Question

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 ___.

Answer 1

Correct Answer: 2

To solve for the difference in resultant intensities at points A and B, we begin by applying the formula for intensity in an interference pattern where two beams of light interfere: I_resultant = I₁ + I₂ + 2√(I₁I₂)cos(δ), where δ is the phase difference.

Given: 

• Intensities: I₁ = I, I₂ = 4I
• Phase difference at point A, δ_A = π/2
• Phase difference at point B, δ_B = π/3

Step 1: Calculate I_resultant at point A:

• Using δ_A = π/2, cos(π/2) = 0
• I_resultant,A = I + 4I + 2√(I*(4I))*0 = 5I

Step 2: Calculate I_resultant at point B:

• Using δ_B = π/3, cos(π/3) = 1/2
• I_resultant,B = I + 4I + 2√(I*(4I))*(1/2)
• Calculate: I_resultant,B = 5I + 2√(4I²)*(1/2) = 5I + 2(2I)*(1/2) = 5I + 2I = 7I

Step 3: Find the difference between the intensities:

• Difference: |I_resultant,B - I_resultant,A| = |7I - 5I| = 2I

The value of x is 2, confirming it falls within the given range (2, 2).

Answer 2

The correct answer is 2
\(I_{R_1}=I_1+I_2+2\sqrt{I_1I_2}\cos⁡\phi\)
\(I_A=I+4I+2\sqrt{I⋅4I}\cos⁡90^∘=5I\)
\(I_B=I+4I+2\sqrt{I⋅4I}\cos⁡60^∘=7I\)
\(I_B–I_A=2I\)
\(\therefore\) value of x is 2