Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are \pi / 2 and \pi / 3 at points A and B respectively. The difference between the resultant intensities at the two points is x I . The value of x will be \underline\hspace2cm .

Wave Optics PYQ — JEE Main 2022 · 27 Jul · Shift 1 · Q65 with Solution

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Correct answer: 2

For two coherent beams of intensities $I_1$ and $I_2$ , the resultant intensity is

I_r = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi

where $\phi$ is the phase difference.

Here,

I_1 = I, \qquad I_2 = 4I

So,

\sqrt{I_1 I_2} = \sqrt{I\cdot 4I} = 2I

Hence,

I_r = I + 4I + 2(2I)\cos\phi = 5I + 4I\cos\phi

At point $A$ , phase difference is

\phi_A = \frac{\pi}{2}

Therefore,

I_A = 5I + 4I\cos\left(\frac{\pi}{2}\right)

Since $\cos(\pi/2)=0$ ,

I_A = 5I

At point $B$ , phase difference is

\phi_B = \frac{\pi}{3}

Therefore,

I_B = 5I + 4I\cos\left(\frac{\pi}{3}\right)

Since $\cos(\pi/3)=\frac{1}{2}$ ,

I_B = 5I + 4I\left(\frac{1}{2}\right) = 5I + 2I = 7I

I_B - I_A = 7I - 5I = 2I

Given that the difference is $xI$ ,

xI = 2I

So,

x = 2

Stored correct answer = $2$

Our derived answer is also $2$ .