The light from a DFB fiber laser with a linewidth below 1 kHz is coupled into a cavity as can be seen in the figure at the top of the next page. The laser has an output power of P_(0) = 5mW and a wavelength of 1050nm. The laser is exactly in resonance with a longitudinal mode of the cavity, and the laser beam is perfectly mode matched to the transversal mode of the cavity with lens L1. From the beginning, there is vacuum (no absorbing gas) inside the cavity. You can assume that there are no additional losses from the mirrors or other optics, and that the laser has been tuned to the resonance for a long time. (9 p)
a) What is the power P_(out) after the cavity?
b) What is the power circulating, P_(circ) inside the cavity?
c) The laser frequency is slowly increased while monitoring the output power. How much has the laser frequency increased when P_(out) has decreased to half its maximum value?
The laser frequency is tuned back to be in resonance with the cavity, and the output power returns to its initial value. Gas is introduced into the cavity. The transmitted intensity starts to decrease, due to absorption inside the cell. Gas is added until the transmitted power has decreased to 70% of its initial value, and at this point, the gas has an absorption coefficient α = 3*10^(-3) m^(-1). The fluorescence from a 10 mm long part within the cell is imaged onto a photodiode detector with lens L3 and L4, which both have a diameter of 50 mm. All of the light from this 10 mm long part that hits lens L3 is focused onto the detector. The fluorescence is emitted isotropically. Every absorbed laser photon produces one fluorescence photon. The fluorescence is detected on the photodiode, which has a quantum efficiency of 75%
d) How large is the photocurrent i_(out) of the photodiode?
The photodiode is connected to a transimpedance amplifier as in the figure at the bottom of the next page. The transimpedance of the amplifier is R_(F) = 10kΩ.
e) What is the voltage V_(OUT) out from the transimpedance amplifier? You can assume that the operational amplifier is ideal. In a separate experiment, the fiber laser is instantaneously turned off, while the exponential decrease of P_(out) is monitored. This experiment is repeated twice, once without gas (vacuum), and once with the same gas concentration as above. The output power decays as P_(out) = P_(start) e^(-(t)/(τ)). Two values of the lifetime τ are recorded, τ_(vacuum) and τ_(gas). P_(start) can have different values in the two experiments
f) Which of the three statements below do you expect to be true? No explanation is needed.
τ_(vacuum) > τ_(gas)
τ_(vacuum) = τ_(gas)
τ_(vacuum) < τ_(gas)