How Does Optical Fiber Cladding Affect Light Reflection Angles?

  • Thread starter emi_nic
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In summary, the maximum angle that a ray of light may make with the glass-air boundary and still remain trapped in the fibre is determined by the refractive index of the core and cladding. The process of blooming involves coating a lens with a thin film of calcium fluoride to cut out unwanted reflections, with the minimum thickness being determined by the refractive indices and wavelength of the light.
  • #1
emi_nic
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this is all making my head hurt![b(] any help for anything on here will be greatly appriciated!

-An optical fibre is made of a thin strand of glass with a refractive index of 1.5. If the fibre is surrounded by air, what is the maximum angle a ray of light may make with the glass-air boundary and still remain trapped in the fibre?
-The glass cores of optical fibres are norally surrounded by cladding of refractive index between 1.0 and 1.5. Explain whether this makes the maximum angle that a ray of light makes with the glass-cladding boundary bigger or small than the answer to the first question?







.|.|../
.|.|./
.|.|/
.|.|\
.|.|\\
/|.|.\\
...R1R2

(ignore the dots, they just idicate where spaces should be. Between the two vertical lines is the film, before those is the glass. the diagonal lines are the rays)

In order to cut out unwanted reflections and increase the percentage of light which is transmitted from a spectacle lens, a process called blooming is used. This involves a coating the lens with a very thin film of calcium fluoride of refractive index 1.45. The diagram shows tw0 reflected light rays, R1 and R2, of wavelength 580nm which will combine together to the right of the film.
-What is the minimum thickness of film which will give rise to R1 and R2 being in antiphase?
-Suggest why for this thickness of film, other wavelengths will not be completely cut out.

cheers.
 
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  • #2
The minimum thickness of film which will give rise to R1 and R2 being in antiphase is given by the formula (n1-n2)*λ/2, where n1 and n2 are the refractive indices of the mediums on either side of the film and λ is the wavelength of the light. In this case, the refractive indices of the glass and air are 1.0 and 1.45, respectively, and the wavelength is 580 nm. This gives a value of (1.0-1.45) * 580nm/2 = -164.5 nm. This thickness of film will only cut out light rays at the exact wavelength of 580 nm, as any other wavelength would require a different thickness of film. This is because the thickness of the film required to shift the ray 180 degrees out of phase varies depending on the wavelength of the light.
 

1. What is the difference between reflection and refraction?

Reflection occurs when light bounces off the surface of an object, whereas refraction is the bending of light as it passes through a medium, such as air or water.

2. What are some real-life applications of optics and reflections?

Optics and reflections are used in a variety of everyday devices, such as mirrors, eyeglasses, cameras, and telescopes. They are also crucial in medical imaging technologies like X-rays and MRIs.

3. How do mirrors produce reflections?

Mirrors are made of a smooth, highly reflective surface, typically glass with a thin layer of metal coating. When light hits the mirror, it is reflected back at the same angle, creating a clear reflection of the object in front of it.

4. Can reflections be manipulated or controlled?

Yes, reflections can be manipulated using various techniques such as changing the angle of the incident light, using polarizing filters, or using specialized coatings on surfaces to reduce or enhance reflections.

5. What is the law of reflection?

The law of reflection states that the angle of incidence (the angle at which light hits a surface) is equal to the angle of reflection (the angle at which light bounces off the surface). This law is fundamental to understanding how reflections occur.

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