Diffraction grating: arbitrary incident angle

Click For Summary
SUMMARY

This discussion focuses on the treatment of diffraction on a one-dimensional grating when incident light approaches from an arbitrary angle, specifically when both the angle of incidence (i) and the angle gamma are non-zero. The participant notes that while the Kx component remains unchanged with respect to i, the Ky component is influenced by the distance d, which is modified by the cosine of the angle of incidence. The equation presented, d*cos(i)*sin(Θ)=n*λ, encapsulates the relationship between these variables in the context of diffraction.

PREREQUISITES
  • Understanding of diffraction principles in physics
  • Familiarity with one-dimensional grating concepts
  • Knowledge of wave optics and interference patterns
  • Basic mathematical skills for manipulating trigonometric equations
NEXT STEPS
  • Research the mathematical derivation of diffraction patterns for arbitrary angles of incidence
  • Explore advanced topics in wave optics, focusing on grating equations
  • Study the impact of varying incident angles on diffraction efficiency
  • Investigate experimental setups for measuring diffraction patterns from laser-printed gratings
USEFUL FOR

Physicists, optical engineers, and students studying wave optics who are interested in advanced diffraction phenomena and their mathematical treatments.

anandr
Messages
6
Reaction score
2
Diffraction on 1D grating is covered in many physics books. Usually they cover simple case when incident light is coming along the normal to the grating. Sometimes they present slightly more complicated case when incident light is tilted in the plane perpendicular to the stripes (left case on the image below).
Does anybody have an idea how to treat diffraction on a grating when incident light comes from arbitrary direction ( gamma ≠ 0 and i ≠ 0 , right case on the image below )?

xqgps4.jpg

I tried to split the incident wave vector k in two components: kx and kyz.
First part (kx) should diffract same way as in case of normal incidence (diffracted maxima will be symmetrically placed around the k across the grating stripes with diffraction angles Θ defined by d*sin(Θ)=n*λ ).
Second part (kyz) seems to be the case presented in left part of the image (these maxima are also symmetrically placed around the k but this time incident angle i is taken into account so d*(sin(Θ)+sin(i))=n*λ ). After that I just added the resulting diffracted wave vectors for each n to get the final diffracted wave vectors. Is this approach correct?
 
Science news on Phys.org
I just played with a Laser printed grating few days ago.

For the Kx part there should be no change with i.
For the Ky part the only variable that changes with i is d. That should look like:
d*cos(i)*sin(Θ)=n*λ
Now i did suppose that the angle of incidence is changed by rotating the grating along the x and y-axis for ease.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 26 ·
Replies
26
Views
11K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K