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°.

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 $η= NA^2 =(0.26)^2=0.0676$

$${η= P_F}{P_{in}}$$

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

$$∴ P_F=67.6nW$$