A light source generating an optical power output equal to 1 µw is coupled into

an optical fiber with a cross sectional area larger than the active area of the

light source. Determine the power coupled into the fiber. $θ_0$ Equal to 15°.

Mumbai University > Electronics Engineering > Sem7 > Optical Fiber Communication

Marks: 5M

Year: May 2012

Given:

$P_{in}$=1µW

$θ_0$=15°

To find: Power coupled into the fiber $P_{in}$

Solution: sin⁡θ=NA

sin⁡15° = 0.26 = NA

Coupling efficiency $η= NA^2 =(0.26)^2=0.0676$

$$η= \frac{P_F}{P_{in}} $$

Power coupled into the fiber: $P_F= η× P_{in}=0.0676×1μW$

∴ $P_F$=67.6 nW