7. Estimate the critical radius of curvature at which large bending losses occur in the following Step index fiber that exhibit the following parameters: a. A multimode fiber with a core refractive index of 1.500 , a relative refractive index difference of \( 3 \% \) and an operating wavelength of \( 0.82 \mu \mathrm{~m} \); b. An \( 8 \mu \mathrm{~m} \) core diameter single-mode fiber with a core refractive index the same as (a), a relative refractive index difference of \( 0.3 \% \) and an operating wavelength of \( 1.55 \mu \mathrm{~m} \).

To estimate the critical radius of curvature at which large bending losses occur in step-index fibers, we can use the following formula for the critical radius of curvature \( R_c \): \[ R_c = \frac{2 \cdot n_1^2 \cdot \Delta}{\lambda} \] where: - \( n_1 \) is the core refractive index, - \( \Delta \) is the relative refractive index difference (expressed as a decimal), - \( \lambda \) is the operating wavelength. The relative refractive index difference \( \Delta \) can be calculated as: \[ \Delta = \frac{n_1 - n_2}{n_1} \] where \( n_2 \) is the cladding refractive index. However, since we are given \( \Delta \) directly, we can use it in the formula. ### Part (a): Multimode Fiber Given: - Core refractive index \( n_1 = 1.500 \) - Relative refractive index difference \( \Delta = 3\% = 0.03 \) - Operating wavelength \( \lambda = 0.82 \, \mu m = 0.82 \times 10^{-6} \, m \) Now, substituting the values into the formula: \[ R_c = \frac{2 \cdot (1.500)^2 \cdot 0.03}{0.82 \times 10^{-6}} \] Calculating \( n_1^2 \): \[ (1.500)^2 = 2.25 \] Now substituting this value: \[ R_c = \frac{2 \cdot 2.25 \cdot 0.03}{0.82 \times 10^{-6}} = \frac{0.135}{0.82 \times 10^{-6}} \approx 1.64634 \times 10^{5} \, m \] Thus, the critical radius of curvature for the multimode fiber is approximately: \[ R_c \approx 164634 \, m \quad \text{(or about 164.6 km)} \] ### Part (b): Single-Mode Fiber Given: - Core refractive index \( n_1 = 1.500 \) - Relative refractive index difference \( \Delta = 0.3\% = 0.003 \) - Operating wavelength \( \lambda = 1.55 \, \mu m = 1.55 \times 10^{-6} \, m \) Now, substituting the values into the formula: \[ R_c = \frac{2 \cdot (1.500)^2 \cdot 0.003}{1.55 \times 10^{-6}} \] Calculating \( n_1^2 \) again: \[ (1.500)^2 = 2.25 \] Now substituting this value: \[ R_c = \frac{2 \cdot 2.25 \cdot 0.003}{1.55 \times 10^{-6}} = \frac{0.0135}{1.55 \times 10^{-6}} \approx 8.70968 \times 10^{3} \, m \] Thus, the critical radius of curvature for the single-mode fiber is approximately: \[ R_c \approx 8709.68 \, m \quad \text{(or about 8.71 km)} \] ### Summary of Results - **Multimode Fiber Critical Radius of Curvature**: \( \approx 164634 \, m \) (or 164.6 km) - **Single-Mode Fiber Critical Radius of Curvature**: \( \approx 8709.68 \, m \) (or 8.71 km)