Answer:
The angle of refraction at the entrance must be less than 18.34 degrees if total internal reflection is to occur at the upper edge of the fiber.
Q_2 < 18.34 degrees
Explanation:
Given:
- Index of refraction of inner core n_2 = 1.497
- Index of refraction of outer cladding n_4 = 1.421
Find:
Numerically, what is the largest angle (in degrees) a ray will make with respect to the interface of the fiber θmax, and still experience total internal reflection?
Solution:
- We will extrapolate the ray inside further in the fiber. (See attachment - of a schematic diagram similar to the problem at hand).
- In our case, We will call the angle of incidence on the bottom edge of the fiber Q_5. We know n_1= 1.00, n_2= 1.497, and n_4= 1.421.
- We want Q_C for the core-cladding interface, values of Q_2 for which Q_3 is greater than Q_C, and then the corresponding values for Q_1 and Q_5.
- The critical angle is found from:
Q_c = sin^-1 (n_4 / n_2)
- Snell's Law at the entrance gives:
n_1*sin (Q_1)= n_2*sin(Q_2).
- Finally, Q_3 and Q_5 are alternate angles for parallel lines. Therefore the two angles are equal:
Q_3 = Q_5
- Using the definition of the critical angle,
Q_c = sin^-1 (1.421/1.497)
Q_c = 71.664 degrees
- For total internal reflection to occur, Q_3 must be greater than the critical angle. Therefore,
Q_3 = 90 - Q_2 > Q_c
90 - Q_c > Q_2
90 - 71.664 > Q_2
Q_2 < 18.34 degrees.
Hence, The angle of refraction at the entrance must be less than 18.34 degrees if total internal reflection is to occur at the upper edge of the fiber.
