Interference fringes with reflected light

In summary, the Homework statement is that a point source of light (λ = 589 nm) is placed 0.4 mm above the surface of a glass mirror. Interference fringes are observed on a screen 6m away, and the interference is coming from light reflecting off the mirror and light coming directly from the source. Find the spacing of the fringes.
  • #1
kankerfist
35
0

Homework Statement



A point source of light (λ = 589 nm) is placed 0.4 mm above the surface of a glass mirror. Interference fringes are observed on a screen 6m away, and the interference is coming from light reflecting off the mirror and light coming directly from the source. Find the spacing of the fringes

Homework Equations



reflection from the mirror will cause a phase shift of pi

the phase is also shifted due to the path difference:
(Δr)(2pi)/λ

so total phase shift is pi + (Δr)(2pi)/λ

The Attempt at a Solution



I can calculate interference at the point on the screen directly in front of the point source:

pi + (Δr)(2pi)/λ = pi + (0.8 mm)(2 pi) / 589 nm

but I'm not sure where to begin finding the fringe spaces on the screen. I can't come up with a function to relate interference on the screen to the path difference caused by the mirror. Any hints would be appreciated!
 
Physics news on Phys.org
  • #2
By "directly in front of" do you mean the direct ray is horizontal so you have two right triangles to work with? If that is the case you know how to do, then probalby what you need is the law of sines or law of cosines to tackle the non-horizontal cases.
 
  • #3
Well, I made a diagram:

phys.jpg


I thought it might help, but I still haven't even figured out where to start.
 
  • #4
kankerfist said:
Well, I made a diagram:

I thought it might help, but I still haven't even figured out where to start.

I think your diagram is not correct. I interpret the problem to mean the mirror is laying horizontally with the source slightly above it. The direct and reflected rays are nearly horizontal propegating to the wall. The angles of incidence and reflection at the mirror are equal. Drawing the diagram with well spaced lines and larger than expected angles is fine, but I think the .4mm is a vertical distance while the 6m is a horizontal distance.

So, more or less, if you put your source at the left wall and the mirror where you have the source, make the vertical distance from the source to the mirror .4mm and the horizontal distance from the source to the right wall 6m, and make equal angles at the mirror, you have it. Then you need to compare the direct path length to the total (two legs) reflected path length and account for any phase reversals.

Before you get too carried away with the calculations, see if you can think of any connection between this and a double slit problem. Where is the image of the source in the mirror?
 
Last edited:
  • #5
OlderDan said:
I think your diagram is not correct.

You just made me slap my forehead! Thank you kindly for pointing out the obvious :)
 

1. What is the principle behind interference fringes with reflected light?

The principle behind interference fringes with reflected light is the phenomenon of interference, where two or more light waves superimpose on each other. When light is reflected off a surface, it creates two reflected waves that interfere with each other. This interference results in a pattern of bright and dark fringes, which can be seen as bands of light and dark regions on the surface.

2. What factors affect the formation of interference fringes with reflected light?

Several factors can affect the formation of interference fringes with reflected light. These include the angle of incidence of the light, the wavelength of the light, the distance between the reflecting surface and the observer, and the properties of the reflecting surface, such as its smoothness and reflectivity.

3. Can interference fringes with reflected light be observed with all types of light?

No, interference fringes with reflected light can only be observed with coherent light, meaning light waves that have a constant phase relationship with each other. This type of light can be produced by lasers, but not by most natural sources of light, such as the sun or a lightbulb.

4. How are interference fringes with reflected light used in scientific research?

Interference fringes with reflected light are used in a variety of scientific research fields, including optics, astronomy, and microscopy. They can be used to measure the properties of materials, such as their refractive index and thickness, and to study the structure of objects at a microscopic level.

5. Can interference fringes with reflected light be seen with the naked eye?

Yes, interference fringes with reflected light can be seen with the naked eye, although they may be difficult to observe without proper equipment. To make the fringes more visible, scientists often use techniques such as using a monochromatic light source, adjusting the angle of incidence, or using a polarizer to filter out unwanted light.

Similar threads

Replies
4
Views
312
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Other Physics Topics
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Replies
2
Views
685
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
194
  • Quantum Physics
2
Replies
53
Views
4K
Replies
33
Views
2K
  • Quantum Physics
Replies
1
Views
914
Back
Top