Wave Optics - Result Question 14
14. Two slits in Young’s experiment have widths in the ratio $1: 25$. The ratio of intensity at the maxima and minima in the interference pattern, $\frac{I _{\max }}{I _{\min }}$ is:
(a) $\frac{121}{49}$
(b) $\frac{49}{121} \quad$ [2015 RS]
(c) $\frac{4}{9}$
(d) $\frac{9}{4}$
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Answer:
Correct Answer: 14. (d)
Solution:
- (d) The ratio of slits width $=\frac{1}{25}$ (given)
$\therefore \frac{I_1}{I_2}=\frac{25}{1}$
$I \propto A^{2} \Rightarrow \frac{I_1}{I_2}=\frac{A_1^{2}}{A_2^{2}}=\frac{25}{1}$ or $\frac{A_1}{A_2}=\frac{5}{1}$
As we know that,
$\frac{A _{\max }}{A _{\min }}=\frac{A_1+A_2}{A_1-A_2}=\frac{5+1}{5-1}=\frac{6}{4}=\frac{3}{2}$
$\therefore \frac{I _{\max }}{I _{\min }}=\frac{A _{\max }^{2}}{A _{\min }^{2}}=(\frac{3}{2})^{2}=\frac{9}{4}$
In YDSE, the radio of $\frac{I _{\max }}{I _{\min }}$ is maximum when both the sources have same intensity.
