Optics 6 Question 4
5. In an interference experiment, the ratio of amplitudes of coherent waves is $\frac{a _1}{a _2}=\frac{1}{3}$. The ratio of maximum and minimum intensities of fringes will be
(a) 2
(b) 18
(c) 4
(d) 9
(2019 Main, 8 April I)
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Solution:
- In Young’s double slit experiment, ratio of maxima and minima intensity is given by
As, intensity $(I) \propto[\operatorname{amplitude}(a)]^{2}$
$\therefore \quad \frac{I _1}{I _2}={\frac{a _1}{a _2}}^{2}=\frac{1}{3}^{2}=\frac{1}{9}$
So, $\quad \frac{I _{\max }}{I _{\min }}=\frac{\frac{1}{3}+1}{\frac{1}{3}-1}=4: 1$
