An optical fiber consists of a glass core (index of refraction n1) surrounded by a coating (index of refration n2<n1). Suppose a beam of light enters the fiber from air at an angle θ with the fiber axis as shown in attached. (a) Show that the greatest possible value of θ for which a ray can travel down the fiber is given by θ = arcsin(sqrt(n1^2 - n2^2)). (b) If the indices of refraction of the glass and coating are 1.58 and 1.53, respectively, what is this value of the incident angle θ?
na*sin(θa) = n1*sin(θ1)
Critical Angle: θc = arcsin(n1/n2)
The Attempt at a Solution
See attached for my drawing. Any angle greater than θc will cause total internal reflection so,
θo > θc
where θo = 90 - θ1
Substituting for θc
θo > arcsin(n2/n1)
Substituting for θo
90 - θ1 > arcsin(n2/n1)
Substituting for θ1
90 - arcsin(sin(θa)/n1) > arcsin(n2/n1)
Taking sine of the whole equation
1 - sin(θa/n1) > n2/n1
Solving for θa
θa < arcsin(n1-n2)
According to my calculations, this would be the maximum angle θa for which a ray can travel down the fiber. Unfortunately it's not correct.
Where do they get the sqrt(n1^2-n2^2) term?
Thanks in advance, MrMoose