Question

Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to plot the peak closest to 632.8 nm for 4 values of R = 0.99, 0.90, 0.75 and 0.6. For each case find the spectral width δλm, the finesse F and Q using the equations given in the book or in class. How accurate are our equations in predicting δλm? (You may want to use a graphing software for this problem.)

Answer 1

Diode Lasers

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250

microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is

4. The optical gain bandwidth (as measured between half intensity points) will

normally depend on the pumping current (diode current) but for this problem

assume that it is 2 nm.

(a) What is the mode integer m of the peak radiation?

(b) What is the separation between the modes of the cavity? Please express your

answer as Δλ.

(c) How many modes are within the gain band of the laser?

(d) What is the reflection coefficient and reflectance at the ends of the optical

cavity (faces of the InGaAsP crystal)?

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the

beam is a Gaussian beam with the waist located at the output. And the beam

waist size is approximately the x-y dimensions of the cavity.)

Solution:

(a) The wavelength λ of a cavity mode and length L are related by

n

mL

2

λ = , where m is the mode number, and n is the refractive index.

So the mode integer of the peak radiation is

1290

1055.1

10250422

6

6

= ×

××× == −

−

λ

nL

m .

(b) The mode spacing is given by nL

c f 2

=Δ . As

λ

c f = , λ

λ

Δ−=Δ 2

c f .

Therefore, we have nm

nL f

c

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .

(c) Since the optical gain bandwidth is 2nm and the mode spacing is 1.2nm, the

bandwidth could fit in two possible modes.

For mode integer of 1290, nm

m

nL 39.1550

1290

10250422 6

= ××× ==

−

λ

Take m = 1291, nm

m

nL 18.1549

1291

10250422 6

= ××× ==

−

λ

Or take m = 1289, nm

m

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Step-by-step explanation: