Total internal reflection inside a fiber optic cable

In summary, we are given the dimensions and refractive indices of a multi-mode fiber optic cable and asked to find the largest possible angle at which light can enter the fiber and experience total internal reflection. To find this angle, we can use the equation sinθc=n2/n1 and ncladdingθ1=ncoreθ2, where n1 is the refractive index of the cladding, n2 is the refractive index of the core, θ1 is the angle of incidence, and θ2 is the angle of refraction. By setting one of the angles to 90 degrees, we can solve for the other angle and find the largest possible value for θin.
  • #1
aChordate
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Homework Statement



A multi-mode fiber optic cable has a core diameter d = 115 mm, ncore = 1.42, and ncladding = 1.17. What is the largest possible angle θin relative to the surface normal that light can enter the fiber (i.e. pass from air into the fiber core) and experience total internal reflection inside the fiber?

Homework Equations



sinθc=n2/n1

ncladdingθ1=ncoreθ2

The Attempt at a Solution


d = 115 mm
ncore = 1.42
ncladding = 1.17
θin=?

Do I have the correct equations?
 
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  • #2
aChordate said:

Homework Statement



A multi-mode fiber optic cable has a core diameter d = 115 mm, ncore = 1.42, and ncladding = 1.17. What is the largest possible angle θin relative to the surface normal that light can enter the fiber (i.e. pass from air into the fiber core) and experience total internal reflection inside the fiber?

Homework Equations



sinθc=n2/n1

ncladdingθ1=ncoreθ2

The Attempt at a Solution


d = 115 mm
ncore = 1.42
ncladding = 1.17
θin=?

Do I have the correct equations?

Is the end of this optical fibre squared off , like a cylinder, or basically hemispherical, like many of these fires are when created?

Also I like to match subscripts, to avoid losing track of which angle you are calculating, so I would write either:

ncladdingsinθcladding=ncoresinθcore

or

n1sinθ1=n2sinθ2

Having established [written down] that medium 1 was the cladding and material 2 was the core.

(and you seem to have left the sin out - though if the angles involved are small you can approximate sinθ to θ
 
  • #3
ncladdingsinθcladding=ncoresinθcore

1.17sinθcladding=1.42sinθcore

How do I find the angles?
 
  • #4
aChordate said:
ncladdingsinθcladding=ncoresinθcore

1.17sinθcladding=1.42sinθcore

How do I find the angles?

For T.I.R. the angle of refraction is set at 90o so one of those angles is 90o.

Which one do you think that might be.

Hint: if you pick the wrong one, the problem gives no real solution.
 
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  • #5


Yes, you have the correct equations. To find the largest possible angle θin, you can use the equation sinθc = n2/n1, where n1 is the refractive index of air (approximately 1) and n2 is the refractive index of the fiber core. In this case, n2 = 1.42. Plugging in these values, we get sinθc = 1.42/1 = 1.42. However, the maximum value for sinθc is 1, so θc = sin^-1(1) = 90°. This means that the largest possible angle θin is 90°, which is the critical angle for total internal reflection. Any angle larger than 90° would result in the light passing through the fiber core and not experiencing total internal reflection.
 

1. What is total internal reflection inside a fiber optic cable?

Total internal reflection is a phenomenon that occurs when light traveling through a material with a higher refractive index strikes the boundary of a material with a lower refractive index at an angle greater than the critical angle. This causes the light to reflect back into the material instead of passing through it.

2. How does total internal reflection work in fiber optic cables?

In fiber optic cables, a core made of a material with a higher refractive index is surrounded by a cladding material with a lower refractive index. When light travels through the core, it strikes the boundary of the cladding at an angle greater than the critical angle, causing total internal reflection. This allows the light to travel through the cable without losing energy or being affected by external interference.

3. What are the advantages of total internal reflection in fiber optic cables?

Total internal reflection allows for efficient transmission of light through fiber optic cables without loss of energy. It also allows for high-speed communication, as light travels much faster than electricity. Additionally, fiber optic cables are immune to electromagnetic interference, making them a reliable option for transmitting data.

4. Are there any limitations to total internal reflection in fiber optic cables?

One limitation of total internal reflection in fiber optic cables is the phenomenon of modal dispersion, where different modes of light travel at different speeds through the cable, causing distortion in the transmitted signal. This can be mitigated by using single-mode fibers, which have a smaller core and only allow one mode of light to travel through.

5. How is total internal reflection used in other applications besides fiber optic cables?

Total internal reflection is also used in other applications such as prisms, mirrors, and lenses in optical devices. It is also used in the field of medicine for imaging techniques such as endoscopy and microscopy. Additionally, total internal reflection is utilized in the field of telecommunications for signal processing and amplification.

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