The diagrams below illustrate a fiber with a core index of n₁ = 1.5 and a cladding index of n₂ = 1.4.  The critical angle for this fiber will be 69°. (Click here for proof.)  Light striking the top edge of the core with an angle of incidence between 0° and 68° will escape the fiber (Figure A), while light striking the interface with an angle of incidence between 69° and 90° will be trapped (Figure B).  (Note:  Claddings are in general much thicker than cores, but the core has been magnified in these drawings to show clearly the path of light.)
 

From the figures, we can tell that the cut-off angle will be between 28° and 62°, probably closer to 28°. We can calculate the exact number as follows:

	Light hitting the top surface of the fiber at the critical angle, 69°, will make an angle of 21° with the normal to the left edge of the fiber.  Click here for more detailed explanation.
	Using Snell's Law at the entrance of the fiber gives an angle of incidence at the front of the fiber of n0 sin θ0max = n1 sin 21° θ0max = sin-1 (1.5*sin 21°) = 32°
	Light incident on the fiber at any angle greater than the one calculated here will not be totally internally reflected at the top edge of the fiber, so we have found the cut-off angle, θ0max = 32°.

Can we relate the cut-off angle directly to the indices of refraction of a fiber? Continue to the next page to find out!