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

Show Answer

Solution:

  1. $\frac{I _{\max }}{I _{\min }}=\frac{\left(\sqrt{I _1}+\sqrt{I _2}\right)^{2}}{\left(\sqrt{I _1}-\sqrt{I _2}\right)^{2}}={\frac{\sqrt{I _1 / I _2}+1}{\sqrt{I _1 / I _2}-1}}^{2}=9$ (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).



NCERT Chapter Video Solution

Dual Pane