283096
A point source that emits waves uniformly in all directions, produces wavefronts that are
1 spherical
2 elliptical
3 cylindrical
4 planar
Explanation:
: A point source that emits waves uniformly in all directions, produces wavefronts that are spherical.
J and K CET- 2011
WAVE OPTICS
283097
If the wavelength of light is \(4000 \AA\) then the number of waves in \(1 \mathrm{~mm}\) length will be
1 25
2 0.25
3 \(0.25 \times 10^4\)
4 \(25 \times 10^4\)
Explanation:
: Given, \(\lambda=4000 \AA=4000 \times 10^{-10} \mathrm{~m}=4 \times 10^{-}\) \(\mathrm{m}\) And, \(\mathrm{n} \lambda=1 \mathrm{~mm} \Rightarrow \mathrm{n}=\frac{1 \mathrm{~mm}}{\lambda}\) So, number of wave \((\mathrm{n})=\frac{1 \mathrm{~mm}}{\lambda}=\frac{10^{-3}}{4 \times 10^{-7}}\) \(\mathrm{n}=\frac{10^4}{4}=2500\) \(\mathrm{n}=0.25 \times 10^4\)
J and K CET- 2004
WAVE OPTICS
283103
A galaxy is moving away from an observer on earth so that sodium light of wavelength \(5892 \AA\) is observed at \(5896 \AA\). The speed of galaxy is
1 \(306 \mathrm{~km} / \mathrm{s}\)
2 \(204 \mathrm{~km} / \mathrm{s}\)
3 \(185 \mathrm{~km} / \mathrm{s}\)
4 \(158 \mathrm{~km} / \mathrm{s}\)
Explanation:
: Given that, \(\lambda_1=5892 \AA, \lambda_2=5896 \AA\) We know that, speed of galaxy- \(\mathrm{v}=\frac{\mathrm{c} \Delta \lambda}{\lambda_1}\) \(\mathrm{~V}=\frac{3 \times 10^8 \times(5896-5892)}{5892}=\frac{3 \times 10^8 \times 4}{5892}\) \(\mathrm{v}=0.00204 \times 10^8=204 \mathrm{~km} / \mathrm{s}\)
J and K-CET-2014
WAVE OPTICS
283106
If the shift in a star light is towards red end, then
1 the star is approaching the earth
2 the star receding from the earth
3 the apparent frequency is lesser than actual
4 Both (b) and (c)
Explanation:
: Is the star of moving away from us, its waves are effectively stretched out when they reach earth, increasing their wavelength. This shifts the star's spectral lines toward the red end of the spectrum. The red and of the spectrum has a lower pitch or frequency. Hence, the star is receding the earth and the apparent frequency is lesser than actual. So, both option (b) and (c) are correct.
UP CPMT-2012
WAVE OPTICS
283107
Two stars situated at distances of 1 and 10 light years respectively from the earth appear to possess the same brightness. The ratio of their real brightness is
1 \(1: 10\)
2 \(10: 1\)
3 \(1: 100\)
4 \(100: 1\)
Explanation:
: We know that, \(\because \quad \mathrm{I}=\frac{\mathrm{L}}{\mathrm{r}^2}\) I is same, then \(\frac{L_1}{r_1^2}=\frac{L_2}{r_2^2}\) \(\frac{L_1}{L_2}=\frac{r_1^2}{r_2^2}=\left(\frac{1}{10}\right)^2\)So, \(\quad \mathrm{L}_1: \mathrm{L}_2=1: 100\)
283096
A point source that emits waves uniformly in all directions, produces wavefronts that are
1 spherical
2 elliptical
3 cylindrical
4 planar
Explanation:
: A point source that emits waves uniformly in all directions, produces wavefronts that are spherical.
J and K CET- 2011
WAVE OPTICS
283097
If the wavelength of light is \(4000 \AA\) then the number of waves in \(1 \mathrm{~mm}\) length will be
1 25
2 0.25
3 \(0.25 \times 10^4\)
4 \(25 \times 10^4\)
Explanation:
: Given, \(\lambda=4000 \AA=4000 \times 10^{-10} \mathrm{~m}=4 \times 10^{-}\) \(\mathrm{m}\) And, \(\mathrm{n} \lambda=1 \mathrm{~mm} \Rightarrow \mathrm{n}=\frac{1 \mathrm{~mm}}{\lambda}\) So, number of wave \((\mathrm{n})=\frac{1 \mathrm{~mm}}{\lambda}=\frac{10^{-3}}{4 \times 10^{-7}}\) \(\mathrm{n}=\frac{10^4}{4}=2500\) \(\mathrm{n}=0.25 \times 10^4\)
J and K CET- 2004
WAVE OPTICS
283103
A galaxy is moving away from an observer on earth so that sodium light of wavelength \(5892 \AA\) is observed at \(5896 \AA\). The speed of galaxy is
1 \(306 \mathrm{~km} / \mathrm{s}\)
2 \(204 \mathrm{~km} / \mathrm{s}\)
3 \(185 \mathrm{~km} / \mathrm{s}\)
4 \(158 \mathrm{~km} / \mathrm{s}\)
Explanation:
: Given that, \(\lambda_1=5892 \AA, \lambda_2=5896 \AA\) We know that, speed of galaxy- \(\mathrm{v}=\frac{\mathrm{c} \Delta \lambda}{\lambda_1}\) \(\mathrm{~V}=\frac{3 \times 10^8 \times(5896-5892)}{5892}=\frac{3 \times 10^8 \times 4}{5892}\) \(\mathrm{v}=0.00204 \times 10^8=204 \mathrm{~km} / \mathrm{s}\)
J and K-CET-2014
WAVE OPTICS
283106
If the shift in a star light is towards red end, then
1 the star is approaching the earth
2 the star receding from the earth
3 the apparent frequency is lesser than actual
4 Both (b) and (c)
Explanation:
: Is the star of moving away from us, its waves are effectively stretched out when they reach earth, increasing their wavelength. This shifts the star's spectral lines toward the red end of the spectrum. The red and of the spectrum has a lower pitch or frequency. Hence, the star is receding the earth and the apparent frequency is lesser than actual. So, both option (b) and (c) are correct.
UP CPMT-2012
WAVE OPTICS
283107
Two stars situated at distances of 1 and 10 light years respectively from the earth appear to possess the same brightness. The ratio of their real brightness is
1 \(1: 10\)
2 \(10: 1\)
3 \(1: 100\)
4 \(100: 1\)
Explanation:
: We know that, \(\because \quad \mathrm{I}=\frac{\mathrm{L}}{\mathrm{r}^2}\) I is same, then \(\frac{L_1}{r_1^2}=\frac{L_2}{r_2^2}\) \(\frac{L_1}{L_2}=\frac{r_1^2}{r_2^2}=\left(\frac{1}{10}\right)^2\)So, \(\quad \mathrm{L}_1: \mathrm{L}_2=1: 100\)
283096
A point source that emits waves uniformly in all directions, produces wavefronts that are
1 spherical
2 elliptical
3 cylindrical
4 planar
Explanation:
: A point source that emits waves uniformly in all directions, produces wavefronts that are spherical.
J and K CET- 2011
WAVE OPTICS
283097
If the wavelength of light is \(4000 \AA\) then the number of waves in \(1 \mathrm{~mm}\) length will be
1 25
2 0.25
3 \(0.25 \times 10^4\)
4 \(25 \times 10^4\)
Explanation:
: Given, \(\lambda=4000 \AA=4000 \times 10^{-10} \mathrm{~m}=4 \times 10^{-}\) \(\mathrm{m}\) And, \(\mathrm{n} \lambda=1 \mathrm{~mm} \Rightarrow \mathrm{n}=\frac{1 \mathrm{~mm}}{\lambda}\) So, number of wave \((\mathrm{n})=\frac{1 \mathrm{~mm}}{\lambda}=\frac{10^{-3}}{4 \times 10^{-7}}\) \(\mathrm{n}=\frac{10^4}{4}=2500\) \(\mathrm{n}=0.25 \times 10^4\)
J and K CET- 2004
WAVE OPTICS
283103
A galaxy is moving away from an observer on earth so that sodium light of wavelength \(5892 \AA\) is observed at \(5896 \AA\). The speed of galaxy is
1 \(306 \mathrm{~km} / \mathrm{s}\)
2 \(204 \mathrm{~km} / \mathrm{s}\)
3 \(185 \mathrm{~km} / \mathrm{s}\)
4 \(158 \mathrm{~km} / \mathrm{s}\)
Explanation:
: Given that, \(\lambda_1=5892 \AA, \lambda_2=5896 \AA\) We know that, speed of galaxy- \(\mathrm{v}=\frac{\mathrm{c} \Delta \lambda}{\lambda_1}\) \(\mathrm{~V}=\frac{3 \times 10^8 \times(5896-5892)}{5892}=\frac{3 \times 10^8 \times 4}{5892}\) \(\mathrm{v}=0.00204 \times 10^8=204 \mathrm{~km} / \mathrm{s}\)
J and K-CET-2014
WAVE OPTICS
283106
If the shift in a star light is towards red end, then
1 the star is approaching the earth
2 the star receding from the earth
3 the apparent frequency is lesser than actual
4 Both (b) and (c)
Explanation:
: Is the star of moving away from us, its waves are effectively stretched out when they reach earth, increasing their wavelength. This shifts the star's spectral lines toward the red end of the spectrum. The red and of the spectrum has a lower pitch or frequency. Hence, the star is receding the earth and the apparent frequency is lesser than actual. So, both option (b) and (c) are correct.
UP CPMT-2012
WAVE OPTICS
283107
Two stars situated at distances of 1 and 10 light years respectively from the earth appear to possess the same brightness. The ratio of their real brightness is
1 \(1: 10\)
2 \(10: 1\)
3 \(1: 100\)
4 \(100: 1\)
Explanation:
: We know that, \(\because \quad \mathrm{I}=\frac{\mathrm{L}}{\mathrm{r}^2}\) I is same, then \(\frac{L_1}{r_1^2}=\frac{L_2}{r_2^2}\) \(\frac{L_1}{L_2}=\frac{r_1^2}{r_2^2}=\left(\frac{1}{10}\right)^2\)So, \(\quad \mathrm{L}_1: \mathrm{L}_2=1: 100\)
283096
A point source that emits waves uniformly in all directions, produces wavefronts that are
1 spherical
2 elliptical
3 cylindrical
4 planar
Explanation:
: A point source that emits waves uniformly in all directions, produces wavefronts that are spherical.
J and K CET- 2011
WAVE OPTICS
283097
If the wavelength of light is \(4000 \AA\) then the number of waves in \(1 \mathrm{~mm}\) length will be
1 25
2 0.25
3 \(0.25 \times 10^4\)
4 \(25 \times 10^4\)
Explanation:
: Given, \(\lambda=4000 \AA=4000 \times 10^{-10} \mathrm{~m}=4 \times 10^{-}\) \(\mathrm{m}\) And, \(\mathrm{n} \lambda=1 \mathrm{~mm} \Rightarrow \mathrm{n}=\frac{1 \mathrm{~mm}}{\lambda}\) So, number of wave \((\mathrm{n})=\frac{1 \mathrm{~mm}}{\lambda}=\frac{10^{-3}}{4 \times 10^{-7}}\) \(\mathrm{n}=\frac{10^4}{4}=2500\) \(\mathrm{n}=0.25 \times 10^4\)
J and K CET- 2004
WAVE OPTICS
283103
A galaxy is moving away from an observer on earth so that sodium light of wavelength \(5892 \AA\) is observed at \(5896 \AA\). The speed of galaxy is
1 \(306 \mathrm{~km} / \mathrm{s}\)
2 \(204 \mathrm{~km} / \mathrm{s}\)
3 \(185 \mathrm{~km} / \mathrm{s}\)
4 \(158 \mathrm{~km} / \mathrm{s}\)
Explanation:
: Given that, \(\lambda_1=5892 \AA, \lambda_2=5896 \AA\) We know that, speed of galaxy- \(\mathrm{v}=\frac{\mathrm{c} \Delta \lambda}{\lambda_1}\) \(\mathrm{~V}=\frac{3 \times 10^8 \times(5896-5892)}{5892}=\frac{3 \times 10^8 \times 4}{5892}\) \(\mathrm{v}=0.00204 \times 10^8=204 \mathrm{~km} / \mathrm{s}\)
J and K-CET-2014
WAVE OPTICS
283106
If the shift in a star light is towards red end, then
1 the star is approaching the earth
2 the star receding from the earth
3 the apparent frequency is lesser than actual
4 Both (b) and (c)
Explanation:
: Is the star of moving away from us, its waves are effectively stretched out when they reach earth, increasing their wavelength. This shifts the star's spectral lines toward the red end of the spectrum. The red and of the spectrum has a lower pitch or frequency. Hence, the star is receding the earth and the apparent frequency is lesser than actual. So, both option (b) and (c) are correct.
UP CPMT-2012
WAVE OPTICS
283107
Two stars situated at distances of 1 and 10 light years respectively from the earth appear to possess the same brightness. The ratio of their real brightness is
1 \(1: 10\)
2 \(10: 1\)
3 \(1: 100\)
4 \(100: 1\)
Explanation:
: We know that, \(\because \quad \mathrm{I}=\frac{\mathrm{L}}{\mathrm{r}^2}\) I is same, then \(\frac{L_1}{r_1^2}=\frac{L_2}{r_2^2}\) \(\frac{L_1}{L_2}=\frac{r_1^2}{r_2^2}=\left(\frac{1}{10}\right)^2\)So, \(\quad \mathrm{L}_1: \mathrm{L}_2=1: 100\)
283096
A point source that emits waves uniformly in all directions, produces wavefronts that are
1 spherical
2 elliptical
3 cylindrical
4 planar
Explanation:
: A point source that emits waves uniformly in all directions, produces wavefronts that are spherical.
J and K CET- 2011
WAVE OPTICS
283097
If the wavelength of light is \(4000 \AA\) then the number of waves in \(1 \mathrm{~mm}\) length will be
1 25
2 0.25
3 \(0.25 \times 10^4\)
4 \(25 \times 10^4\)
Explanation:
: Given, \(\lambda=4000 \AA=4000 \times 10^{-10} \mathrm{~m}=4 \times 10^{-}\) \(\mathrm{m}\) And, \(\mathrm{n} \lambda=1 \mathrm{~mm} \Rightarrow \mathrm{n}=\frac{1 \mathrm{~mm}}{\lambda}\) So, number of wave \((\mathrm{n})=\frac{1 \mathrm{~mm}}{\lambda}=\frac{10^{-3}}{4 \times 10^{-7}}\) \(\mathrm{n}=\frac{10^4}{4}=2500\) \(\mathrm{n}=0.25 \times 10^4\)
J and K CET- 2004
WAVE OPTICS
283103
A galaxy is moving away from an observer on earth so that sodium light of wavelength \(5892 \AA\) is observed at \(5896 \AA\). The speed of galaxy is
1 \(306 \mathrm{~km} / \mathrm{s}\)
2 \(204 \mathrm{~km} / \mathrm{s}\)
3 \(185 \mathrm{~km} / \mathrm{s}\)
4 \(158 \mathrm{~km} / \mathrm{s}\)
Explanation:
: Given that, \(\lambda_1=5892 \AA, \lambda_2=5896 \AA\) We know that, speed of galaxy- \(\mathrm{v}=\frac{\mathrm{c} \Delta \lambda}{\lambda_1}\) \(\mathrm{~V}=\frac{3 \times 10^8 \times(5896-5892)}{5892}=\frac{3 \times 10^8 \times 4}{5892}\) \(\mathrm{v}=0.00204 \times 10^8=204 \mathrm{~km} / \mathrm{s}\)
J and K-CET-2014
WAVE OPTICS
283106
If the shift in a star light is towards red end, then
1 the star is approaching the earth
2 the star receding from the earth
3 the apparent frequency is lesser than actual
4 Both (b) and (c)
Explanation:
: Is the star of moving away from us, its waves are effectively stretched out when they reach earth, increasing their wavelength. This shifts the star's spectral lines toward the red end of the spectrum. The red and of the spectrum has a lower pitch or frequency. Hence, the star is receding the earth and the apparent frequency is lesser than actual. So, both option (b) and (c) are correct.
UP CPMT-2012
WAVE OPTICS
283107
Two stars situated at distances of 1 and 10 light years respectively from the earth appear to possess the same brightness. The ratio of their real brightness is
1 \(1: 10\)
2 \(10: 1\)
3 \(1: 100\)
4 \(100: 1\)
Explanation:
: We know that, \(\because \quad \mathrm{I}=\frac{\mathrm{L}}{\mathrm{r}^2}\) I is same, then \(\frac{L_1}{r_1^2}=\frac{L_2}{r_2^2}\) \(\frac{L_1}{L_2}=\frac{r_1^2}{r_2^2}=\left(\frac{1}{10}\right)^2\)So, \(\quad \mathrm{L}_1: \mathrm{L}_2=1: 100\)
