1. In an optical fiber the refractive indexes of core and cladding are 1.46 and 1.45 respectively. Find:
Refractive index of the core (\( n_1 \)) = 1.46
Refractive index of the cladding (\( n_2 \)) = 1.45
1) The critical angle is given by:
\[ \sin \theta_c = \frac{n_2}{n_1} \] \[ \sin \theta_c = \frac{1.45}{1.46} \] \[ \theta_c = \sin^{-1} \left( \frac{1.45}{1.46} \right) \] \[ \theta_c = 83.1^\circ \]2) The numerical aperture is given by:
\[ N.A. = \frac{\sqrt{n_1^2 - n_2^2}}{n_0} \] Assuming \( n_0 = 1 \) (air), \[ N.A. = \sqrt{1.46^2 - 1.45^2} \] \[ N.A. = 0.171 \]3) The angle of acceptance is related to the numerical aperture by:
\[ \alpha_{max} = \sin^{-1} (N.A.) \] \[ \alpha_{max} = \sin^{-1} (0.171) \] \[ \alpha_{max} = 9.8^\circ \]2. An optical fiber has a core material of refractive index of 1.55 and cladding material refractive index of 1.50. The light is launched into it in air. Calculate the numerical aperture?
Refractive index of the core (\( n_1 \)) = 1.55
Refractive index of the cladding (\( n_2 \)) = 1.50
The numerical aperture is given by:
\[ N.A. = \frac{\sqrt{n_1^2 - n_2^2}}{n_0} \] Assuming \( n_0 = 1 \) (air), \[ N.A. = \frac{\sqrt{1.55^2 - 1.50^2}}{1} \] \[ N.A. = 0.39 \]The angle of acceptance is related to the numerical aperture by:
\[ \alpha_{max} = \sin^{-1} (N.A.) \] \[ \alpha_{max} = \sin^{-1} (0.391) \] \[ \alpha_{max} = 23^\circ \]3. Calculate the fractional index change for a given optical fiber if the refractive indices of the core and cladding are 1.563 and 1.498, respectively
The fractional index change (\(\Delta\)) for an optical fiber is given by:
\[ \Delta = \frac{n_1 - n_2}{n_1} \]Refractive index of the core (\( n_1 \)) = 1.563
Refractive index of the cladding (\( n_2 \)) = 1.498
\[ \Delta = \frac{1.563 - 1.498}{1.563} \] \[ \Delta = 0.0416 \]