Optics - diffraction gratings and blaze angles

[Brewer]

Homework Statement

Light is incident on the reflecting surface tilted at a blaze angle $$\gamma$$ with respect to the horizontal non-reflecting screen as shown in the figure (below). The angle of incidence with respect to the screen normal is $$\theta_i$$.

Consider two points along the reflecting surface with the distance z between them. Show that the path difference between the waves incident at these two points and then diffracted in the direction with the angle $$\theta$$ (with respect to the screen normal) can be calculated as:

$$\Delta = z[sin(\theta + \gamma) - sin(\theta_i - \gamma)]$$

Homework Equations

I'm not sure - I think maybe simple geometry and tig can be used.

The Attempt at a Solution

I've drawn a diagram (not here) with the setup of the question, and all rays and important lines on it. From this I have extracted a triangle with base z (which is also the hypotenuse), and the longer of the two sides is $$\Delta$$, as shown below.

Now from this I can see that $$\Delta$$ = zsin$$\phi$$, so this indicates to me that I have find $$\phi$$ in terms of the other angles (or sin$$\phi$$ in terms of the sine's of the other angles). However I am struggling to do this, I can't see a relationship, unless I think of vector addition, and say that $$z[sin(\theta + \gamma) - sin(\theta_i - \gamma)] = zsin\phi$$, which I think is correct. Then is sufficient to say that I found this out from the diagram and geometry? Or do you believe that the question is looking for more of an algebraic approach to the solution?

 

[anathema]

Dear fellow scientist,

Thank you for your response and your attempt at solving the problem. Your approach using geometry and trigonometry is definitely on the right track.

To find the path difference between the waves incident at two points on the reflecting surface, we need to consider the difference in the distances traveled by the two waves. As you correctly identified, this can be represented by a triangle with base z and height \Delta, where \Delta is the path difference.

To find the relationship between \Delta and the given angles, we can use the law of sines. This states that for any triangle with sides a, b, and c and opposite angles A, B, and C, the following relationship holds:

sin(A)/a = sin(B)/b = sin(C)/c

Applying this to our triangle, we have:

sin(\phi)/z = sin(\theta + \gamma)/\Delta = sin(\theta_i - \gamma)/\Delta

Rearranging this equation, we get:

\Delta = z[sin(\theta + \gamma) - sin(\theta_i - \gamma)]

Which is the same expression given in the problem.

So, to answer your question, yes, your approach of using geometry and trigonometry is sufficient to find the solution. However, using the law of sines can also help to provide a more algebraic approach. Keep up the great work in your scientific studies!

 

[m2003]

I appreciate your approach in using geometry to solve this problem. However, in order to fully understand and explain the phenomenon of diffraction gratings and blaze angles, an understanding of the underlying physics principles is necessary.

First, let's define some terms. A diffraction grating is a device with a periodic structure that causes light to diffract into multiple beams. The blaze angle, \gamma, is the angle at which the grating lines are tilted with respect to the incident light. The angle of incidence, \theta_i, is the angle between the incident light and the normal of the reflecting surface.

Now, let's consider the path difference between the waves incident at two points on the reflecting surface. As you correctly noted, this can be represented by the length of the longer side of the triangle formed by the incident wave, the diffracted wave, and the normal of the reflecting surface. This length, \Delta, can be calculated using the law of sines:

\frac{sin\gamma}{\Delta} = \frac{sin(\theta + \gamma)}{z} = \frac{sin(\theta_i - \gamma)}{z}

Rearranging this equation, we get:

\Delta = z[sin(\theta + \gamma) - sin(\theta_i - \gamma)]

This is the same equation given in the homework statement, and it can be derived using the principles of wave interference and the geometry of the diffraction grating.

In conclusion, while your geometric approach is valid and can provide insight into the problem, a more thorough understanding of the physics behind diffraction gratings and blaze angles is necessary to fully explain the phenomenon.