A) 1 : 4
B) 4 : 1
C) 3 : 2
D) 2 : 3
Correct Answer: B
Solution :
Given: \[{{I}_{1}}:{{I}_{2}}=1:9\] But, intensity\[I\propto {{a}^{2}}\] \[\therefore \] \[a_{1}^{2}.a_{2}^{2}=1:9\] \[{{a}_{1}}.{{a}_{2}}=1:3\] Let \[{{a}_{1}}=a,{{a}_{2}}=3a\] Therefore, \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\] \[={{\left( \frac{4a}{2a} \right)}^{2}}=\frac{4}{1}\]
You need to login to perform this action.
You will be redirected in
3 sec
