Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
In Young's double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringe as 9. This implies (1) the intensities at the screen due to the two slits are 5 and 4 units (2) the intensities at the screen due to the two slits are 4 and 1 units (3) the amplitude ratio is 3 (4) the amplitude ratio is 2A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: C
Solution :
In case of interference \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi \] For\[I\]to be maximum \[\therefore \]\[{{I}_{\max }}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}\] and\[{{I}_{\min }}={{I}_{1}}+{{I}_{2}}-2\sqrt{{{I}_{1}}{{I}_{2}}}={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\] According to given problem \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}}{{{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}}=\frac{9}{1}\] Or \[\frac{\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}}}{\sqrt{{{I}_{2}}}-\sqrt{{{i}_{2}}}}=\frac{3}{1}\] By componendo and dividendo theorem \[\frac{\sqrt{{{I}_{1}}}}{\sqrt{{{I}_{2}}}}=\frac{3+1}{3-1}\]or\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{4}{1}=4\] For a wave\[I\propto {{A}^{2}}\] \[\frac{{{I}_{1}}}{{{I}_{2}}}={{\left[ \frac{{{A}_{1}}}{{{A}_{2}}} \right]}^{2}}\] \[{{\left[ \frac{{{A}_{1}}}{{{A}_{2}}} \right]}^{2}}=4\] \[\frac{{{A}_{1}}}{{{A}_{2}}}=2\]
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