| \[{{S}_{1}}\]and \[{{S}_{2}}\] are two coherent current sources of radiations separated by distance \[100.25\lambda \],where \[\lambda \] Is the wavelength of radiation. \[{{S}_{1}}\] Leads \[{{S}_{2}}\]in phase by\[\pi \]/2. A and B are two points on the line joining \[{{S}_{1}}\]and \[{{S}_{2}}\] as shown in figure. |
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| The ratio of amplitudes of source \[{{S}_{1}}\]and \[{{S}_{2}}\]are in the ratio \[~1:2\]then the ratio of intensity at A to that of \[\operatorname{B}\left( \frac{{{I}_{A}}}{{{I}_{B}}} \right)\]is |
A) \[9:1\]
B) \[3:1\]
C) \[1:3\]
D) \[1:9\]
Correct Answer: D
Solution :
| For interference at A: |
| \[{{S}_{2}}\]Is behind of \[{{S}_{1}}\] by a distance of \[100\lambda +\frac{\lambda }{4}\](equal to phase difference\[\pi /2\]). |
| Further \[{{S}_{2}}\]lags \[{{S}_{1}}\]by\[\pi /2\]. Hence the waves from \[{{S}_{1}}\]and \[{{S}_{2}}\]interference at B with a phase difference of |
| \[~200.5\pi +0.5\pi =201\pi =\pi .\] |
| Hence the net amplitude at A is \[2a-a=a\] |
| For interference at B: |
| \[{{S}_{2}}\]is ahead of \[{{S}_{1}}\]by a distance of \[100\lambda +\frac{\lambda }{4}\] |
| (Equal to phase difference\[\frac{\pi }{2}\]). |
| Further \[{{S}_{2}}\]lags \[{{S}_{1}}\]by \[\frac{\pi }{2}\]. Hence the waves from \[{{S}_{1}}\]and \[{{S}_{2}}\]interference at B with a phase difference of \[200.5\pi -0.5\pi =200\pi \]\[=0\pi \] |
| Hence, the net amplitude at A is \[2a+a=3a.\] |
| Hence \[\left( \frac{{{I}_{A}}}{{{I}_{B}}} \right)={{\left( \frac{a}{3a} \right)}^{2}}=\frac{1}{9}\] |
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