Wave Optics - Result Question 14

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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:

======= ####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:

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/wave-optics/wave-optics—result-question-14.md (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:

  1. (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.