Why Does Light Diminish with a Phase Difference of 5λ?

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Homework Help Overview

The discussion revolves around the behavior of light waves with a phase difference of 5 wavelengths (λ) and the implications of this phase difference on their interference. The subject area includes wave optics, specifically focusing on coherent light sources and diffraction phenomena.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the interpretation of the original poster's (OP) assertion regarding the phase difference and its effect on light intensity. There are questions about the exact wording of the problem and the implications of a path length difference of 2 µm being equivalent to 5 wavelengths. Some participants also discuss the arithmetic involved in applying the diffraction grating equation and question the correctness of the OP's calculations.

Discussion Status

The discussion is active, with participants providing feedback on the OP's interpretations and calculations. There is a recognition of potential errors in the OP's explanation regarding phase relationships, and some participants are attempting to clarify the mathematical approach used in the second question. Multiple interpretations of the problem are being explored, and guidance is being offered without reaching a consensus.

Contextual Notes

There are indications that the original problem may have been translated, leading to potential misunderstandings. Participants are also noting the absence of diagrams that could aid in clarifying the situation described in the questions.

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Homework Statement
1. Two coherent waves traveling along paths 2 µm apart collide at a point. Will the light with a wavelength of 400 nm at that point be amplified or diminished?
2. A diffraction grating is perpendicularly incident on light of 760 nm. A first-order diffraction maximum occurs at an angle of 10 degrees. How many strokes are there per millimetre of the diffraction grating?
Relevant Equations
d sin(θ) = mλ ; N = 1 / d
1 exercise: The light at that point will diminish.

Explanation:
Since the waves are coherent, their phase difference depends on the difference in distance travelled.
A path difference of 2 µm is equal to 5 wavelengths (λ = 400 nm).
When the phase difference of the waves is 5λ, they are in opposite phase and are attenuated when added together.

2 exercise:
We use the diffraction grating equation:
d sin(θ) = mλ

d is the width of the grid stroke
θ is the angle of the diffraction maximum
m is the order number of the diffraction maximum (in the first order m = 1)
λ is the wavelength of light

d = λ / (m sin(θ)) = 760 nm / (1 * sin(10°)) = 4374 nm
N = 1 / d = 1 / (4374 nm * 10^(-9) m/nm) = 1200 mm^(-1)

Answer: 1200 strokes?
I need help with figuring out if this is the correct way to solve these problems and if my answers are correct. I would be grateful for any type of assistance.
 
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Is question 1 quoted exactly? It doesn't make much sense to me. Your interpretation seems to be that "2##\mu##m apart" means that the waves are travelling in the same direction and in the same place, but in that case, what does it mean that they collide? Is there a diagram you can share? Or an exact quote of the text?

For the second question your process seems correct but your arithmetic has gone wrong somewhere. Your number for ##d## seems to be off by a little; your number for ##N## is off by a lot.
 
Ibix said:
Is question 1 quoted exactly? It doesn't make much sense to me. Your interpretation seems to be that "2##\mu##m apart" means that the waves are travelling in the same direction and in the same place, but in that case, what does it mean that they collide? Is there a diagram you can share? Or an exact quote of the text?
Judging from the explanation provided by OP,
femaleinstemneedhelp said:
Explanation:
Since the waves are coherent, their phase difference depends on the difference in distance travelled.
A path difference of 2 µm is equal to 5 wavelengths (λ = 400 nm).
"2 μm apart" means "a path length difference of 2 μm." OP found this path length difference to be 5λ which is correct. However, OP's assertion
femaleinstemneedhelp said:
When the phase difference of the waves is 5λ, they are in opposite phase and are attenuated when added together.
is incorrect. It would be correct if the path length difference were ½(5λ) i.e. half-integer wavelengths.
 
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Ibix said:
Is question 1 quoted exactly? It doesn't make much sense to me. Your interpretation seems to be that "2##\mu##m apart" means that the waves are travelling in the same direction and in the same place, but in that case, what does it mean that they collide? Is there a diagram you can share? Or an exact quote of the text?

For the second question your process seems correct but your arithmetic has gone wrong somewhere. Your number for ##d## seems to be off by a little; your number for ##N## is off by a lot.
First of all thanks for replying. English isn't my first language, both the questions are translated from my native language so I apologise if it doesn't make sense. No diagram was provided for both questions. Regarding the second question, judging from the reply, it seems that everything is correct except for the actual math part?
 
kuruman said:
Judging from the explanation provided by OP,

"2 μm apart" means "a path length difference of 2 μm." OP found this path length difference to be 5λ which is correct. However, OP's assertion

is incorrect. It would be correct if the path length difference were ½(5λ) i.e. half-integer wavelengths.
Thanks for replying. Did you mean that the whole answer is incorrect or is it only problems with the explanation?
 
femaleinstemneedhelp said:
Regarding the second question, judging from the reply, it seems that everything is correct except for the actual math part?
Yes.
 
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femaleinstemneedhelp said:
Thanks for replying. Did you mean that the whole answer is incorrect or is it only problems with the explanation?
Your conclusion "When the phase difference of the waves is 5λ, they are in opposite phase and are attenuated when added together" is incorrect. When the wavefront advances by one wavelength, the phase advances by 2π; when the wavefront advances by one-half wavelength, the phase advances by π.

In part 2 your algebra is faulty. If $$d~\sin\theta =m~\lambda~,$$is $$d=\frac{\lambda}{m~\sin\theta}~??$$
 

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