Question

If the ratio of maximum and minimum intensities in an interference pattern is $36:1$ then the ratio of amplitudes of two interfering waves will be

• A. $5:7$
• B. $7:4$
• C. $4:7$
• D. $7:5$

Answer 1

The correct option is D $7:5$

Let the amplitudes of the two interfering waves be $A_1,A_2$.

Thus $I_{max}=A_1^2+A_2^2+2A_1{A}_{2}cos{0}^{\circ }=(A_1+A_2{)}^2$

$I_{min}=A_1^2+A_2^2+2A_1{A}_{2}cos{180}^{\circ }=(A_1-A_2{)}^2$

Thus $\frac{I_{max}}{I_{min}}=\frac{(A_1+A_2{)}^2}{(A_1-A_2{)}^2}=36$

$⟹\frac{A_1}{A_2}=\frac{7}{5}$