NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII10:WAVE OPTICS
367799
The equations of displacement of two waves are given as \(y_{1}=10 \sin (3 \pi t+\pi / 3)\), \(y_{2}=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\), then what is the ratio of their amplitude?
1 None of these
2 \(1: 1\)
3 \(2: 1\)
4 \(1: 2\)
Explanation:
The given waves are \({y_1} = 10\sin \left( {3\pi t + \frac{\pi }{3}} \right)\) \({y_2} = 10\left( {\frac{1}{2}\sin 3\pi t + \frac{{\sqrt 3 }}{2}\cos 3\pi t} \right)\) \( = 10\left( {\cos 60^\circ \sin 3\pi t + \sin 60^\circ \cos 3\pi t} \right)\) \( = 10\sin \left( {3\pi t + 60^\circ } \right)\) \( \Rightarrow \frac{a}{b} = \frac{{10}}{{10}} = 1:1\)
PHXII10:WAVE OPTICS
367800
Two monochromatic light waves of amplitudes \(3A\) and \(2A\) interfering at a point have a phase difference of \(60^\circ \). The intensity at that point will be proportional to
1 \(5{A^2}\)
2 \(13{A^2}\)
3 \(7{A^2}\)
4 \(19{A^2}\)
Explanation:
Here, \({A_1} = 3A,{A_2} = 2A,\phi = 60^\circ \) The resultant amplitude at a point is \(R = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } \) \( = \sqrt {{{\left( {3A} \right)}^2} + {{\left( {2A} \right)}^2} + 2 \times 3A \times 2A \times \cos 60^\circ } \) \( = \sqrt {9{A^2} + 4{A^2} + 6{A^2}} = A\sqrt {19} \) As intensity \( \propto {\left( {{\rm{Amplitude}}} \right)^2}\) Therefore, intensity at the same point is \(I \propto 19{A^2}\)
KCET - 2011
PHXII10:WAVE OPTICS
367801
In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern
1 The intensity of maxima decreases and the minima has zero intensity
2 The intensity of maxima increases and the minima has zero intensity
3 The intensities of both the maxima and minima increases
4 The intensity of maxima decreases and that of the minima increases
Explanation:
In interference between waves of equal amplitude \(a\), the minimum intensity is zero and the maximum intensity is proportional to \(4{a^2}\). For waves of unequal amplitudes \(a\) and \(2a\), the minimum intensity is non- zero and the maximum intensity is proportional to \({\left( {9a} \right)^2}\) which is greater than \(4{a^2}\).
PHXII10:WAVE OPTICS
367802
If the ratio of the intensity of two coherent sources is 4 then the visibility \({\left[\left(I_{\text {max }}-I_{\text {min }}\right) /\left(I_{\text {max }}+I_{\text {min }}\right)\right]}\) of the fringes is
367799
The equations of displacement of two waves are given as \(y_{1}=10 \sin (3 \pi t+\pi / 3)\), \(y_{2}=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\), then what is the ratio of their amplitude?
1 None of these
2 \(1: 1\)
3 \(2: 1\)
4 \(1: 2\)
Explanation:
The given waves are \({y_1} = 10\sin \left( {3\pi t + \frac{\pi }{3}} \right)\) \({y_2} = 10\left( {\frac{1}{2}\sin 3\pi t + \frac{{\sqrt 3 }}{2}\cos 3\pi t} \right)\) \( = 10\left( {\cos 60^\circ \sin 3\pi t + \sin 60^\circ \cos 3\pi t} \right)\) \( = 10\sin \left( {3\pi t + 60^\circ } \right)\) \( \Rightarrow \frac{a}{b} = \frac{{10}}{{10}} = 1:1\)
PHXII10:WAVE OPTICS
367800
Two monochromatic light waves of amplitudes \(3A\) and \(2A\) interfering at a point have a phase difference of \(60^\circ \). The intensity at that point will be proportional to
1 \(5{A^2}\)
2 \(13{A^2}\)
3 \(7{A^2}\)
4 \(19{A^2}\)
Explanation:
Here, \({A_1} = 3A,{A_2} = 2A,\phi = 60^\circ \) The resultant amplitude at a point is \(R = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } \) \( = \sqrt {{{\left( {3A} \right)}^2} + {{\left( {2A} \right)}^2} + 2 \times 3A \times 2A \times \cos 60^\circ } \) \( = \sqrt {9{A^2} + 4{A^2} + 6{A^2}} = A\sqrt {19} \) As intensity \( \propto {\left( {{\rm{Amplitude}}} \right)^2}\) Therefore, intensity at the same point is \(I \propto 19{A^2}\)
KCET - 2011
PHXII10:WAVE OPTICS
367801
In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern
1 The intensity of maxima decreases and the minima has zero intensity
2 The intensity of maxima increases and the minima has zero intensity
3 The intensities of both the maxima and minima increases
4 The intensity of maxima decreases and that of the minima increases
Explanation:
In interference between waves of equal amplitude \(a\), the minimum intensity is zero and the maximum intensity is proportional to \(4{a^2}\). For waves of unequal amplitudes \(a\) and \(2a\), the minimum intensity is non- zero and the maximum intensity is proportional to \({\left( {9a} \right)^2}\) which is greater than \(4{a^2}\).
PHXII10:WAVE OPTICS
367802
If the ratio of the intensity of two coherent sources is 4 then the visibility \({\left[\left(I_{\text {max }}-I_{\text {min }}\right) /\left(I_{\text {max }}+I_{\text {min }}\right)\right]}\) of the fringes is
367799
The equations of displacement of two waves are given as \(y_{1}=10 \sin (3 \pi t+\pi / 3)\), \(y_{2}=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\), then what is the ratio of their amplitude?
1 None of these
2 \(1: 1\)
3 \(2: 1\)
4 \(1: 2\)
Explanation:
The given waves are \({y_1} = 10\sin \left( {3\pi t + \frac{\pi }{3}} \right)\) \({y_2} = 10\left( {\frac{1}{2}\sin 3\pi t + \frac{{\sqrt 3 }}{2}\cos 3\pi t} \right)\) \( = 10\left( {\cos 60^\circ \sin 3\pi t + \sin 60^\circ \cos 3\pi t} \right)\) \( = 10\sin \left( {3\pi t + 60^\circ } \right)\) \( \Rightarrow \frac{a}{b} = \frac{{10}}{{10}} = 1:1\)
PHXII10:WAVE OPTICS
367800
Two monochromatic light waves of amplitudes \(3A\) and \(2A\) interfering at a point have a phase difference of \(60^\circ \). The intensity at that point will be proportional to
1 \(5{A^2}\)
2 \(13{A^2}\)
3 \(7{A^2}\)
4 \(19{A^2}\)
Explanation:
Here, \({A_1} = 3A,{A_2} = 2A,\phi = 60^\circ \) The resultant amplitude at a point is \(R = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } \) \( = \sqrt {{{\left( {3A} \right)}^2} + {{\left( {2A} \right)}^2} + 2 \times 3A \times 2A \times \cos 60^\circ } \) \( = \sqrt {9{A^2} + 4{A^2} + 6{A^2}} = A\sqrt {19} \) As intensity \( \propto {\left( {{\rm{Amplitude}}} \right)^2}\) Therefore, intensity at the same point is \(I \propto 19{A^2}\)
KCET - 2011
PHXII10:WAVE OPTICS
367801
In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern
1 The intensity of maxima decreases and the minima has zero intensity
2 The intensity of maxima increases and the minima has zero intensity
3 The intensities of both the maxima and minima increases
4 The intensity of maxima decreases and that of the minima increases
Explanation:
In interference between waves of equal amplitude \(a\), the minimum intensity is zero and the maximum intensity is proportional to \(4{a^2}\). For waves of unequal amplitudes \(a\) and \(2a\), the minimum intensity is non- zero and the maximum intensity is proportional to \({\left( {9a} \right)^2}\) which is greater than \(4{a^2}\).
PHXII10:WAVE OPTICS
367802
If the ratio of the intensity of two coherent sources is 4 then the visibility \({\left[\left(I_{\text {max }}-I_{\text {min }}\right) /\left(I_{\text {max }}+I_{\text {min }}\right)\right]}\) of the fringes is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII10:WAVE OPTICS
367799
The equations of displacement of two waves are given as \(y_{1}=10 \sin (3 \pi t+\pi / 3)\), \(y_{2}=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\), then what is the ratio of their amplitude?
1 None of these
2 \(1: 1\)
3 \(2: 1\)
4 \(1: 2\)
Explanation:
The given waves are \({y_1} = 10\sin \left( {3\pi t + \frac{\pi }{3}} \right)\) \({y_2} = 10\left( {\frac{1}{2}\sin 3\pi t + \frac{{\sqrt 3 }}{2}\cos 3\pi t} \right)\) \( = 10\left( {\cos 60^\circ \sin 3\pi t + \sin 60^\circ \cos 3\pi t} \right)\) \( = 10\sin \left( {3\pi t + 60^\circ } \right)\) \( \Rightarrow \frac{a}{b} = \frac{{10}}{{10}} = 1:1\)
PHXII10:WAVE OPTICS
367800
Two monochromatic light waves of amplitudes \(3A\) and \(2A\) interfering at a point have a phase difference of \(60^\circ \). The intensity at that point will be proportional to
1 \(5{A^2}\)
2 \(13{A^2}\)
3 \(7{A^2}\)
4 \(19{A^2}\)
Explanation:
Here, \({A_1} = 3A,{A_2} = 2A,\phi = 60^\circ \) The resultant amplitude at a point is \(R = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } \) \( = \sqrt {{{\left( {3A} \right)}^2} + {{\left( {2A} \right)}^2} + 2 \times 3A \times 2A \times \cos 60^\circ } \) \( = \sqrt {9{A^2} + 4{A^2} + 6{A^2}} = A\sqrt {19} \) As intensity \( \propto {\left( {{\rm{Amplitude}}} \right)^2}\) Therefore, intensity at the same point is \(I \propto 19{A^2}\)
KCET - 2011
PHXII10:WAVE OPTICS
367801
In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern
1 The intensity of maxima decreases and the minima has zero intensity
2 The intensity of maxima increases and the minima has zero intensity
3 The intensities of both the maxima and minima increases
4 The intensity of maxima decreases and that of the minima increases
Explanation:
In interference between waves of equal amplitude \(a\), the minimum intensity is zero and the maximum intensity is proportional to \(4{a^2}\). For waves of unequal amplitudes \(a\) and \(2a\), the minimum intensity is non- zero and the maximum intensity is proportional to \({\left( {9a} \right)^2}\) which is greater than \(4{a^2}\).
PHXII10:WAVE OPTICS
367802
If the ratio of the intensity of two coherent sources is 4 then the visibility \({\left[\left(I_{\text {max }}-I_{\text {min }}\right) /\left(I_{\text {max }}+I_{\text {min }}\right)\right]}\) of the fringes is