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 π/2 and π/3 at points A and B respectively. The difference between the resultant intensities at the two points is xI. The value of x will be ________. Numerical answer type. Work it out, then open the answer.

Answer: 2. IR = I₁ + I₂ + 2√(I₁I₂)cos φ = 5I + 4Icos φ At A ( φ = π/2): IA = 5I At B ( φ = π/3): IB = 5I + 2I = 7I Difference = 2I, so x = 2.

Two beams of light having intensities I and 4I interfere to produce a… | JEE Main 2022 (27 Jul, Shift 1)

Q72JEE Main 2022 · 27 JUL · SHIFT 1Medium

Superposition, Coherence & Intensity

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 $xI$ . The value of $x$ will be ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

$I_R = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\phi = 5I + 4I\cos\phi$
At A ( $\phi = \pi/2$ ): $I_A = 5I$
At B ( $\phi = \pi/3$ ): $I_B = 5I + 2I = 7I$
Difference $= 2I$ , so $x = 2$ .

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