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Interference fringes are produced on a screen by using two light sources of intensities ‘I’ and ‘9I’. The phase difference between the beams is $\dfrac{\pi }{2}$ at point P and $\pi $ at point Q on the screen. The difference between the resultant intensities at point P and Q is:
A. 2I
B. 4I
C. 6I
D. 8I

Last updated date: 20th Jun 2024
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Hint: We will use the formula for resultant intensities to find the intensities at points P and Q using the phase difference given in the question and then use that to find the difference between both of them as is asked in the question.

Formula used:
Resultant intensity of two waves
${{I}_{t}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \theta $, here θ is the phase difference between the two waves.

Complete step by step answer:
First, we will insert the values of $I_1$ and $I_2$ in the formula to get the formula that can be used directly in this question. Here $I_1$ = I and $I_2$ = 9I.
So, ${{I}_{t}}=I+9I+2\sqrt{9{{I}^{2}}}\cos \theta =10I+6I\cos \theta $
Now we will put in values of θ in the equation to get the intensity at both the points. First, we will find the intensities at point P. Here, there is a phase difference of $\dfrac{\pi }{2}$.
As $\cos \dfrac{\pi }{2}=0$
So, the resultant intensity at point P will be 10I.
Now at point Q, phase difference is $\pi $.
And now as $\cos \pi =-1$
So, the resultant intensity at point Q will be 4I. The difference between the intensities at point P and Q will be 6I.

Hence, the correct option is C, i.e. 6I.

The maximum intensity exists at a point where the phase difference is zero and minimum intensity exists at point where the phase difference is $\pi $, just like point Q in this case. Alternatively, we can take the sum of their amplitudes as intensities are the square of amplitudes. We would get the same equation as we got here but that will be a longer and more tedious approach. Therefore, it will be better to solve the question using this method only.