Question

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

• 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

Answer 1

Answer: (D)

the amplitude ratio is 2

Answer 2

$\frac{I_{max}}{I_{min}}=\frac{(\sqrt {I_1}+\sqrt {I_2})^2}{(\sqrt {I_1}-\sqrt {I_2})^2}=\bigg(\frac{\sqrt{I_1 / I_2}+1}{\sqrt{I_1 / I_2}-1}\bigg)^2=9$ (Given)
Solving this, we have $\frac{I_1}{I_2}=4 \, \, but \, \, \, I \propto A^2$
\therefore \hspace25mm \frac{A_1}{A_2}=2
$\therefore$ Correct options are (b) and (d).