A problem involving dark fringes & light wave & interference.

In summary, the problem involves two pieces of glass with an index of refraction of 1.5 touching at one edge and separated by a thin stretched fiber at the other edge. When light of wavelength 500nm is shone from above, 100 dark fringes are counted. Replacing the air between the glass with water (n=1.33) changes the film, resulting in a different number of observed dark fringes. The correct answer is a) 133, as the film is not the glass itself but the material between the glass. The thickness of the film is determined by the gap between the two pieces of glass, and if this gap is too large the phenomenon will not occur.
  • #1
erinec
31
0

Homework Statement


Two pieces of optically flat glass (n=1.5) touch at one edge and are separated by a thin stretched fibre at the opposite edge. Light of wavelength 500nm is shone from above and 100 dark fringes are counted. The air between the glass is now replaced by water (n=1.33). How many dark fringes are now observed?

a) 133
b) 75
c) 150
d) 100
e) 67

Homework Equations


2nt = (m+0.5)(lambda)
.. I think?

The Attempt at a Solution


I assumed that since the index of refraction for both water AND air are smaller than the index of refraction for glass, it does not make a difference. So I chose d)100. However, the answer is supposed to be a) 133.
 
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  • #2
Hi erinec,

erinec said:

Homework Statement


Two pieces of optically flat glass (n=1.5) touch at one edge and are separated by a thin stretched fibre at the opposite edge. Light of wavelength 500nm is shone from above and 100 dark fringes are counted. The air between the glass is now replaced by water (n=1.33). How many dark fringes are now observed?

a) 133
b) 75
c) 150
d) 100
e) 67


Homework Equations


2nt = (m+0.5)(lambda)
.. I think?

No, I do not believe this is the correct equation. There is one phase reversal here, so the other equation will give the destructive interference conditions.

The Attempt at a Solution


I assumed that since the index of refraction for both water AND air are smaller than the index of refraction for glass, it does not make a difference. So I chose d)100. However, the answer is supposed to be a) 133.

In this problem the film is not the glass, it is whatever is between the glass. So when you change from air to water, you change the n value in the destructive interference equation.
 
  • #3
Hi alphysicist,

Thanks a lot for your insight.

alphysicist said:
In this problem the film is not the glass, it is whatever is between the glass. So when you change from air to water, you change the n value in the destructive interference equation.

Would you be so kind enough to explain why the film is whatever is between the glass and not the glass?

I appreciate your help very much.
 
  • #4
erinec said:
Hi alphysicist,

Thanks a lot for your insight.



Would you be so kind enough to explain why the film is whatever is between the glass and not the glass?

I appreciate your help very much.

When we talk abou a "thin film" here we mean a material whose parallel boundaries are close enough so that light reflecting off the boundaries can interfere.

Here you have a thick piece of glass on the bottom, a thick piece of glass on top, and a very thin space that is either filled with air or water. So the air or water will be the thin film.

(The space has a triangular outline, since the glass pieces touch at one end and are separated by only a wire at the other end. But it's a very thin triangle! and it is the triangular shape that makes you have a series of lines--because the thickness of the film changes as you move from one end of the space to the other.)
 
  • #5
I see.
So if the gap is too large, it won't work right? (Just making sure.)

Thank you! :biggrin:
 

1. What is a dark fringe in relation to light waves?

A dark fringe is a region of reduced or no intensity in an interference pattern formed by overlapping light waves. It occurs when light waves from different sources or parts of the same source cancel each other out due to their opposite phases.

2. How is interference involved in this problem?

Interference is the phenomenon where two or more waves combine to form a resultant wave. In this problem, the interference of light waves is responsible for the formation of dark fringes due to the cancellation of light waves with opposite phases.

3. What causes the formation of dark fringes in an interference pattern?

Dark fringes are formed when the crests of one wave align with the troughs of another wave, resulting in complete destructive interference. This occurs when the path difference between the two waves is equal to half the wavelength of light.

4. How are dark fringes and light waves related?

Dark fringes are a result of the interference of light waves. The interaction of light waves with opposite phases leads to the formation of dark fringes in an interference pattern.

5. Can dark fringes be observed in everyday life?

Yes, dark fringes can be observed in everyday life. They can be seen in the patterns formed by reflected or transmitted light, such as the colors seen on soap bubbles, the iridescence of a CD, or the rainbow patterns on oil spills.

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