Question 259: Waves Optics

Question

Interference fringes are produced on the screen by using two light sources of intensities I and 9I. The phase difference between the beams is $\pi / 2$ at point P and $\pi$ at point Q on the screen. The difference between the resultant intensities at points P and Q is $\left(\cos 90^{\circ}=0, \cos \pi=-1\right)$

Accepted Answer

Given:
The intensities of the two sources are $I_1 = I$ and $I_2 = 9I$.
Phase difference at point $ P $ is $\dfrac{\pi}{2}$.
Phase difference at point $ Q $ is $\pi$.
$\cos 90^\circ = 0, \cos \pi = -1$.
Step 1: Resultant Intensity Formula
For two sources with intensities $I_1$ and $I_2$, and phase difference $\phi$, the resultant intensity is:
$ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi $
Step 2: Intensity at P ($\phi_P = \frac{\pi}{2}$)
$ I_P = I + 9I + 2\sqrt{I \cdot 9I} \cos \left(\frac{\pi}{2}ight) $
$ I_P = 10I + 2 	imes 3I 	imes 0 $
$ I_P = 10I + 0 = 10I $
Step 3: Intensity at Q ($\phi_Q = \pi$)
$ I_Q = I + 9I + 2\sqrt{I \cdot 9I} \cos \pi $
$ I_Q = 10I + 2 	imes 3I 	imes (-1) $
$ I_Q = 10I - 6I = 4I $
Step 4: Difference in Intensities
$ I_P - I_Q = 10I - 4I = 6I $
Final Answer:
$ \boxed{6I} $
Correct Option:
Option C

Answer

2 I

Answer

4 I

Answer

6 I

Answer

8 I

Waves Optics

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