How many bright fringes are seen in the reflected light

In summary: Substitute l = 15 cm and fringe width from above equation to find the number of fringes. In summary, the problem involves two optically flat plates separated by a wire with an air gap ranging from 0 to 0.210 mm. The plates are illuminated with light of wavelength 570.0 nm. By comparing similar triangles, one can find the value of x and substitute it into the equation 2t = (m + ½)λ to find the fringe width between two bright fringes. The number of fringes can then be calculated by dividing the length of the plates by the fringe width.
  • #1
msk172
22
0

Homework Statement



Two optically flat plates of glass are separated at one end by a wire of diameter 0.210 mm; at the other end they touch. Thus, the air gap between the plates has a thickness ranging from 0 to 0.210 mm. The plates are 15.0 cm long and are illuminated from above with light of wavelength 570.0 nm. How many bright fringes are seen in the reflected light?

Homework Equations



2t = (m + ½)λ (constructive)
m = 0, 1, 2, …
t = thickness of the air wedge
λ = wavelength in vacuum (air)

The Attempt at a Solution



Pretty darn clueless on this one. I don't think the above equation is even proper. I'm lost on what to do with the plethora of values they give you, and can't understand how to fit them into ANY of the equations I've been mulling through. Since we are looking for bright spots, then clearly we are dealing with constructive interference, but I'm quite lost after that. Any help greatly appreciated. Thanks in advance!
 
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  • #2
msk172 said:

Homework Statement



Two optically flat plates of glass are separated at one end by a wire of diameter 0.210 mm; at the other end they touch. Thus, the air gap between the plates has a thickness ranging from 0 to 0.210 mm. The plates are 15.0 cm long and are illuminated from above with light of wavelength 570.0 nm. How many bright fringes are seen in the reflected light?

Homework Equations



2t = (m + ½)λ (constructive)
m = 0, 1, 2, …
t = thickness of the air wedge
λ = wavelength in vacuum (air)

At a point distance x from the point of contact of the plates can be found by comparing similar triangles formed at x and at extreme end.
So t/x = h/l. In the problem h = 0.210mm and l = 15 cm.
Substitute the value t in 2t = (m + ½)λ.
Find the value x for m and m + 1. From that you can find the fringe width.
 
  • #3
rl.bhat said:
At a point distance x from the point of contact of the plates can be found by comparing similar triangles formed at x and at extreme end.
So t/x = h/l. In the problem h = 0.210mm and l = 15 cm.
Substitute the value t in 2t = (m + ½)λ.
Find the value x for m and m + 1. From that you can find the fringe width.

I am trying like heck to follow the explanation you've provided, but have not gotten anywhere. I see what you mean about substituting the 2t equation in for t, but don't know where to go from there. Perhaps you could simply re-word in a fashion that would be slightly easier to follow? I do greatly appreciate your feedback!
 
  • #4
Plus, you reference fringe width, but the question asks for number of fringes seen, not what their width is. Perhaps these two are connected, but if they are I am not seeing it...
 
  • #5
So t/x = h/l. In the problem h = 0.210mm and l = 15 cm.
Substitute the value t in 2t = (m + ½)λ.

The equation reduces to x = l/2*h*(m + ½)λ
When m = 1, you get fringe width between two bright fringes.
Number of fringes = l/fringe width.
 

1. How is the number of bright fringes in reflected light determined?

The number of bright fringes in reflected light is determined by the number of wavelengths of the incident light that fit into the distance between the reflecting surface and the observer. This is known as the path difference.

2. What factors affect the number of bright fringes in reflected light?

The number of bright fringes in reflected light is affected by the wavelength of the incident light, the distance between the reflecting surface and the observer, and the angle of incidence of the light.

3. Can the number of bright fringes in reflected light be changed?

Yes, the number of bright fringes in reflected light can be changed by altering the factors that affect it. For example, changing the distance between the reflecting surface and the observer or the angle of incidence of the light can change the number of fringes seen.

4. How does the number of bright fringes in reflected light relate to the concept of interference?

The number of bright fringes in reflected light is a result of the interference of the incident light waves. When two or more waves overlap, they can either enhance or cancel each other out, resulting in a pattern of bright and dark fringes.

5. Is the number of bright fringes in reflected light always the same for every reflecting surface?

No, the number of bright fringes in reflected light can vary depending on the properties of the reflecting surface, such as its smoothness and material. Different surfaces can produce different patterns of interference and therefore result in a different number of fringes.

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