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Light was producing interference have their amplitudes in the ratio 3:2. The intensity ratio of maximum and minimum of interference fringes is
- a)36:1
- b)9:4
- c)25:1
- d)6:4
Correct answer is option 'C'. Can you explain this answer?
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Light was producing interference have their amplitudes in the ratio 3:...
Explanation:
When light waves produce interference, the resulting interference pattern consists of a series of bright and dark fringes. The intensity of these fringes depends on the amplitude of the interfering waves.
Given that the amplitudes of the interfering waves are in the ratio 3:2, let's assume the amplitudes to be 3A and 2A, respectively.
Intensity and Amplitude:
The intensity of a wave is directly proportional to the square of its amplitude. Therefore, the intensity of the first wave is (3A)^2 = 9A^2, and the intensity of the second wave is (2A)^2 = 4A^2.
Intensity Ratio:
To find the intensity ratio of the maximum and minimum fringes, we need to consider the superposition of the waves at different points in the interference pattern.
At the constructive interference points (maximum fringes), the amplitudes of the two waves add up, resulting in a net amplitude of 3A + 2A = 5A. The intensity at these points is proportional to the square of the amplitude, (5A)^2 = 25A^2.
At the destructive interference points (minimum fringes), the amplitudes of the two waves cancel each other out, resulting in a net amplitude of 3A - 2A = A. The intensity at these points is proportional to the square of the amplitude, (A)^2 = A^2.
Therefore, the intensity ratio of the maximum and minimum fringes is (25A^2)/(A^2) = 25:1.
Hence, the correct answer is option 'C' - 25:1.
When light waves produce interference, the resulting interference pattern consists of a series of bright and dark fringes. The intensity of these fringes depends on the amplitude of the interfering waves.
Given that the amplitudes of the interfering waves are in the ratio 3:2, let's assume the amplitudes to be 3A and 2A, respectively.
Intensity and Amplitude:
The intensity of a wave is directly proportional to the square of its amplitude. Therefore, the intensity of the first wave is (3A)^2 = 9A^2, and the intensity of the second wave is (2A)^2 = 4A^2.
Intensity Ratio:
To find the intensity ratio of the maximum and minimum fringes, we need to consider the superposition of the waves at different points in the interference pattern.
At the constructive interference points (maximum fringes), the amplitudes of the two waves add up, resulting in a net amplitude of 3A + 2A = 5A. The intensity at these points is proportional to the square of the amplitude, (5A)^2 = 25A^2.
At the destructive interference points (minimum fringes), the amplitudes of the two waves cancel each other out, resulting in a net amplitude of 3A - 2A = A. The intensity at these points is proportional to the square of the amplitude, (A)^2 = A^2.
Therefore, the intensity ratio of the maximum and minimum fringes is (25A^2)/(A^2) = 25:1.
Hence, the correct answer is option 'C' - 25:1.
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Light was producing interference have their amplitudes in the ratio 3:...
Imax÷Imin=(A1+A2)^2÷(A1-A2)^2
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Light was producing interference have their amplitudes in the ratio 3:2. The intensity ratio of maximum and minimum of interference fringes isa) 36:1 b) 9:4 c) 25:1 d) 6:4 Correct answer is option 'C'. Can you explain this answer?
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