Interference of Waves
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PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength \(\lambda \), the path difference should be

1 \(\left( {2n + 1} \right)\frac{\lambda }{2}\)
2 \(\left( {2n - 1} \right)\frac{\lambda }{2}\)
3 \(\left( {2n + 1} \right)\frac{\lambda }{4}\)
4 None of these
PHXII10:WAVE OPTICS

367781 The two waves represented by \({y_{1}=a \sin (\omega t)}\) and \({y_{2}=b \cos (\omega t)}\) have a phase difference of

1 0
2 \({\dfrac{\pi}{2}}\)
3 \({\pi}\)
4 \({\dfrac{\pi}{4}}\)
PHXII10:WAVE OPTICS

367783 Let \({S_1}\,{\rm{and}}\,{S_2}\) be two sources as shown in the figure.
Which of the following statement \((s)\) is/are correct?
I.
\({S_2}P - {S_1}P = 3\lambda \)
II.
Waves from \({S_1}\) arrives exactly three cycles earlier than waves from \({S_2}\)
III.
At \(P\) waves from \({S_1}{\rm{ and }}{S_2}\) are in phase.
supporting img

1 I, II and III
2 I and III
3 II and III
4 I and II
PHXII10:WAVE OPTICS

367784 The \(YDSE\) apparatus is as shown in the figure below. The condition for point \(P\) to be a dark fringe is \({\rm{(}}\lambda {\rm{ = }}\) wavelength of light waves)
supporting img

1 \(\left( {{l_1} + {l_2}} \right) - \left( {{l_2} + {l_3}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
2 \(\left( {{l_1} + {l_3}} \right) - \left( {{l_2} + {l_4}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
3 \(\left( {{l_1} - {l_3}} \right) - \left( {{l_2} - {l_4}} \right) = n\lambda \)
4 \(\left( {{l_1} - {l_2}} \right) - \left( {{l_3} - {l_4}} \right) = n\lambda \)
PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength \(\lambda\) in vacuum are in phase initially. Then the first ray travels a path \(l_{1}\) through a medium of refractive index \(n_{1}\) while the second ray travels a path of length \(l_{2}\) through a medium of refractive index \(n_{2}\). The two waves are then combined to observe interference. The phase difference between the two waves is

1 \(\dfrac{2 \pi}{\lambda}\left(l_{2}-l_{1}\right)\)
2 \(\dfrac{2 \pi}{\lambda}\left(n_{1} l_{1}-n_{2} l_{2}\right)\)
3 \(\dfrac{2 \pi}{\lambda}\left(n_{2} l_{2}-n_{1} l_{1}\right)\)
4 \(\dfrac{2 \pi}{\lambda}\left(\dfrac{l_{1}}{n_{1}}-\dfrac{l_{2}}{n_{2}}\right)\)
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PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength \(\lambda \), the path difference should be

1 \(\left( {2n + 1} \right)\frac{\lambda }{2}\)
2 \(\left( {2n - 1} \right)\frac{\lambda }{2}\)
3 \(\left( {2n + 1} \right)\frac{\lambda }{4}\)
4 None of these
PHXII10:WAVE OPTICS

367781 The two waves represented by \({y_{1}=a \sin (\omega t)}\) and \({y_{2}=b \cos (\omega t)}\) have a phase difference of

1 0
2 \({\dfrac{\pi}{2}}\)
3 \({\pi}\)
4 \({\dfrac{\pi}{4}}\)
PHXII10:WAVE OPTICS

367783 Let \({S_1}\,{\rm{and}}\,{S_2}\) be two sources as shown in the figure.
Which of the following statement \((s)\) is/are correct?
I.
\({S_2}P - {S_1}P = 3\lambda \)
II.
Waves from \({S_1}\) arrives exactly three cycles earlier than waves from \({S_2}\)
III.
At \(P\) waves from \({S_1}{\rm{ and }}{S_2}\) are in phase.
supporting img

1 I, II and III
2 I and III
3 II and III
4 I and II
PHXII10:WAVE OPTICS

367784 The \(YDSE\) apparatus is as shown in the figure below. The condition for point \(P\) to be a dark fringe is \({\rm{(}}\lambda {\rm{ = }}\) wavelength of light waves)
supporting img

1 \(\left( {{l_1} + {l_2}} \right) - \left( {{l_2} + {l_3}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
2 \(\left( {{l_1} + {l_3}} \right) - \left( {{l_2} + {l_4}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
3 \(\left( {{l_1} - {l_3}} \right) - \left( {{l_2} - {l_4}} \right) = n\lambda \)
4 \(\left( {{l_1} - {l_2}} \right) - \left( {{l_3} - {l_4}} \right) = n\lambda \)
PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength \(\lambda\) in vacuum are in phase initially. Then the first ray travels a path \(l_{1}\) through a medium of refractive index \(n_{1}\) while the second ray travels a path of length \(l_{2}\) through a medium of refractive index \(n_{2}\). The two waves are then combined to observe interference. The phase difference between the two waves is

1 \(\dfrac{2 \pi}{\lambda}\left(l_{2}-l_{1}\right)\)
2 \(\dfrac{2 \pi}{\lambda}\left(n_{1} l_{1}-n_{2} l_{2}\right)\)
3 \(\dfrac{2 \pi}{\lambda}\left(n_{2} l_{2}-n_{1} l_{1}\right)\)
4 \(\dfrac{2 \pi}{\lambda}\left(\dfrac{l_{1}}{n_{1}}-\dfrac{l_{2}}{n_{2}}\right)\)
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PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength \(\lambda \), the path difference should be

1 \(\left( {2n + 1} \right)\frac{\lambda }{2}\)
2 \(\left( {2n - 1} \right)\frac{\lambda }{2}\)
3 \(\left( {2n + 1} \right)\frac{\lambda }{4}\)
4 None of these
PHXII10:WAVE OPTICS

367781 The two waves represented by \({y_{1}=a \sin (\omega t)}\) and \({y_{2}=b \cos (\omega t)}\) have a phase difference of

1 0
2 \({\dfrac{\pi}{2}}\)
3 \({\pi}\)
4 \({\dfrac{\pi}{4}}\)
PHXII10:WAVE OPTICS

367783 Let \({S_1}\,{\rm{and}}\,{S_2}\) be two sources as shown in the figure.
Which of the following statement \((s)\) is/are correct?
I.
\({S_2}P - {S_1}P = 3\lambda \)
II.
Waves from \({S_1}\) arrives exactly three cycles earlier than waves from \({S_2}\)
III.
At \(P\) waves from \({S_1}{\rm{ and }}{S_2}\) are in phase.
supporting img

1 I, II and III
2 I and III
3 II and III
4 I and II
PHXII10:WAVE OPTICS

367784 The \(YDSE\) apparatus is as shown in the figure below. The condition for point \(P\) to be a dark fringe is \({\rm{(}}\lambda {\rm{ = }}\) wavelength of light waves)
supporting img

1 \(\left( {{l_1} + {l_2}} \right) - \left( {{l_2} + {l_3}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
2 \(\left( {{l_1} + {l_3}} \right) - \left( {{l_2} + {l_4}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
3 \(\left( {{l_1} - {l_3}} \right) - \left( {{l_2} - {l_4}} \right) = n\lambda \)
4 \(\left( {{l_1} - {l_2}} \right) - \left( {{l_3} - {l_4}} \right) = n\lambda \)
PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength \(\lambda\) in vacuum are in phase initially. Then the first ray travels a path \(l_{1}\) through a medium of refractive index \(n_{1}\) while the second ray travels a path of length \(l_{2}\) through a medium of refractive index \(n_{2}\). The two waves are then combined to observe interference. The phase difference between the two waves is

1 \(\dfrac{2 \pi}{\lambda}\left(l_{2}-l_{1}\right)\)
2 \(\dfrac{2 \pi}{\lambda}\left(n_{1} l_{1}-n_{2} l_{2}\right)\)
3 \(\dfrac{2 \pi}{\lambda}\left(n_{2} l_{2}-n_{1} l_{1}\right)\)
4 \(\dfrac{2 \pi}{\lambda}\left(\dfrac{l_{1}}{n_{1}}-\dfrac{l_{2}}{n_{2}}\right)\)
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PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength \(\lambda \), the path difference should be

1 \(\left( {2n + 1} \right)\frac{\lambda }{2}\)
2 \(\left( {2n - 1} \right)\frac{\lambda }{2}\)
3 \(\left( {2n + 1} \right)\frac{\lambda }{4}\)
4 None of these
PHXII10:WAVE OPTICS

367781 The two waves represented by \({y_{1}=a \sin (\omega t)}\) and \({y_{2}=b \cos (\omega t)}\) have a phase difference of

1 0
2 \({\dfrac{\pi}{2}}\)
3 \({\pi}\)
4 \({\dfrac{\pi}{4}}\)
PHXII10:WAVE OPTICS

367783 Let \({S_1}\,{\rm{and}}\,{S_2}\) be two sources as shown in the figure.
Which of the following statement \((s)\) is/are correct?
I.
\({S_2}P - {S_1}P = 3\lambda \)
II.
Waves from \({S_1}\) arrives exactly three cycles earlier than waves from \({S_2}\)
III.
At \(P\) waves from \({S_1}{\rm{ and }}{S_2}\) are in phase.
supporting img

1 I, II and III
2 I and III
3 II and III
4 I and II
PHXII10:WAVE OPTICS

367784 The \(YDSE\) apparatus is as shown in the figure below. The condition for point \(P\) to be a dark fringe is \({\rm{(}}\lambda {\rm{ = }}\) wavelength of light waves)
supporting img

1 \(\left( {{l_1} + {l_2}} \right) - \left( {{l_2} + {l_3}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
2 \(\left( {{l_1} + {l_3}} \right) - \left( {{l_2} + {l_4}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
3 \(\left( {{l_1} - {l_3}} \right) - \left( {{l_2} - {l_4}} \right) = n\lambda \)
4 \(\left( {{l_1} - {l_2}} \right) - \left( {{l_3} - {l_4}} \right) = n\lambda \)
PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength \(\lambda\) in vacuum are in phase initially. Then the first ray travels a path \(l_{1}\) through a medium of refractive index \(n_{1}\) while the second ray travels a path of length \(l_{2}\) through a medium of refractive index \(n_{2}\). The two waves are then combined to observe interference. The phase difference between the two waves is

1 \(\dfrac{2 \pi}{\lambda}\left(l_{2}-l_{1}\right)\)
2 \(\dfrac{2 \pi}{\lambda}\left(n_{1} l_{1}-n_{2} l_{2}\right)\)
3 \(\dfrac{2 \pi}{\lambda}\left(n_{2} l_{2}-n_{1} l_{1}\right)\)
4 \(\dfrac{2 \pi}{\lambda}\left(\dfrac{l_{1}}{n_{1}}-\dfrac{l_{2}}{n_{2}}\right)\)
NEET DIGITAL LIBRARY
PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength \(\lambda \), the path difference should be

1 \(\left( {2n + 1} \right)\frac{\lambda }{2}\)
2 \(\left( {2n - 1} \right)\frac{\lambda }{2}\)
3 \(\left( {2n + 1} \right)\frac{\lambda }{4}\)
4 None of these
PHXII10:WAVE OPTICS

367781 The two waves represented by \({y_{1}=a \sin (\omega t)}\) and \({y_{2}=b \cos (\omega t)}\) have a phase difference of

1 0
2 \({\dfrac{\pi}{2}}\)
3 \({\pi}\)
4 \({\dfrac{\pi}{4}}\)
PHXII10:WAVE OPTICS

367783 Let \({S_1}\,{\rm{and}}\,{S_2}\) be two sources as shown in the figure.
Which of the following statement \((s)\) is/are correct?
I.
\({S_2}P - {S_1}P = 3\lambda \)
II.
Waves from \({S_1}\) arrives exactly three cycles earlier than waves from \({S_2}\)
III.
At \(P\) waves from \({S_1}{\rm{ and }}{S_2}\) are in phase.
supporting img

1 I, II and III
2 I and III
3 II and III
4 I and II
PHXII10:WAVE OPTICS

367784 The \(YDSE\) apparatus is as shown in the figure below. The condition for point \(P\) to be a dark fringe is \({\rm{(}}\lambda {\rm{ = }}\) wavelength of light waves)
supporting img

1 \(\left( {{l_1} + {l_2}} \right) - \left( {{l_2} + {l_3}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
2 \(\left( {{l_1} + {l_3}} \right) - \left( {{l_2} + {l_4}} \right) = \frac{{\left( {2n - 1} \right)\lambda }}{2}\)
3 \(\left( {{l_1} - {l_3}} \right) - \left( {{l_2} - {l_4}} \right) = n\lambda \)
4 \(\left( {{l_1} - {l_2}} \right) - \left( {{l_3} - {l_4}} \right) = n\lambda \)
PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength \(\lambda\) in vacuum are in phase initially. Then the first ray travels a path \(l_{1}\) through a medium of refractive index \(n_{1}\) while the second ray travels a path of length \(l_{2}\) through a medium of refractive index \(n_{2}\). The two waves are then combined to observe interference. The phase difference between the two waves is

1 \(\dfrac{2 \pi}{\lambda}\left(l_{2}-l_{1}\right)\)
2 \(\dfrac{2 \pi}{\lambda}\left(n_{1} l_{1}-n_{2} l_{2}\right)\)
3 \(\dfrac{2 \pi}{\lambda}\left(n_{2} l_{2}-n_{1} l_{1}\right)\)
4 \(\dfrac{2 \pi}{\lambda}\left(\dfrac{l_{1}}{n_{1}}-\dfrac{l_{2}}{n_{2}}\right)\)