Optics Ques 156
- 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 $2$
Show Answer
Answer:
Correct Answer: 156.(b,d)
Solution:
Formula:
For Destructive interference :
- $\frac{I _{\max }}{I _{\min }}=\frac{\left(\sqrt{I _1}+\sqrt{I _2}\right)^{2}}{\left(\sqrt{I _1}-\sqrt{I _2}\right)^{2}}={\left(\frac{\sqrt{I _1 / I _2}+1}{\sqrt{I _1 / I _2}-1}\right)}^{2}=9$ (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).
BETA
