# Homework Help: Calculating path length difference - help!

1. Feb 12, 2013

### Enthusiast94

1. The problem statement, all variables and given/known data

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))

2. Relevant equations
m.lambda=d sin(theta)

3. 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)

#### Attached Files:

• ###### 123.jpg
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2. Feb 12, 2013

### tiny-tim

Welcome to PF!

Hi Enthusiast94! Welcome to PF!
isn't there a path difference on both sides?
Yes.

3. Feb 12, 2013

### TSny

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

4. Feb 12, 2013

### Enthusiast94

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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• ###### 20130212_204311.jpg
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5. Feb 12, 2013

### TSny

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

6. Feb 12, 2013

### Enthusiast94

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

7. Feb 12, 2013

### tiny-tim

No, but you need two right-angled triangles …

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

draw it carefully, and you'll see!

8. Feb 12, 2013

### TSny

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?

9. Feb 12, 2013

### Enthusiast94

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

#### Attached Files:

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10. Feb 12, 2013

### tiny-tim

are you deliberately trying to confuse yourself?

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!
i'm not sure what you mean

11. Feb 12, 2013

### Enthusiast94

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. Feb 12, 2013

### tiny-tim

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. Feb 12, 2013

### Enthusiast94

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

Anyway, thanks for your help.

14. Feb 12, 2013

### TSny

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.
You should be able to use geometry to determine the magnitudes of the angles in your right triangles.

Last edited: Feb 12, 2013
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