Optics 6 Question 39
41. In the Young’s double slit experiment, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 9 . This implies that
(1982, 2M)
(a) the intensities at the screen due to the two slits are 5 units and 4 units respectively
(b) the intensities at the screen due to the two slits are 4 units and 1 unit respectively
(c) the amplitude ratio is 3
(d) the amplitude ratio is 2
Fill in the Blanks
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Solution:
- $\frac{I _{\max }}{I _{\min }}=\frac{\left(\sqrt{I _1}+\sqrt{I _2}\right)^{2}}{\left(\sqrt{I _1}-\sqrt{I _2}\right)^{2}}={\frac{\sqrt{I _1 / I _2}+1}{\sqrt{I _1 / I _2}-1}}^{2}=9$ (Given)
Solving this, we have $\frac{I _1}{I _2}=4$ but $I \propto A^{2}$
$\therefore \quad \frac{A _1}{A _2}=2$
$\therefore$ Correct options are (b) and (d).
