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The ratio of maximum intensity to minimum intensity due to interference of two light waves of amplitudes `a_(1) and a_(2)` is
A. `(I_("max"))/(I_("min")) =((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))`
B. `(I_("max"))/(I_("min")) =(a_(1)+a_(2))/(a_(1)-a_(2))`
C. `(I_("max"))/(I_("min")) =(a_(1)^(2)+a_(2)^(2))/(a_(1)^(2)-a_(2)^(2))`
D. none of these

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1 Answers

Correct Answer - A
`I= a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2) cos (alpha_(1)-alpha_(2))`
Now, `(I_("max"))/(I_("min"))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))`

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