Calculating path length difference - help

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Homework Help Overview

The discussion revolves around the calculation of path length differences in the context of constructive interference involving a diffraction grating. The original poster presents a scenario with a narrow light beam and a specific wavelength, seeking to understand the derivation of an equation related to path differences.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the geometry of the problem, questioning the correctness of the drawn triangles and the angles involved. There is discussion about the path differences on both sides of the setup and whether the triangles can be considered right triangles. Some participants suggest alternative diagrams to clarify the situation.

Discussion Status

The conversation is active, with participants providing feedback on the original poster's diagrams and reasoning. There is a focus on ensuring the geometric representations are accurate to facilitate understanding of the path differences. Multiple interpretations of the triangle configurations are being explored.

Contextual Notes

Participants are addressing potential misconceptions regarding the angles in the drawn diagrams and the implications for deriving the equation. There is an emphasis on the necessity of right triangles for the calculations involved.

Enthusiast94
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Homework Statement



A narrow parallel light beam of wavelength 632.8 nm is incident upon a diffraction
ruler (i.e. the light reflects off of the grating rather than passing through it). The
ruler has 600 lines per mm.

Show that the expression for the condition for constructive interference is
m.lambda = d(cos(alpha) - cos(beta))

Homework Equations


m.lambda=d sin(theta)

The Attempt at a Solution


So the right angle triangle has been formed and path difference (green part) = d.cos(alpha). So, how do I get (- d cos (beta)) in my equation. Also, is the path length I shaded correct?
(Diagram attached)
 

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Welcome to PF!

Hi Enthusiast94! Welcome to PF! :smile:
Enthusiast94 said:
So the right angle triangle has been formed and path difference (green part) = d.cos(alpha). So, how do I get (- d cos (beta)) in my equation

isn't there a path difference on both sides? :wink:
Also, is the path length I shaded correct?

Yes. :smile:
 
Enthusiast94 said:
Also, is the path length I shaded correct?
(Diagram attached)

It appears that you marked the angle between an incoming ray and an outgoing ray as perpendicular. Is that necessarily so?
 
tiny-tim said:
Hi Enthusiast94! Welcome to PF! :smile:


isn't there a path difference on both sides? :wink:


Yes. :smile:

TSny said:
It appears that you marked the angle between an incoming ray and an outgoing ray as perpendicular. Is that necessarily so?


Using the new diagram(attached), I've got r1 = dcos(alpha) and r2 = dsin(beta). So now, delta r = r1 - r2 = d(cos alpha - cos beta).

Does this make sense?
 

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I would still question whether or not your triangle is a right triangle.
 
TSny said:
I would still question whether or not your triangle is a right triangle.

Could the given equation be derived w/o considering it as a right triangle? If so, please explain.
 
Enthusiast94 said:
Could the given equation be derived w/o considering it as a right triangle? If so, please explain.

No, but you need two right-angled triangles …

(the one triangle you've drawn can't be right-angled, can it? :wink:)

draw it carefully, and you'll see! :smile:​
 
Enthusiast94 said:
Could the given equation be derived w/o considering it as a right triangle? If so, please explain.

Try another drawing where the angle in your triangle is not a right angle. Can you still find the path differences by constructing other triangles that are right triangles?
 
tiny-tim said:
No, but you need two right-angled triangles …

(the one triangle you've drawn can't be right-angled, can it? :wink:)

draw it carefully, and you'll see! :smile:​

TSny said:
Try another drawing where the angle in your triangle is not a right angle. Can you still find the path differences by constructing other triangles that are right triangles?

How about this?
But, now the other two angles wouldn't be alpha/beta. :/
 

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  • #10
Enthusiast94 said:
How about this?

are you deliberately trying to confuse yourself? :redface:

the point of a diagram is to help you understand the situation

draw it again, and try to make the right-angles look like right-angles! :smile:​
But, now the other two angles wouldn't be alpha/beta. :/

i'm not sure what you mean :confused:
 
  • #11
tiny-tim said:
are you deliberately trying to confuse yourself? :redface:

the point of a diagram is to help you understand the situation

draw it again, and try to make the right-angles look like right-angles! :smile:​
i'm not sure what you mean :confused:
What went wrong in the last diagram? Please be a bit more specific...

As you said, I'll need two right triangles, so I drew a perp from the point at which it strikes to the opposite ray.
 
  • #12
Enthusiast94 said:
What went wrong in the last diagram?

those are obviously not right-angles!

you really are kidding yourself if you think that drawings like that will reliably steer you towards the right answers!
 
  • #13
tiny-tim said:
those are obviously not right-angles!

you really are kidding yourself if you think that drawings like that will reliably steer you towards the right answers!

It's not about the drawing, it's about sending a message. : )

Anyway, thanks for your help.
 
  • #14
Enthusiast94 said:
How about this?

I think you're getting closer. But as tiny-tim mentioned, your dotted lines were not drawn very carefully. Note how the angles that you marked as right angles look quite a bit larger than 90 degrees.
But, now the other two angles wouldn't be alpha/beta. :/
You should be able to use geometry to determine the magnitudes of the angles in your right triangles.
 
Last edited:

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