Question

Define numerical aperture and angle of acceptance. An optical fiber has a numerical
aperture of 0.20 and cladding refractive index of 1.59. Calculate the refractive index
of the core material and the acceptance angle of the fibre in water whose refractive
index is 1.33.

Answer 1

Answer on Question #60471, Physics / Optics

Define numerical aperture and angle of acceptance. An optical fiber has a numerical aperture of 0.20 and cladding refractive index of 1.59. Calculate the refractive index of the core material and the acceptance angle of the fibre in water whose refractive index is 1.33.

Solution:

Given:

$$\mathrm{NA} = 0.2$$ $$n_2 = 1.59$$

The Numerical Aperture (NA) is a measure of how much light can be collected by an optical system such as an optical fibre or a microscope lens.

The NA of any glass combination may be calculated as follows:

$$NA = \sqrt{n_1^2 - n_2^2}$$

where $n_1$ = the index of refraction of the core glass, and $n_2$ = the index of refraction of the cladding glass.

Thus,

$$n_1 = \sqrt{NA^2 + n_2^2} = \sqrt{0.2^2 + 1.59^2} = 1.602 \approx 1.60$$

The NA is related to the acceptance angle $\alpha$, which indicates the size of a cone of light that can be accepted by the fibre.

$$NA = n_0 \sin \alpha$$

where $n_0$ is refractive index of medium outside the fiber. For water $n_0 = 1.33$.

Thus,

$$\alpha = \sin^{-1} \left(\frac{NA}{n_0}\right) = \sin^{-1} \left(\frac{0.2}{1.33}\right) = 8.649{}^\circ \approx 8.65{}^\circ$$

Output: $n_1 = 1.60$; $\alpha = 8.65{}^\circ$.

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