367807
Two waves of intensity \(I\) undergo Interference. The maximum inensity obtained is
1 \(4I\)
2 \(I/2\)
3 \(I\)
4 \(2I\)
Explanation:
For maximum intensity \(\phi = 0^\circ \) \(\therefore I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \) \({I_{\max }} = I + I + 2\sqrt {I\,\,I} \cos 0^\circ = 4I\)
PHXII10:WAVE OPTICS
367808
When two coherent monochromatic light beams of intensites \(I\) and \(9I\) are superimposed, what are the maximum and minimum possible intensities in the resulting beams?
367809
Two coherent waves are represented by \({y_1} = {a_1}\cos \omega t\) and \({y_2} = {a_2}\,\,\sin \omega t\), superimposed on each other. The resultant intensity is propostional to
367810
If the ratio of the intensities of two waves producing interference is \(49: 16\), then the ratio of the resultant maximum intensity to minimum intensity will be
1 \(11: 3\)
2 \(49: 16\)
3 \(121: 9\)
4 \(7: 4\)
Explanation:
Let the intesities of two waves be \(I_{1}\) and \(I_{2}\). \(I_{1}: I_{2}=49: 16\) (Given) \(\Rightarrow \dfrac{I_{\max }}{I_{\text {min }}}=\left(\dfrac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}+\sqrt{I_{2}}}\right)^{2}\) \(\Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\left(\dfrac{\sqrt{49}+\sqrt{16}}{\sqrt{49}-\sqrt{16}}\right)^{2}\) \(\begin{aligned}& \Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\left(\dfrac{7+4}{7-4}\right)^{2} \\& \Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\dfrac{121}{9} \\& \Rightarrow I_{\text {max }}: I_{\text {min }}=121: 9\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII10:WAVE OPTICS
367807
Two waves of intensity \(I\) undergo Interference. The maximum inensity obtained is
1 \(4I\)
2 \(I/2\)
3 \(I\)
4 \(2I\)
Explanation:
For maximum intensity \(\phi = 0^\circ \) \(\therefore I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \) \({I_{\max }} = I + I + 2\sqrt {I\,\,I} \cos 0^\circ = 4I\)
PHXII10:WAVE OPTICS
367808
When two coherent monochromatic light beams of intensites \(I\) and \(9I\) are superimposed, what are the maximum and minimum possible intensities in the resulting beams?
367809
Two coherent waves are represented by \({y_1} = {a_1}\cos \omega t\) and \({y_2} = {a_2}\,\,\sin \omega t\), superimposed on each other. The resultant intensity is propostional to
367810
If the ratio of the intensities of two waves producing interference is \(49: 16\), then the ratio of the resultant maximum intensity to minimum intensity will be
1 \(11: 3\)
2 \(49: 16\)
3 \(121: 9\)
4 \(7: 4\)
Explanation:
Let the intesities of two waves be \(I_{1}\) and \(I_{2}\). \(I_{1}: I_{2}=49: 16\) (Given) \(\Rightarrow \dfrac{I_{\max }}{I_{\text {min }}}=\left(\dfrac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}+\sqrt{I_{2}}}\right)^{2}\) \(\Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\left(\dfrac{\sqrt{49}+\sqrt{16}}{\sqrt{49}-\sqrt{16}}\right)^{2}\) \(\begin{aligned}& \Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\left(\dfrac{7+4}{7-4}\right)^{2} \\& \Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\dfrac{121}{9} \\& \Rightarrow I_{\text {max }}: I_{\text {min }}=121: 9\end{aligned}\)
367807
Two waves of intensity \(I\) undergo Interference. The maximum inensity obtained is
1 \(4I\)
2 \(I/2\)
3 \(I\)
4 \(2I\)
Explanation:
For maximum intensity \(\phi = 0^\circ \) \(\therefore I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \) \({I_{\max }} = I + I + 2\sqrt {I\,\,I} \cos 0^\circ = 4I\)
PHXII10:WAVE OPTICS
367808
When two coherent monochromatic light beams of intensites \(I\) and \(9I\) are superimposed, what are the maximum and minimum possible intensities in the resulting beams?
367809
Two coherent waves are represented by \({y_1} = {a_1}\cos \omega t\) and \({y_2} = {a_2}\,\,\sin \omega t\), superimposed on each other. The resultant intensity is propostional to
367810
If the ratio of the intensities of two waves producing interference is \(49: 16\), then the ratio of the resultant maximum intensity to minimum intensity will be
1 \(11: 3\)
2 \(49: 16\)
3 \(121: 9\)
4 \(7: 4\)
Explanation:
Let the intesities of two waves be \(I_{1}\) and \(I_{2}\). \(I_{1}: I_{2}=49: 16\) (Given) \(\Rightarrow \dfrac{I_{\max }}{I_{\text {min }}}=\left(\dfrac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}+\sqrt{I_{2}}}\right)^{2}\) \(\Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\left(\dfrac{\sqrt{49}+\sqrt{16}}{\sqrt{49}-\sqrt{16}}\right)^{2}\) \(\begin{aligned}& \Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\left(\dfrac{7+4}{7-4}\right)^{2} \\& \Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\dfrac{121}{9} \\& \Rightarrow I_{\text {max }}: I_{\text {min }}=121: 9\end{aligned}\)
367807
Two waves of intensity \(I\) undergo Interference. The maximum inensity obtained is
1 \(4I\)
2 \(I/2\)
3 \(I\)
4 \(2I\)
Explanation:
For maximum intensity \(\phi = 0^\circ \) \(\therefore I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \) \({I_{\max }} = I + I + 2\sqrt {I\,\,I} \cos 0^\circ = 4I\)
PHXII10:WAVE OPTICS
367808
When two coherent monochromatic light beams of intensites \(I\) and \(9I\) are superimposed, what are the maximum and minimum possible intensities in the resulting beams?
367809
Two coherent waves are represented by \({y_1} = {a_1}\cos \omega t\) and \({y_2} = {a_2}\,\,\sin \omega t\), superimposed on each other. The resultant intensity is propostional to
367810
If the ratio of the intensities of two waves producing interference is \(49: 16\), then the ratio of the resultant maximum intensity to minimum intensity will be
1 \(11: 3\)
2 \(49: 16\)
3 \(121: 9\)
4 \(7: 4\)
Explanation:
Let the intesities of two waves be \(I_{1}\) and \(I_{2}\). \(I_{1}: I_{2}=49: 16\) (Given) \(\Rightarrow \dfrac{I_{\max }}{I_{\text {min }}}=\left(\dfrac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}+\sqrt{I_{2}}}\right)^{2}\) \(\Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\left(\dfrac{\sqrt{49}+\sqrt{16}}{\sqrt{49}-\sqrt{16}}\right)^{2}\) \(\begin{aligned}& \Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\left(\dfrac{7+4}{7-4}\right)^{2} \\& \Rightarrow \dfrac{I_{\text {max }}}{I_{\text {min }}}=\dfrac{121}{9} \\& \Rightarrow I_{\text {max }}: I_{\text {min }}=121: 9\end{aligned}\)
