Question

4-3: For a glass (n= 1.5)/quarts (n= 1.38) interface and an angle of incidence of 35 degrees, determine the angle of refraction.

4-4: Determine the critical angle for the fiber described in problems 4-3.

4-5: Determine the acceptance angle for the fiber described in problems 4-3.

4-6: Determine the numerical angle for the fiber described in problems 4-3.

Answer 1

4.3) 1.5 * sine (35 degrees) = 1.38 * sine (Θr) => Θr = 38.57 deg

4.4) The critical angle for the given interface is just

sin^-1(n2/n1)
= sin^-1(1.38/1.5)
= sin^-1(0.92)
= 66.9 degrees.

4.5)

Numerical Aperture (NA): NA is the light gathering ability or capacity of an optical fiber. More the NA. the more efficient will be fiber. It is also known as figure of merit.

NA is related to refractive index of core (n1), cladding (n2) and outside medium (n0) as

NA = √n12 – n22/n0

If the medium is air then n0 =1, then

NA = √n12 – n22 = 0.59

Acceptance angle (θ): It is the maximum angle made by the light ray with the fiber axis, so that light can propagate through the fiber after total internal reflection.

Relation NA and acceptance angle:

NA = Sin θ = 0.59

θ = 36 deg

4.6) Numerical Aperture (NA) = 0.59