A) 4 : 9
B) 3 : 4
C) 1 : 25
D) 25 : 1
Correct Answer: D
Solution :
The intensity of interfering beam is given by \[I=k{{a}^{2}}\]where a is amplitude. \[\therefore \] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{a}_{1}}^{2}}{{{a}_{2}}^{2}}\] Given, \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{9}{4}\] \[\therefore \] \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{2}\] \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{({{a}_{1}}+{{a}_{2}})}^{2}}}{{{({{a}_{1}}-{{a}_{2}})}^{2}}}\] \[=\frac{{{\left( \frac{3}{2}{{a}_{2}}+{{a}_{2}} \right)}^{2}}}{{{\left( \frac{3}{2}{{a}_{2}}-{{a}_{2}} \right)}^{2}}}=\frac{25}{1}\]
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