Question 286: Waves Optics

Question

Considering interference between two sources of intensities ' I ' and ' 4 I ', the intensity at a point where the phase difference is $\pi$ is $(\cos \pi=-1)$

Accepted Answer

The intensity at a point due to interference between two sources can be found using the formula for resultant intensity:
$$ I_{	ext{resultant}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi $$
where:
$ I_1 $ and $ I_2 $ are the intensities of the two sources.
$\phi$ is the phase difference between the waves.
In this case:
$ I_1 = I $
$ I_2 = 4I $
$\phi = \pi$ (thus, $\cos \pi = -1$)
Substituting these values into the formula:
$$ I_{	ext{resultant}} = I + 4I + 2\sqrt{I \cdot 4I} \cdot (-1) $$
Calculating each term:
$ I + 4I = 5I $
$ 2\sqrt{I \cdot 4I} = 2\sqrt{4I^2} = 2 	imes 2I = 4I $
Plug these results back into the equation:
$$ I_{	ext{resultant}} = 5I - 4I $$
Thus, the resultant intensity at the point where the phase difference is $\pi$ is:
$$ I_{	ext{resultant}} = I $$
Therefore, the correct answer is:
Option A: I

Answer

I

Answer

4I

Answer

5I

Answer

3I

Waves Optics

Considering interference between two sources of intensities ' I ' and ' 4 I ', the intensity at a point where the phase difference is \pi is (\cos \pi=-1)

Choose the correct answer: