water is \(4500 \AA\). Then the speed of light in water will be
1 \(5 \times 10^{14} \mathrm{~m} / \mathrm{s}\)
2 \(2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
3 \(4.0 \times 10^8 \mathrm{~m} / \mathrm{s}\)
4 zero
Explanation:
: Given, \(\lambda_{\text {air }}=6000 \AA=6000 \times 10^{-10} \mathrm{~m}\) \(\lambda_{\text {water }}=4500 \AA=4500 \times 10^{-10} \mathrm{~m}\) Speed of light in air, \(\mathrm{c}=\mu \lambda_{\text {air }}\) Speed of light in water \(\mathrm{v}=\mu \lambda_{\text {water }}\) Dividing equation (ii) \& (i), we get - \(\frac{\mathrm{v}}{\mathrm{c}} =\frac{\mu \lambda_{\text {water }}}{\mu \lambda_{\text {air }}}=\frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(\mathrm{v} =\mathrm{c} \times \frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(=3 \times 10^8 \times \frac{4500 \times 10^{-10}}{6000 \times 10^{-10}}\) \(=2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Manipal UGET-2017
WAVE OPTICS
283081
A galaxy moves with respect to the earth so that sodium line of \(589.0 \mathrm{~nm}\) is observed at \(589.6 \mathrm{~nm}\). The speed of the galaxy is
1 \(300 \mathrm{~km} \mathrm{~s}^{-1}\)
2 \(306 \mathrm{~km} \mathrm{~s}^{-1}\)
3 \(400 \mathrm{~km} \mathrm{~s}^{-1}\)
4 \(406 \mathrm{~km} \mathrm{~s}^{-1}\)
Explanation:
: Wavelength of sodium line \(\left(\lambda_1\right)=589 \mathrm{~nm}\) Wavelength of sodium line observed \(\left(\lambda_2\right)=589.6 \mathrm{~nm}\) Change in wavelength \((\Delta \lambda)=\lambda_2-\lambda_1\) \(=589.6-589=0.6 \mathrm{~nm}\) Velocity of galaxy \((\mathrm{v})=\mathrm{c} \times \frac{\Delta \lambda}{\lambda_1}\) \(=\frac{3 \times 10^8 \times 0.6 \times 10^{-9}}{589 \times 10^{-9}}\) \(=0.00306 \times 10^8\) \(=306 \times 10^3 \mathrm{~m} / \mathrm{s}\) \(=306 \mathrm{~km} / \mathrm{s}\)
283060
The phase difference between the following two waves \(y_2\) and \(y_1\) is \(\mathrm{y}_1=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) ; \mathrm{y}_2=\mathrm{b} \cos \left(\omega \mathrm{t}-\mathrm{kx}+\frac{\pi}{3}\right)\)
283065
What is the wavelength of light of frequency \(5 \times 10^{14} \mathrm{~Hz}\) in glass of refractive index 1.5? velocity of light of air \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
1 \(3600 \AA\)
2 \(4500 \AA\)
3 \(4000 \AA\)
4 \(5000 \AA\)
Explanation:
: Given, \(\mathrm{f}=5 \times 10^{14} \mathrm{~Hz}\), Refractive index, \(\mu=1.5\) Velocity of light in air (c) \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) Since, frequency does not change with the change in medium, \(\therefore\) Frequency in vacuum \(=\) frequency in medium \(\frac{\lambda_{\text {vacuum }}}{\mu}=\lambda_{\text {medium }}\) \(\lambda_{\text {medium }}=\frac{\mathrm{c} / \mathrm{f}}{\mu}=\frac{\frac{3 \times 10^8}{5 \times 10^{14}}}{1.5}\) \(=0.4 \times 10^{-6} \mathrm{~m}\) \(=4000 \times 10^{-10} \mathrm{~m}=4000 \AA\)
MHT-CET 2020
WAVE OPTICS
283070
A light wave has a frequency of \(4 \times \mathbf{1 0}^{14} \mathrm{~Hz}\) and a wavelength of \(5 \times 10^{-7} \mathrm{~m}\) in a medium. The refractive index of the medium is
283079
For a colour of light the wavelength for air is \(6000 \AA\) and water is \(4500 \AA\). Then the speed of light in water will be
1 \(5 \times 10^{14} \mathrm{~m} / \mathrm{s}\)
2 \(2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
3 \(4.0 \times 10^8 \mathrm{~m} / \mathrm{s}\)
4 zero
Explanation:
: Given, \(\lambda_{\text {air }}=6000 \AA=6000 \times 10^{-10} \mathrm{~m}\) \(\lambda_{\text {water }}=4500 \AA=4500 \times 10^{-10} \mathrm{~m}\) Speed of light in air, \(\mathrm{c}=\mu \lambda_{\text {air }}\) Speed of light in water \(\mathrm{v}=\mu \lambda_{\text {water }}\) Dividing equation (ii) \& (i), we get - \(\frac{\mathrm{v}}{\mathrm{c}} =\frac{\mu \lambda_{\text {water }}}{\mu \lambda_{\text {air }}}=\frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(\mathrm{v} =\mathrm{c} \times \frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(=3 \times 10^8 \times \frac{4500 \times 10^{-10}}{6000 \times 10^{-10}}\) \(=2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Manipal UGET-2017
WAVE OPTICS
283081
A galaxy moves with respect to the earth so that sodium line of \(589.0 \mathrm{~nm}\) is observed at \(589.6 \mathrm{~nm}\). The speed of the galaxy is
1 \(300 \mathrm{~km} \mathrm{~s}^{-1}\)
2 \(306 \mathrm{~km} \mathrm{~s}^{-1}\)
3 \(400 \mathrm{~km} \mathrm{~s}^{-1}\)
4 \(406 \mathrm{~km} \mathrm{~s}^{-1}\)
Explanation:
: Wavelength of sodium line \(\left(\lambda_1\right)=589 \mathrm{~nm}\) Wavelength of sodium line observed \(\left(\lambda_2\right)=589.6 \mathrm{~nm}\) Change in wavelength \((\Delta \lambda)=\lambda_2-\lambda_1\) \(=589.6-589=0.6 \mathrm{~nm}\) Velocity of galaxy \((\mathrm{v})=\mathrm{c} \times \frac{\Delta \lambda}{\lambda_1}\) \(=\frac{3 \times 10^8 \times 0.6 \times 10^{-9}}{589 \times 10^{-9}}\) \(=0.00306 \times 10^8\) \(=306 \times 10^3 \mathrm{~m} / \mathrm{s}\) \(=306 \mathrm{~km} / \mathrm{s}\)
283060
The phase difference between the following two waves \(y_2\) and \(y_1\) is \(\mathrm{y}_1=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) ; \mathrm{y}_2=\mathrm{b} \cos \left(\omega \mathrm{t}-\mathrm{kx}+\frac{\pi}{3}\right)\)
283065
What is the wavelength of light of frequency \(5 \times 10^{14} \mathrm{~Hz}\) in glass of refractive index 1.5? velocity of light of air \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
1 \(3600 \AA\)
2 \(4500 \AA\)
3 \(4000 \AA\)
4 \(5000 \AA\)
Explanation:
: Given, \(\mathrm{f}=5 \times 10^{14} \mathrm{~Hz}\), Refractive index, \(\mu=1.5\) Velocity of light in air (c) \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) Since, frequency does not change with the change in medium, \(\therefore\) Frequency in vacuum \(=\) frequency in medium \(\frac{\lambda_{\text {vacuum }}}{\mu}=\lambda_{\text {medium }}\) \(\lambda_{\text {medium }}=\frac{\mathrm{c} / \mathrm{f}}{\mu}=\frac{\frac{3 \times 10^8}{5 \times 10^{14}}}{1.5}\) \(=0.4 \times 10^{-6} \mathrm{~m}\) \(=4000 \times 10^{-10} \mathrm{~m}=4000 \AA\)
MHT-CET 2020
WAVE OPTICS
283070
A light wave has a frequency of \(4 \times \mathbf{1 0}^{14} \mathrm{~Hz}\) and a wavelength of \(5 \times 10^{-7} \mathrm{~m}\) in a medium. The refractive index of the medium is
283079
For a colour of light the wavelength for air is \(6000 \AA\) and water is \(4500 \AA\). Then the speed of light in water will be
1 \(5 \times 10^{14} \mathrm{~m} / \mathrm{s}\)
2 \(2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
3 \(4.0 \times 10^8 \mathrm{~m} / \mathrm{s}\)
4 zero
Explanation:
: Given, \(\lambda_{\text {air }}=6000 \AA=6000 \times 10^{-10} \mathrm{~m}\) \(\lambda_{\text {water }}=4500 \AA=4500 \times 10^{-10} \mathrm{~m}\) Speed of light in air, \(\mathrm{c}=\mu \lambda_{\text {air }}\) Speed of light in water \(\mathrm{v}=\mu \lambda_{\text {water }}\) Dividing equation (ii) \& (i), we get - \(\frac{\mathrm{v}}{\mathrm{c}} =\frac{\mu \lambda_{\text {water }}}{\mu \lambda_{\text {air }}}=\frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(\mathrm{v} =\mathrm{c} \times \frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(=3 \times 10^8 \times \frac{4500 \times 10^{-10}}{6000 \times 10^{-10}}\) \(=2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Manipal UGET-2017
WAVE OPTICS
283081
A galaxy moves with respect to the earth so that sodium line of \(589.0 \mathrm{~nm}\) is observed at \(589.6 \mathrm{~nm}\). The speed of the galaxy is
1 \(300 \mathrm{~km} \mathrm{~s}^{-1}\)
2 \(306 \mathrm{~km} \mathrm{~s}^{-1}\)
3 \(400 \mathrm{~km} \mathrm{~s}^{-1}\)
4 \(406 \mathrm{~km} \mathrm{~s}^{-1}\)
Explanation:
: Wavelength of sodium line \(\left(\lambda_1\right)=589 \mathrm{~nm}\) Wavelength of sodium line observed \(\left(\lambda_2\right)=589.6 \mathrm{~nm}\) Change in wavelength \((\Delta \lambda)=\lambda_2-\lambda_1\) \(=589.6-589=0.6 \mathrm{~nm}\) Velocity of galaxy \((\mathrm{v})=\mathrm{c} \times \frac{\Delta \lambda}{\lambda_1}\) \(=\frac{3 \times 10^8 \times 0.6 \times 10^{-9}}{589 \times 10^{-9}}\) \(=0.00306 \times 10^8\) \(=306 \times 10^3 \mathrm{~m} / \mathrm{s}\) \(=306 \mathrm{~km} / \mathrm{s}\)
283060
The phase difference between the following two waves \(y_2\) and \(y_1\) is \(\mathrm{y}_1=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) ; \mathrm{y}_2=\mathrm{b} \cos \left(\omega \mathrm{t}-\mathrm{kx}+\frac{\pi}{3}\right)\)
283065
What is the wavelength of light of frequency \(5 \times 10^{14} \mathrm{~Hz}\) in glass of refractive index 1.5? velocity of light of air \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
1 \(3600 \AA\)
2 \(4500 \AA\)
3 \(4000 \AA\)
4 \(5000 \AA\)
Explanation:
: Given, \(\mathrm{f}=5 \times 10^{14} \mathrm{~Hz}\), Refractive index, \(\mu=1.5\) Velocity of light in air (c) \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) Since, frequency does not change with the change in medium, \(\therefore\) Frequency in vacuum \(=\) frequency in medium \(\frac{\lambda_{\text {vacuum }}}{\mu}=\lambda_{\text {medium }}\) \(\lambda_{\text {medium }}=\frac{\mathrm{c} / \mathrm{f}}{\mu}=\frac{\frac{3 \times 10^8}{5 \times 10^{14}}}{1.5}\) \(=0.4 \times 10^{-6} \mathrm{~m}\) \(=4000 \times 10^{-10} \mathrm{~m}=4000 \AA\)
MHT-CET 2020
WAVE OPTICS
283070
A light wave has a frequency of \(4 \times \mathbf{1 0}^{14} \mathrm{~Hz}\) and a wavelength of \(5 \times 10^{-7} \mathrm{~m}\) in a medium. The refractive index of the medium is
283079
For a colour of light the wavelength for air is \(6000 \AA\) and water is \(4500 \AA\). Then the speed of light in water will be
1 \(5 \times 10^{14} \mathrm{~m} / \mathrm{s}\)
2 \(2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
3 \(4.0 \times 10^8 \mathrm{~m} / \mathrm{s}\)
4 zero
Explanation:
: Given, \(\lambda_{\text {air }}=6000 \AA=6000 \times 10^{-10} \mathrm{~m}\) \(\lambda_{\text {water }}=4500 \AA=4500 \times 10^{-10} \mathrm{~m}\) Speed of light in air, \(\mathrm{c}=\mu \lambda_{\text {air }}\) Speed of light in water \(\mathrm{v}=\mu \lambda_{\text {water }}\) Dividing equation (ii) \& (i), we get - \(\frac{\mathrm{v}}{\mathrm{c}} =\frac{\mu \lambda_{\text {water }}}{\mu \lambda_{\text {air }}}=\frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(\mathrm{v} =\mathrm{c} \times \frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(=3 \times 10^8 \times \frac{4500 \times 10^{-10}}{6000 \times 10^{-10}}\) \(=2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Manipal UGET-2017
WAVE OPTICS
283081
A galaxy moves with respect to the earth so that sodium line of \(589.0 \mathrm{~nm}\) is observed at \(589.6 \mathrm{~nm}\). The speed of the galaxy is
1 \(300 \mathrm{~km} \mathrm{~s}^{-1}\)
2 \(306 \mathrm{~km} \mathrm{~s}^{-1}\)
3 \(400 \mathrm{~km} \mathrm{~s}^{-1}\)
4 \(406 \mathrm{~km} \mathrm{~s}^{-1}\)
Explanation:
: Wavelength of sodium line \(\left(\lambda_1\right)=589 \mathrm{~nm}\) Wavelength of sodium line observed \(\left(\lambda_2\right)=589.6 \mathrm{~nm}\) Change in wavelength \((\Delta \lambda)=\lambda_2-\lambda_1\) \(=589.6-589=0.6 \mathrm{~nm}\) Velocity of galaxy \((\mathrm{v})=\mathrm{c} \times \frac{\Delta \lambda}{\lambda_1}\) \(=\frac{3 \times 10^8 \times 0.6 \times 10^{-9}}{589 \times 10^{-9}}\) \(=0.00306 \times 10^8\) \(=306 \times 10^3 \mathrm{~m} / \mathrm{s}\) \(=306 \mathrm{~km} / \mathrm{s}\)
283060
The phase difference between the following two waves \(y_2\) and \(y_1\) is \(\mathrm{y}_1=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) ; \mathrm{y}_2=\mathrm{b} \cos \left(\omega \mathrm{t}-\mathrm{kx}+\frac{\pi}{3}\right)\)
283065
What is the wavelength of light of frequency \(5 \times 10^{14} \mathrm{~Hz}\) in glass of refractive index 1.5? velocity of light of air \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
1 \(3600 \AA\)
2 \(4500 \AA\)
3 \(4000 \AA\)
4 \(5000 \AA\)
Explanation:
: Given, \(\mathrm{f}=5 \times 10^{14} \mathrm{~Hz}\), Refractive index, \(\mu=1.5\) Velocity of light in air (c) \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) Since, frequency does not change with the change in medium, \(\therefore\) Frequency in vacuum \(=\) frequency in medium \(\frac{\lambda_{\text {vacuum }}}{\mu}=\lambda_{\text {medium }}\) \(\lambda_{\text {medium }}=\frac{\mathrm{c} / \mathrm{f}}{\mu}=\frac{\frac{3 \times 10^8}{5 \times 10^{14}}}{1.5}\) \(=0.4 \times 10^{-6} \mathrm{~m}\) \(=4000 \times 10^{-10} \mathrm{~m}=4000 \AA\)
MHT-CET 2020
WAVE OPTICS
283070
A light wave has a frequency of \(4 \times \mathbf{1 0}^{14} \mathrm{~Hz}\) and a wavelength of \(5 \times 10^{-7} \mathrm{~m}\) in a medium. The refractive index of the medium is
283079
For a colour of light the wavelength for air is \(6000 \AA\) and water is \(4500 \AA\). Then the speed of light in water will be
1 \(5 \times 10^{14} \mathrm{~m} / \mathrm{s}\)
2 \(2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
3 \(4.0 \times 10^8 \mathrm{~m} / \mathrm{s}\)
4 zero
Explanation:
: Given, \(\lambda_{\text {air }}=6000 \AA=6000 \times 10^{-10} \mathrm{~m}\) \(\lambda_{\text {water }}=4500 \AA=4500 \times 10^{-10} \mathrm{~m}\) Speed of light in air, \(\mathrm{c}=\mu \lambda_{\text {air }}\) Speed of light in water \(\mathrm{v}=\mu \lambda_{\text {water }}\) Dividing equation (ii) \& (i), we get - \(\frac{\mathrm{v}}{\mathrm{c}} =\frac{\mu \lambda_{\text {water }}}{\mu \lambda_{\text {air }}}=\frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(\mathrm{v} =\mathrm{c} \times \frac{\lambda_{\text {water }}}{\lambda_{\text {air }}}\) \(=3 \times 10^8 \times \frac{4500 \times 10^{-10}}{6000 \times 10^{-10}}\) \(=2.25 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Manipal UGET-2017
WAVE OPTICS
283081
A galaxy moves with respect to the earth so that sodium line of \(589.0 \mathrm{~nm}\) is observed at \(589.6 \mathrm{~nm}\). The speed of the galaxy is
1 \(300 \mathrm{~km} \mathrm{~s}^{-1}\)
2 \(306 \mathrm{~km} \mathrm{~s}^{-1}\)
3 \(400 \mathrm{~km} \mathrm{~s}^{-1}\)
4 \(406 \mathrm{~km} \mathrm{~s}^{-1}\)
Explanation:
: Wavelength of sodium line \(\left(\lambda_1\right)=589 \mathrm{~nm}\) Wavelength of sodium line observed \(\left(\lambda_2\right)=589.6 \mathrm{~nm}\) Change in wavelength \((\Delta \lambda)=\lambda_2-\lambda_1\) \(=589.6-589=0.6 \mathrm{~nm}\) Velocity of galaxy \((\mathrm{v})=\mathrm{c} \times \frac{\Delta \lambda}{\lambda_1}\) \(=\frac{3 \times 10^8 \times 0.6 \times 10^{-9}}{589 \times 10^{-9}}\) \(=0.00306 \times 10^8\) \(=306 \times 10^3 \mathrm{~m} / \mathrm{s}\) \(=306 \mathrm{~km} / \mathrm{s}\)
