A) 49 : 1
B) 25 : 7
C) 10 : 9
D) 4 : 3
Correct Answer: A
Solution :
Key Idea: At points where the two interfering waves meet in the same phase the resultant intensity is maximum. The resulting intensity at any point depends upon the phase difference \[(o|)\]between the two waves at that point \[I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \text{o}\text{ }\!\!|\!\!\text{ }\] For maximum intensity or constructive interference \[{{I}_{\max }}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\] \[={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}=K{{({{a}_{1}}+{{a}_{2}})}^{2}}\] (Since, intensity oc amplitude square.) For destructive interference or minimum intensity, \[{{I}_{\min }}={{I}_{1}}+{{I}_{2}}-2\sqrt{{{I}_{1}}{{I}_{2}}}\] \[={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}\] \[K={{({{a}_{1}}-{{a}_{2}})}^{2}}\] Given, \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{9}{16}=\frac{a_{1}^{2}}{a_{2}^{2}}\]You need to login to perform this action.
You will be redirected in
3 sec