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Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are:
- a)5I and I
- b)5I and 3I
- c)9I and I
- d)9I and 3I
Correct answer is option 'C'. Can you explain this answer?
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Two coherent monochromatic light beams of intensities I and 4I are sup...
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Two coherent monochromatic light beams of intensities I and 4I are sup...
Given: Two coherent monochromatic light beams of intensities I and 4I are superposed.
To find: The maximum and minimum possible intensities in the resulting beam.
Solution:
When two coherent beams of light are superposed, the resultant intensity is given by:
I_r = I_1 + I_2 + 2√(I_1I_2)cosΦ
where I_1 and I_2 are the intensities of the two beams, Φ is the phase difference between them.
Now, in our case, the intensities of the two beams are I and 4I.
Therefore, the resultant intensity is given by:
I_r = I + 4I + 2√(I×4I)cosΦ
I_r = 5I + 4IcosΦ
Maximum and minimum intensities:
The maximum value of cosΦ is +1, which occurs when the two waves are in phase.
So, the maximum intensity of the resulting beam is:
I_max = 5I + 4IcosΦ = 5I + 4I×1 = 9I
The minimum value of cosΦ is -1, which occurs when the two waves are in opposition (180° out of phase).
So, the minimum intensity of the resulting beam is:
I_min = 5I + 4IcosΦ = 5I + 4I×(-1) = I
Therefore, the maximum and minimum possible intensities in the resulting beam are 9I and I, respectively.
Hence, option (c) is correct.
To find: The maximum and minimum possible intensities in the resulting beam.
Solution:
When two coherent beams of light are superposed, the resultant intensity is given by:
I_r = I_1 + I_2 + 2√(I_1I_2)cosΦ
where I_1 and I_2 are the intensities of the two beams, Φ is the phase difference between them.
Now, in our case, the intensities of the two beams are I and 4I.
Therefore, the resultant intensity is given by:
I_r = I + 4I + 2√(I×4I)cosΦ
I_r = 5I + 4IcosΦ
Maximum and minimum intensities:
The maximum value of cosΦ is +1, which occurs when the two waves are in phase.
So, the maximum intensity of the resulting beam is:
I_max = 5I + 4IcosΦ = 5I + 4I×1 = 9I
The minimum value of cosΦ is -1, which occurs when the two waves are in opposition (180° out of phase).
So, the minimum intensity of the resulting beam is:
I_min = 5I + 4IcosΦ = 5I + 4I×(-1) = I
Therefore, the maximum and minimum possible intensities in the resulting beam are 9I and I, respectively.
Hence, option (c) is correct.
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Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are:a)5I and Ib)5I and 3Ic)9I and Id)9I and 3ICorrect answer is option 'C'. Can you explain this answer?
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