Q46P
Question
Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica for the cladding and silica doped with germanium for the core. (a) What is the critical angle for light traveling in the core and reflecting at the interface with the cladding material? (b) The numerical aperture (NA) is defined as the angle of incidence at the flat end of the cable for which light is incident on the core–cladding interface at angle ucrit (Fig. P33.46). Show that (c) What is the value of ui for ?
Answer
Step-by-Step Solution
Verifiedthe critical angle at interface will be 81.79°
In the triangle ABC We have the 0c so the value of 9,. will be:
the critical angle of incidence is
The critical angle is the angle of incidence where the angle of refraction is 90°. The light must travel from an optically more dense medium to an optically less dense medium.
(a) For 0a.. at point A apply the Snell's Law.
Hence, the critical angle at interface will be 81.79°
(b) For the value of index of refraction 77,1 and 11.2, We have the 911 at point A as below. Apply the Snell's Law.
In the triangle ABC We have the 0c so the value of 9,. will be:
Apply the Snell's Law to get the 9,:
\
(c) Use the result of part (b) of this question We have:
Therefore the critical angle of incidence is \