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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?
Most Upvoted Answer
Light was producing interference have their amplitudes in the ratio 3:...
Interference of Light
Interference of light is a phenomenon that occurs when two or more waves of light are superimposed on each other. The waves can either reinforce or cancel each other out, resulting in a pattern of bright and dark fringes.
Amplitude Ratio
In this problem, it is given that the amplitudes of the interfering waves are in the ratio 3:2. This means that if the amplitude of one wave is 3 units, the amplitude of the other wave is 2 units.
Intensity Ratio
The intensity of light is proportional to the square of its amplitude. Therefore, if the amplitude ratio is 3:2, the intensity ratio will be (3^2):(2^2), which simplifies to 9:4.
Maximum and Minimum Intensity
The maximum intensity occurs when the two waves are in phase and reinforce each other, while the minimum intensity occurs when they are out of phase and cancel each other out. The intensity of the interference fringes varies between the maximum and minimum values.
Intensity Ratio of Maxima and Minima
The intensity ratio of the maxima and minima of the interference fringes is given by (Imax/Imin), where Imax is the maximum intensity and Imin is the minimum intensity. From the previous calculations, we know that Imax/Imin = 9:4. Therefore, the correct option is (c) 25:1.
Conclusion
The intensity ratio of the maxima and minima of the interference fringes depends on the amplitude ratio of the interfering waves. In this problem, the amplitude ratio was given as 3:2, which led to an intensity ratio of 9:4 and a final answer of 25:1 for the ratio of maxima and minima.
Interference of light is a phenomenon that occurs when two or more waves of light are superimposed on each other. The waves can either reinforce or cancel each other out, resulting in a pattern of bright and dark fringes.
Amplitude Ratio
In this problem, it is given that the amplitudes of the interfering waves are in the ratio 3:2. This means that if the amplitude of one wave is 3 units, the amplitude of the other wave is 2 units.
Intensity Ratio
The intensity of light is proportional to the square of its amplitude. Therefore, if the amplitude ratio is 3:2, the intensity ratio will be (3^2):(2^2), which simplifies to 9:4.
Maximum and Minimum Intensity
The maximum intensity occurs when the two waves are in phase and reinforce each other, while the minimum intensity occurs when they are out of phase and cancel each other out. The intensity of the interference fringes varies between the maximum and minimum values.
Intensity Ratio of Maxima and Minima
The intensity ratio of the maxima and minima of the interference fringes is given by (Imax/Imin), where Imax is the maximum intensity and Imin is the minimum intensity. From the previous calculations, we know that Imax/Imin = 9:4. Therefore, the correct option is (c) 25:1.
Conclusion
The intensity ratio of the maxima and minima of the interference fringes depends on the amplitude ratio of the interfering waves. In this problem, the amplitude ratio was given as 3:2, which led to an intensity ratio of 9:4 and a final answer of 25:1 for the ratio of maxima and minima.
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Light was producing interference have their amplitudes in the ratio 3:...
Solution:
Given, the ratio of amplitudes of light producing interference is 3:2.
Let the amplitudes of the two waves be A and B, such that A/B = 3/2.
The intensity of light is proportional to the square of the amplitude, so the ratio of intensities will be (A/B)^2 = (3/2)^2 = 9/4.
The interference pattern formed will consist of alternate bright and dark fringes, with the maximum intensity at the bright fringes and minimum intensity at the dark fringes.
Let I_max be the intensity at the bright fringes and I_min be the intensity at the dark fringes.
The intensity at any point on the interference pattern can be given by the formula:
I = I_max + I_min + 2√(I_maxI_min)cosΦ
where Φ is the phase difference between the two waves.
At the bright fringes, Φ = 2nπ, where n is an integer, so cosΦ = 1.
At the dark fringes, Φ = (2n+1)π, where n is an integer, so cosΦ = -1.
Therefore, the ratio of maximum and minimum intensities is:
(I_max/I_min) = (I_max + I_min + 2√(I_maxI_min))/(I_max + I_min - 2√(I_maxI_min))
At the bright fringes, I = I_max, so the above equation becomes:
(I_max/I_min) = (1 + √(I_max/I_min))/(1 - √(I_max/I_min))
Substituting the value of (I_max/I_min) = 9/4, we get:
(I_max/I_min) = (1 + 3/2)/(1 - 3/2) = 5/1
Therefore, the ratio of maximum and minimum intensities is 25:1.
Hence, the correct option is (c) 25:1.
Given, the ratio of amplitudes of light producing interference is 3:2.
Let the amplitudes of the two waves be A and B, such that A/B = 3/2.
The intensity of light is proportional to the square of the amplitude, so the ratio of intensities will be (A/B)^2 = (3/2)^2 = 9/4.
The interference pattern formed will consist of alternate bright and dark fringes, with the maximum intensity at the bright fringes and minimum intensity at the dark fringes.
Let I_max be the intensity at the bright fringes and I_min be the intensity at the dark fringes.
The intensity at any point on the interference pattern can be given by the formula:
I = I_max + I_min + 2√(I_maxI_min)cosΦ
where Φ is the phase difference between the two waves.
At the bright fringes, Φ = 2nπ, where n is an integer, so cosΦ = 1.
At the dark fringes, Φ = (2n+1)π, where n is an integer, so cosΦ = -1.
Therefore, the ratio of maximum and minimum intensities is:
(I_max/I_min) = (I_max + I_min + 2√(I_maxI_min))/(I_max + I_min - 2√(I_maxI_min))
At the bright fringes, I = I_max, so the above equation becomes:
(I_max/I_min) = (1 + √(I_max/I_min))/(1 - √(I_max/I_min))
Substituting the value of (I_max/I_min) = 9/4, we get:
(I_max/I_min) = (1 + 3/2)/(1 - 3/2) = 5/1
Therefore, the ratio of maximum and minimum intensities is 25:1.
Hence, the correct option is (c) 25:1.
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