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The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point P and π at point Q. Then the difference between the resultant intensities at P and Q will be:
- a)2I
- b)6I
- c)5I
- d)7I
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The two light beams having intensities I and 9I interfere to produce a...
The resultant intensity is:
= 10I - 6 I = 4I
∴ IP - IQ = 10I - 4I = 6I
= 10I - 6 I = 4I
∴ IP - IQ = 10I - 4I = 6I
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Community Answer
The two light beams having intensities I and 9I interfere to produce a...
To determine the phase difference between the two light beams, we need to consider the conditions for constructive and destructive interference.
For constructive interference, the phase difference between the two beams should be a multiple of 2π. This means that the path difference between the two beams should be an integer number of wavelengths.
For destructive interference, the phase difference between the two beams should be an odd multiple of π. This means that the path difference between the two beams should be an odd number of half-wavelengths.
Given that the intensities of the two beams are I and 9I, we can assume that the first beam has an intensity of I and the second beam has an intensity of 9I.
The intensity of the interference pattern is given by the equation:
Intensity = 4I * cos^2(Δφ/2)
Where Δφ is the phase difference between the two beams.
Since the intensity of the interference pattern is proportional to the square of the cosine of half the phase difference, we can determine the phase difference by taking the inverse cosine of the square root of the intensity ratio:
Δφ = 2 * arccos(sqrt(I/9I))
Simplifying this equation, we get:
Δφ = 2 * arccos(sqrt(1/9))
Δφ = 2 * arccos(1/3)
Using a calculator, we find that:
Δφ ≈ 1.23 radians
Therefore, the phase difference between the two beams is approximately 1.23 radians.
For constructive interference, the phase difference between the two beams should be a multiple of 2π. This means that the path difference between the two beams should be an integer number of wavelengths.
For destructive interference, the phase difference between the two beams should be an odd multiple of π. This means that the path difference between the two beams should be an odd number of half-wavelengths.
Given that the intensities of the two beams are I and 9I, we can assume that the first beam has an intensity of I and the second beam has an intensity of 9I.
The intensity of the interference pattern is given by the equation:
Intensity = 4I * cos^2(Δφ/2)
Where Δφ is the phase difference between the two beams.
Since the intensity of the interference pattern is proportional to the square of the cosine of half the phase difference, we can determine the phase difference by taking the inverse cosine of the square root of the intensity ratio:
Δφ = 2 * arccos(sqrt(I/9I))
Simplifying this equation, we get:
Δφ = 2 * arccos(sqrt(1/9))
Δφ = 2 * arccos(1/3)
Using a calculator, we find that:
Δφ ≈ 1.23 radians
Therefore, the phase difference between the two beams is approximately 1.23 radians.
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The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2at point P and πat point Q. Then the difference between the resultant intensities at P and Q will be:a)2Ib)6Ic)5Id)7ICorrect answer is option 'B'. Can you explain this answer?
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