A) \[3:1\]
B) \[4:1\]
C) \[2:1\]
D) \[9:1\]
Correct Answer: B
Solution :
For bright fringe, the resultant amplitude is given by \[=3a+a=4a\] For dark fringe the resultant amplitude is given by \[=3a-a=2a\] \[\therefore \] \[Since,\text{ }intensity=K\text{ (}amplitude{{)}^{2}}\] \[\therefore \] \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(4a)}^{2}}}{{{(2a)}^{2}}}=\frac{4}{1}\]You need to login to perform this action.
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