Answer to Question #150471 in Optics for Debajit Bora

Question

Using simple ray theory, describe the mechanism for the transmission of light within
an optical fiber. Briefly discuss with the aid of a suitable diagram what is meant by
the acceptance angle for an optical fiber. Show how this is related to the fiber
numerical aperture and the refractive indices for the fiber core and cladding.
An optical fiber has a numerical aperture of 0.20 and a cladding refractive index of
1.59. Determine:
(a) the acceptance angle for the fiber in water which has a refractive index of 1.33;
(b) the critical angle at the core–cladding interface.
Comment on any assumptions made about the fiber.

Answer 1

Using simple ray theory, describe the mechanism for the transmission of light within an optical fiber.

Lightray is launched into a fiber as shown in Fig. The incident ray I₁ enters the fiber at the angle θa. I₁ is refracted upon entering the fiber and is transmitted to the core-cladding interface. The ray then strikes the core-cladding interface at the critical angle θc. I₁ is totally reflected back into the core and continues to propagate along with the fiber. The incident ray I₂ enters the fiber at an angle greater than θa. Again, I₂ is refracted upon entering the fiber and is transmitted to the core-cladding interface. I₂ strikes the core-cladding interface at an angle less than the critical angle θc. I₂ is refracted into the cladding and is eventually lost. The light ray incident on the fiber core must be within the acceptance cone defined by the angle θa.

Briefly discuss with the aid of a suitable diagram what is meant by the acceptance angle for an optical fiber. Show how this is related to the fiber numerical aperture and the refractive indices for the fiber core and cladding.

θa is defined as the acceptance angle. The acceptance angle θa is the maximum angle to the axis of the fiber that light entering the fiber is propagated. The value of the angle of acceptance θa depends on fiber properties and transmission conditions. The acceptance angle is related to the refractive indices of the core, cladding, and medium surrounding the fiber. The numerical aperture (NA) is a measurement of the ability of an optical fiber to capture light. The NA is also used to define the acceptance cone of an optical fiber.

Fig. shows the relationship between the acceptance angle and the refractive indices. The index of refraction of the fiber core is n₁. The index of refraction of the fiber cladding is n₂. The index of refraction of the surrounding medium is n₀. By using Snell’s law and basic trigonometric relationships, the NA of the fiber is:

$NA = n_0 \times sinθ_a = (n_1^2-n_2^2)^{1/2}$

Optical fiber has a numerical aperture of 0.20 and a cladding refractive index of 1.59. Determine:

(a) the acceptance angle for the fiber in water which has a refractive index of 1.33;

(b) the critical angle at the core–cladding interface.

Comment on any assumptions made about the fiber.

a) We know the relation between acceptance angle and numerical aperture is given as:

$NA = n_0 \times sinα \\ sinα = \frac{NA}{n_0} \\ α = sin^{-1}(\frac{NA}{n_0}) \\ α = sin^{-1}(\frac{0.2}{1.33}) \\ α = sin^{-1}(0.15) \\ α = 8.65º$

So, the acceptance angle in water is 8.65º

b) To find the critical angle at the core-cladding interface we need to know the refractive index of the core.

$NA = \sqrt{n_1^2-n_2^2} \\ (NA)^2 = n_1^2-n_2^2 \\ (NA)^2 + n_2^2 = n_1^2\\ n_1 = \sqrt{(NA)^2 + n_2^2} \\ n_1 = \sqrt{(0.2)^2 + (1.59)^2} \\ n_1 = \sqrt{2.5681} \\ n_1 = 1.602$

So, the refractive index of core is 1.602

Let the critical angle at the core-cladding interface be θc

$sinθ_c = \frac{n_2}{n_1} \\ θ_c = sin^{-1}(\frac{n_2}{n_1}) \\ θ_c = sin^{-1}(\frac{1.59}{1.602}) \\ θ_c = sin^{-1}(0.9925) \\ θ_c = 82.978º$

So, the critical angle for the core-cladding interface = 82.978º