Optics Ques 8
- In a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is 1 $\frac{1}{8}$ th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to
(2019 Main, 11 Jan I)
(a) 0.80
(b) 0.74
(c) 0.94
(d) 0.85
Show Answer
Answer:
Correct Answer: 8.( d )
Solution:
Formula:
- Let intensity of each wave is $I_0$.
Then, intensity at the centre of bright fringe will be $4 I_0$.
Given, path difference, $\Delta x=\lambda / 8$
$\therefore$ Phase difference, $\phi=\Delta x \times \frac{2 \pi}{\lambda}$
$\Rightarrow \quad \phi=\frac{\lambda}{8} \times \frac{2 \pi}{\lambda}$ or $\phi=\pi / 4$
Intensity of light at this point,
$ I^{\prime}=I_0+I_0+2 I_0 \cos (\pi / 4)=2 I_0+\sqrt{2} I_0=3.41 I_0 $
Now,
$ \frac{I^{\prime}}{4 I_0}=\frac{3.41}{4}=0.85 $
BETA
