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ExamsJEE AdvancedPhysics

In an interference pattern, the intensity is given by I = I0 * sin²(theta). At theta = 30 deg, the measured intensity is I = 5 +/- 0.002 W/m², and I0 = 20 W/m². If the percentage error in angle theta is (n/pi)*sqrt(3)*10^(-2) percent, find the value of n.

  1. 1
  2. 2
  3. 3
  4. 4

Correct answer: 4

Solution

Differentiating I = I0*sin²(theta) gives dI = I0*sin(2*theta)*d(theta). With I0=20, theta=30 deg, dI=0.002, we get d(theta) = 0.002/(20*sin(60 deg)) = 0.002/(20*(sqrt(3)/2)) = 0.002/(10*sqrt(3)). Percentage error = (d(theta)/theta)*100 = (0.002/(10*sqrt(3))) / (pi/6) * 100 = (0.002*6)/(10*sqrt(3)*pi) * 100 = 0.012/(10*sqrt(3)*pi)*100 = 0.12/(sqrt(3)*pi) = (2/pi)*(sqrt(3)/20)... Let me redo carefully.

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