ECE4853
OptoElectronics Problem Set #1
Vers 3.2
5-1. The capture angle S is the angle defined by the receiving
aperture, which in the case of optical fibers is determined in terms
of the refractive indices of the optical fiber core n1 and of the
cladding around the core n2. These refractive indices define the
numerical aperture (NA) of the fiber. The capture angle is twice
the acceptance angle 0.
(a) Determine the capture angle S and
(b) The numerical aperture (NA)
for an optical fiber with n1 = 1.975 and n2 = 1.972
Note that very slight differences in the refractive index are sufficient to capture radiation in a cylindrical
dielectric fiber for a reasonable capture angle, usually on the order of 10o.
Answers: S = 12.49o, NA = 0.109
5-2. The emission angle S is that angle for which the radiance drops to
0.5B0, where B0 is the radiance along the axis.
(a) Find S for a source which has non-Lambertian radiation pattern
B = B0cos100
(b) Find the fraction of radiation captured by a cylindrical fiber
having capture angle 16o from a laser source with S = 10o.
5-3. An alien spacecraft crash-lands on the main MSU quadrangle, and
after dispatching the insectoid aliens that are in the spacecraft its weapons systems are evaluated by the
fiber optics class in a desperate effort to understand the technology before the earth is invaded. It is
determined that the main firing system is constructed of triple arrays of DHDFB lasers of dissonant
wavelengths, each of which illuminated a spot 1.25mm in diameter at a distance of 2.5cm. Determine the
non-Lambertian index m if 50% of the radiant power falls within this spot.
Answer m = 2218
5-4. A cylindrical step fiber having NA = 0.150 is placed directly on
top of a source, as shown. Neglecting surface reflection determine the
fraction of the radiant power captured by the fiber if the source is
(a) Lambertian (m = 1) such as for an LED
(b) Non-lambertian with m = 200 such as for a laser.
5-5. The M31A cricket laser garden guard robot (GGR) has binocular DHDFB laser diodes made from
6H SiC and designed to give insects the lethal evil-eye in the blue-visible region of the spectrum. The
laser diodes also serve as a communication link to the garden base station. The bandgap for this form of
polymeric SiC is 2.9eV. Determine the wavelength ph of the radiation and the thermal bandwidth ,
assuming operating temperature of the laser diode to be 330oK.
Answers: ph = 0.4234m,  = 13.57nm
5.6 The active region of a ph = 1.56 m LED is 0.4m thick and D = 54 m
in diameter. The overall quantum efficiency is  = 0.2. The fiber core is 18
m in diameter and has NA = 0.150. Assuming Lambertian source what
current density is required to deliver 2.0 W to the fiber core.
5-7. A GaAs LED is used as the transmitter for a PCM network. A current of 8mA corresponds to logic
high and 0 mA corresponds to logic low. The quantum efficiency of the LED is 0.12. What optical
power results for the logic high state?
Answer: Pout = 1.398 mW
5-8. An LED for which I = 50mA generates an optical output at wavelength = 1.46m with total
quantum efficiency = .02. When reverse-biased the diode has reverse saturation current 2.5 pA.
Calculate the power efficiency of the diode, e.g. the optical power produced by the diode divided by the
electrical power input. Does it surprise you that the power efficiency is greater than the quantum
efficiency?
5-9. The external quantum efficiency of the Burrus LED is degraded by reflection losses at the dielectric
interfaces between (a) the LED and the external medium and (b) between the external medium and the
fiber. Assume transmittance T = 1 – R, where R = reflectance and that the total transmittance is the
product of the transmittances for the two interfaces. The indices of refraction are 3.42 for the LED
material and 1.42 for the laser core material, respectively. Determine the LED-to-fiber transmittance
factor when the medium between them is
(a) Air (refractive index n2 = 1.0)
(b) An index-matching fluid with optimal refractive index
n2 = 2.21
(c) An index-matching fluid with less-than optimal
refractive index n2 = 1.65
Answers: (a) T13 = 0.6791 (b) T13 = 0.9086 (c) T13 = 0.8731