# What is the critical angle of an optical fiber when cladding is added of n = 1.49?

The critical angle of a fiber of core (for example, Flint glass) of refractive index 1,62 and cladding of index 1,49 should be 66,89°.

Considering the fact that on the interface between the two media the incoming ray (inside the core) must be refracted parallel to the interface we can use Snell's Law to calculate the angle at which this phenomenon occurs.

The angle of incidence (q1) inside the core will be the critical one and the angle of refraction (q2, inside the cladding) will be at least 90°.

From Snell's Law:

n1 sen (q1) = n2 sem (q2)

But

sen(q2)=sen(90°)=1

and

q1=arcsen(n2/n1)=arcsen(1,49/1,62)=66,89°

Using the same idea (and considering some geometry of triangles) you can now evaluate the angle of acceptance of the fiber, i.e., the angle of incidence of the light ray with the front surface of the glass core (so that the ray enters the core at the right inclination to produce internally this critical angle).

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The critical angle (( \theta_c )) for an optical fiber, when light moves from the core to the cladding, can be calculated using Snell's Law for total internal reflection, given as ( n_1 \sin(\theta_c) = n_2 \sin(90^\circ) ), where ( n_1 ) is the refractive index of the core, ( n_2 ) is the refractive index of the cladding, and ( \sin(90^\circ) = 1 ).

Given ( n_2 = 1.49 ) for the cladding, and assuming the core has a higher refractive index which is necessary for total internal reflection but not specified in your question, the equation simplifies to ( \sin(\theta_c) = \frac{n_2}{n_1} ).

Without the core's refractive index (( n_1 )), we cannot calculate a numerical value for ( \theta_c ). Typically, the core's refractive index is slightly higher than that of the cladding for optical fibers to function correctly. If you have the core's refractive index, you can calculate the critical angle using the formula provided.

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