# Interference in Thin Waves

1. Jun 9, 2015

### rockerdude1210

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

One microscope slide is placed on top of another with their left edges in contact and a human hair under the right edge of the upper slide. As a result, a wedge of air exists between the slides. An interference pattern results when monochromatic light is incident on the wedge. What kind of fringe is seen at the left edges of the slides?
A. a dark fringe
B. a bright fringe
C. a circular fringe
D. impossible to determine

2. Relevant equations
2nt=mλ and 2nt=(m+0.5)λ

3. The attempt at a solution

A is the correct answer but I honestly have no idea how to approach this.

2. Jun 10, 2015

### andrevdh

How are the interference fringes created?

3. Jun 10, 2015

### rockerdude1210

Aren't they created when the reflections of light interact by constructive and destructive interference? Which is caused by some light rays moving a further or lesser distance.

4. Jun 10, 2015

### andrevdh

Yes, but where are the light reflected from?

5. Jun 10, 2015

### rockerdude1210

Is the light reflected from the surfaces of the slides?

6. Jun 11, 2015

### andrevdh

Yes, but more correctly at the top and bottom of the air-glass interface.

7. Jun 12, 2015

### rockerdude1210

Oh okay.

So how would you be able to tell using that information whether dark and bright fringes were produced?

8. Jun 15, 2015

### andrevdh

The two beams are combined in the eye and they interfere with each other.
What do you think would happen it two beams from the left hand side
of the wedge of air would interfere with each other?

9. Jun 17, 2015

### rockerdude1210

Oh it makes sense now. They would interfere destructively at the end

10. Jun 17, 2015

### Wily Willy

I think it is important to note that when light reflects off of the surface of a material that is "less optically dense" it does not undergo a phase change. However, if it does reflect off the surface of a more optically dense material then it does undergo a phase change of 180 degrees. As the distance between the second air-glass interface and the third air-glass interface approaches zero, the phase difference of the light reflected from them approaches 180 degrees. Thereby the left side appears darker than the right.

11. Jun 18, 2015

### andrevdh

Here are several videos which demonstrates the reflections of a wave (pulse) at the intersection
of a high and a low density medium. The first set is where two strings meet, one with
a high and the other with a low linear density. Below these are simulations of a light
wave reflected at a glass-air interface. Look carefully and you would see the difference
in the reflection when the wave (pulse) is reflected as it travels first from less dense to
the more dense medium and then from the more dense to the less dense medium.
http://www.animations.physics.unsw.edu.au/jw/light/reflection-and-phases.html

Last edited: Jun 18, 2015
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