In Young's double-slit experiment, the intensities of the two coherent sources are in the ratio $4:1$. The ratio of maximum to minimum intensity in the resulting interference pattern is:
A$5 : 3$
B$9 : 1$
C$25 : 1$
D$3 : 1$
Answer & Solution
Correct answer: B. $9 : 1$
Amplitudes go as $\sqrt{I}$, so $a_1 : a_2 = 2 : 1$. $I_{max} \propto (a_1 + a_2)^2 = 9$; $I_{min} \propto (a_1 - a_2)^2 = 1$. Ratio = 9 : 1. (Used Young-experiment Q in the textbook: if $I_{max}/I_{min} = 25/1$, then $(a_1+a_2)/(a_1-a_2) = 5$ ⇒ $a_1 : a_2 = 3 : 2$, $I_1 : I_2 = 9 : 4$.)
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