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# Two coherent sources hence intensities in the ratio $81 : 4$. The ratio $I_{max} \;and\; I_{min}$ of the interfering light is

$(a)\;9:2 \\ (b)\;121:49 \\ (c)\;85:77 \\ (d)16:7$

Can you answer this question?

Given,
$\large\frac{I_1}{I_2}= \large\frac{{a_1}^2}{{a_2}^2}$
$\large\frac{I_1}{I_2}=\frac{81}{4}$
$\qquad= \large\frac{9^2}{2^2}$
$\therefore \large\frac{a_1}{a_2}=\frac{9}{2}$
$\large\frac{I_{max}}{I_{min}}=\frac{(a_1+a_2)^2}{(a_1-a_2)^2}$
$\qquad= \large\frac{(a+2)^2}{(a-2)}$
$\qquad= \large\frac{121}{49}$
Hence b is the correct answer.
answered Jan 21, 2014 by