283258
In the case of light waves from two coherent sources \(S_1\) and \(S_2\), there will be constructive interference at an arbitrary point \(P\), if the path difference \(S_1 P-S_2 P\) is
1 \(\left(n+\frac{1}{2}\right) \lambda\)
2 \(\mathrm{n} \lambda\)
3 \(\left(\mathrm{n}-\frac{1}{2}\right) \lambda\)
4 \(\frac{\lambda}{2}\)
Explanation:
: Constructive interference, Path difference \(=\) Integral multiple of \(\lambda\) \(\left(\mathrm{S}_1 \mathrm{P}-\mathrm{S}_2 \mathrm{P}\right)=\mathrm{n} \lambda\)
J and K CET- 2011
WAVE OPTICS
283259
Two coherent sources of intensity ratio \(1: 4\) produce an interference pattern. The fringe visibility will be
283262
Two coherent monochromatic lights make constructive interference when their phase difference is
1 \(\frac{3}{2} \pi\)
2 \(2 \pi\)
3 \(\pi\)
4 \(\frac{\pi}{2}\)
Explanation:
: In case of constructive interference, Phase difference \((\phi)=2 \mathrm{n} \pi\) \(\phi=2 \pi\) \((\because \mathrm{n}=1)\)
WB JEE-2007
WAVE OPTICS
283267
The maximum constructive interference of 2 waves cannot occur if the phase difference is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(5 \pi\)
4 0
Explanation:
: Intensity of two interfering waves on the screen is given by \(\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}\) Where, \(\phi=\) phase difference \(\therefore\) For constructive interference, \(\phi=0,2 \pi, 4 \pi \ldots \ldots . .\)We get only even multiples of \(\pi\), so maximum intensity corresponds to constructive interference, \(\mathrm{I}=4 \mathrm{I}_0\). So \(\phi\) can not be \(5 \pi\).
SRM JEE - 2016
WAVE OPTICS
283269
Interference was observed in an interference chamber when air was present. Now, the chamber is evacuated and if the same light is used, a careful observation will show
1 No interference
2 Interference with dark bands
3 Interference with dark bands
4 Interference in which breadth of the fringe will slightly increased
Explanation:
: A chamber filled with air, if this air is evacuated then, \(\because\) Fringe width- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)As \(\lambda\) increases then fringe width also increases.
283258
In the case of light waves from two coherent sources \(S_1\) and \(S_2\), there will be constructive interference at an arbitrary point \(P\), if the path difference \(S_1 P-S_2 P\) is
1 \(\left(n+\frac{1}{2}\right) \lambda\)
2 \(\mathrm{n} \lambda\)
3 \(\left(\mathrm{n}-\frac{1}{2}\right) \lambda\)
4 \(\frac{\lambda}{2}\)
Explanation:
: Constructive interference, Path difference \(=\) Integral multiple of \(\lambda\) \(\left(\mathrm{S}_1 \mathrm{P}-\mathrm{S}_2 \mathrm{P}\right)=\mathrm{n} \lambda\)
J and K CET- 2011
WAVE OPTICS
283259
Two coherent sources of intensity ratio \(1: 4\) produce an interference pattern. The fringe visibility will be
283262
Two coherent monochromatic lights make constructive interference when their phase difference is
1 \(\frac{3}{2} \pi\)
2 \(2 \pi\)
3 \(\pi\)
4 \(\frac{\pi}{2}\)
Explanation:
: In case of constructive interference, Phase difference \((\phi)=2 \mathrm{n} \pi\) \(\phi=2 \pi\) \((\because \mathrm{n}=1)\)
WB JEE-2007
WAVE OPTICS
283267
The maximum constructive interference of 2 waves cannot occur if the phase difference is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(5 \pi\)
4 0
Explanation:
: Intensity of two interfering waves on the screen is given by \(\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}\) Where, \(\phi=\) phase difference \(\therefore\) For constructive interference, \(\phi=0,2 \pi, 4 \pi \ldots \ldots . .\)We get only even multiples of \(\pi\), so maximum intensity corresponds to constructive interference, \(\mathrm{I}=4 \mathrm{I}_0\). So \(\phi\) can not be \(5 \pi\).
SRM JEE - 2016
WAVE OPTICS
283269
Interference was observed in an interference chamber when air was present. Now, the chamber is evacuated and if the same light is used, a careful observation will show
1 No interference
2 Interference with dark bands
3 Interference with dark bands
4 Interference in which breadth of the fringe will slightly increased
Explanation:
: A chamber filled with air, if this air is evacuated then, \(\because\) Fringe width- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)As \(\lambda\) increases then fringe width also increases.
283258
In the case of light waves from two coherent sources \(S_1\) and \(S_2\), there will be constructive interference at an arbitrary point \(P\), if the path difference \(S_1 P-S_2 P\) is
1 \(\left(n+\frac{1}{2}\right) \lambda\)
2 \(\mathrm{n} \lambda\)
3 \(\left(\mathrm{n}-\frac{1}{2}\right) \lambda\)
4 \(\frac{\lambda}{2}\)
Explanation:
: Constructive interference, Path difference \(=\) Integral multiple of \(\lambda\) \(\left(\mathrm{S}_1 \mathrm{P}-\mathrm{S}_2 \mathrm{P}\right)=\mathrm{n} \lambda\)
J and K CET- 2011
WAVE OPTICS
283259
Two coherent sources of intensity ratio \(1: 4\) produce an interference pattern. The fringe visibility will be
283262
Two coherent monochromatic lights make constructive interference when their phase difference is
1 \(\frac{3}{2} \pi\)
2 \(2 \pi\)
3 \(\pi\)
4 \(\frac{\pi}{2}\)
Explanation:
: In case of constructive interference, Phase difference \((\phi)=2 \mathrm{n} \pi\) \(\phi=2 \pi\) \((\because \mathrm{n}=1)\)
WB JEE-2007
WAVE OPTICS
283267
The maximum constructive interference of 2 waves cannot occur if the phase difference is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(5 \pi\)
4 0
Explanation:
: Intensity of two interfering waves on the screen is given by \(\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}\) Where, \(\phi=\) phase difference \(\therefore\) For constructive interference, \(\phi=0,2 \pi, 4 \pi \ldots \ldots . .\)We get only even multiples of \(\pi\), so maximum intensity corresponds to constructive interference, \(\mathrm{I}=4 \mathrm{I}_0\). So \(\phi\) can not be \(5 \pi\).
SRM JEE - 2016
WAVE OPTICS
283269
Interference was observed in an interference chamber when air was present. Now, the chamber is evacuated and if the same light is used, a careful observation will show
1 No interference
2 Interference with dark bands
3 Interference with dark bands
4 Interference in which breadth of the fringe will slightly increased
Explanation:
: A chamber filled with air, if this air is evacuated then, \(\because\) Fringe width- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)As \(\lambda\) increases then fringe width also increases.
283258
In the case of light waves from two coherent sources \(S_1\) and \(S_2\), there will be constructive interference at an arbitrary point \(P\), if the path difference \(S_1 P-S_2 P\) is
1 \(\left(n+\frac{1}{2}\right) \lambda\)
2 \(\mathrm{n} \lambda\)
3 \(\left(\mathrm{n}-\frac{1}{2}\right) \lambda\)
4 \(\frac{\lambda}{2}\)
Explanation:
: Constructive interference, Path difference \(=\) Integral multiple of \(\lambda\) \(\left(\mathrm{S}_1 \mathrm{P}-\mathrm{S}_2 \mathrm{P}\right)=\mathrm{n} \lambda\)
J and K CET- 2011
WAVE OPTICS
283259
Two coherent sources of intensity ratio \(1: 4\) produce an interference pattern. The fringe visibility will be
283262
Two coherent monochromatic lights make constructive interference when their phase difference is
1 \(\frac{3}{2} \pi\)
2 \(2 \pi\)
3 \(\pi\)
4 \(\frac{\pi}{2}\)
Explanation:
: In case of constructive interference, Phase difference \((\phi)=2 \mathrm{n} \pi\) \(\phi=2 \pi\) \((\because \mathrm{n}=1)\)
WB JEE-2007
WAVE OPTICS
283267
The maximum constructive interference of 2 waves cannot occur if the phase difference is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(5 \pi\)
4 0
Explanation:
: Intensity of two interfering waves on the screen is given by \(\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}\) Where, \(\phi=\) phase difference \(\therefore\) For constructive interference, \(\phi=0,2 \pi, 4 \pi \ldots \ldots . .\)We get only even multiples of \(\pi\), so maximum intensity corresponds to constructive interference, \(\mathrm{I}=4 \mathrm{I}_0\). So \(\phi\) can not be \(5 \pi\).
SRM JEE - 2016
WAVE OPTICS
283269
Interference was observed in an interference chamber when air was present. Now, the chamber is evacuated and if the same light is used, a careful observation will show
1 No interference
2 Interference with dark bands
3 Interference with dark bands
4 Interference in which breadth of the fringe will slightly increased
Explanation:
: A chamber filled with air, if this air is evacuated then, \(\because\) Fringe width- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)As \(\lambda\) increases then fringe width also increases.
283258
In the case of light waves from two coherent sources \(S_1\) and \(S_2\), there will be constructive interference at an arbitrary point \(P\), if the path difference \(S_1 P-S_2 P\) is
1 \(\left(n+\frac{1}{2}\right) \lambda\)
2 \(\mathrm{n} \lambda\)
3 \(\left(\mathrm{n}-\frac{1}{2}\right) \lambda\)
4 \(\frac{\lambda}{2}\)
Explanation:
: Constructive interference, Path difference \(=\) Integral multiple of \(\lambda\) \(\left(\mathrm{S}_1 \mathrm{P}-\mathrm{S}_2 \mathrm{P}\right)=\mathrm{n} \lambda\)
J and K CET- 2011
WAVE OPTICS
283259
Two coherent sources of intensity ratio \(1: 4\) produce an interference pattern. The fringe visibility will be
283262
Two coherent monochromatic lights make constructive interference when their phase difference is
1 \(\frac{3}{2} \pi\)
2 \(2 \pi\)
3 \(\pi\)
4 \(\frac{\pi}{2}\)
Explanation:
: In case of constructive interference, Phase difference \((\phi)=2 \mathrm{n} \pi\) \(\phi=2 \pi\) \((\because \mathrm{n}=1)\)
WB JEE-2007
WAVE OPTICS
283267
The maximum constructive interference of 2 waves cannot occur if the phase difference is
1 \(2 \pi\)
2 \(4 \pi\)
3 \(5 \pi\)
4 0
Explanation:
: Intensity of two interfering waves on the screen is given by \(\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}\) Where, \(\phi=\) phase difference \(\therefore\) For constructive interference, \(\phi=0,2 \pi, 4 \pi \ldots \ldots . .\)We get only even multiples of \(\pi\), so maximum intensity corresponds to constructive interference, \(\mathrm{I}=4 \mathrm{I}_0\). So \(\phi\) can not be \(5 \pi\).
SRM JEE - 2016
WAVE OPTICS
283269
Interference was observed in an interference chamber when air was present. Now, the chamber is evacuated and if the same light is used, a careful observation will show
1 No interference
2 Interference with dark bands
3 Interference with dark bands
4 Interference in which breadth of the fringe will slightly increased
Explanation:
: A chamber filled with air, if this air is evacuated then, \(\because\) Fringe width- \(\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)As \(\lambda\) increases then fringe width also increases.
