3.3.2.3
Worked example: calculating the angle of incidence at an optical fibre entry
Refraction & Total Internal Reflection — AQA A-Level Physics
Worked Example
Light enters the end of a step-index fibre and refracts along the core-cladding boundary. Speed of light in core = \(2.027 \times 10^{8}\) m/s. Speed of light in cladding = \(2.055 \times 10^{8}\) m/s. Calculate the angle of incidence at the fibre entry.
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1
Calculate refractive indices
$$n_{\text{core}} = \frac{c}{c_{\text{core}}} = \frac{3.00 \times 10^{8}}{2.027 \times 10^{8}} = 1.48$$
$$n_{\text{clad}} = \frac{c}{c_{\text{clad}}} = \frac{3.00 \times 10^{8}}{2.055 \times 10^{8}} = 1.46$$2
Find the critical angleThe angle of incidence at which the refracted ray travels along the boundary (angle of refractionThe change in direction of a wave as it passes from one medium to another, caused by a change in wave speed. = 90 degrees). For angles greater than this, total internal reflection occurs. at core-cladding boundary
$$\sin \theta_c = \frac{n_{\text{clad}}}{n_{\text{core}}} = \frac{1.46}{1.48}$$
$$\theta_c = \sin^{-1}(0.9865) = 80.6^{\circ}$$3
Find the angle of refractionThe change in direction of a wave as it passes from one medium to another, caused by a change in wave speed. at the fibre entry
At the entry face, the refracted ray hits the core-cladding boundary at the critical angleThe angle of incidence at which the refracted ray travels along the boundary (angle of refractionThe change in direction of a wave as it passes from one medium to another, caused by a change in wave speed. = 90 degrees). For angles greater than this, total internal reflection occurs.. From geometry:
$$\theta_r = 90^{\circ} - 80.6^{\circ} = 9.4^{\circ}$$4
Apply Snell's law at the entry face
Air to core:
$$n_{\text{air}} \sin \theta = n_{\text{core}} \sin \theta_r$$ $$\sin \theta = 1.48 \times \sin 9.4^{\circ} = 1.48 \times 0.1633 = 0.2417$$ $$\theta = \sin^{-1}(0.2417) = 14.0^{\circ}$$Answer
$\theta = 14.0^{\circ}$
Related:Diffraction