Please help me on as-level physics PHYA2. Got exam tomorrow! :)?
I have a question from the past paper. I can not find the paper from the internet so I will have to explain.
BTW, the paper is june 2009 and the question is Q6aii)
Q) An optical fiber used for communications has a core of refractive index 1.55 which is surrounded by cladding of refractive index 1.45.
6a) critical angle is 63.9 deg and incident angle is 60 deg.
Note: Incidence substance is the core.
Q6aii) state why the light ray enters the cladding at Q.
Note: Q is the cladding boundary.
Answer: I know that the incident angle is smaller than the critical angle. (so refraction should occur). However, the incident substance (the core) has a larger refractive index than the cladding, (so TIR should occur).
I just want to know why does refraction occurs if the incident substance (core) has larger refractive index but the incident angle is smaller than the critical angle.
For TIR(total internal reflection) to occur,
1) incident angle should exceed the critical angle. (which it does not)
2) Incident substance (core) has larger refractive index than the cladding. (which is does).
So, why still refraction occurs and not reflection?
For total internal reflection, why does both criteria (1 and 2) have to be met?
- 8 years agoFavorite Answer
Quite simple: refraction occurs because the incident angle (60 deg) is LESS than the critical angle.
The bit about the core having a larger refractice index than the cladding is a requirement for TIR to occur at all. If the core had a small refractive index, TIR could NEVER occur.
Because the critical angle (x) is derived from the refraction law as sin x = n2 / n1
since sin x will always be less than 1, n1 must be bigger than n2 or there is no solution to the equation.