Question

5. The ends of a 10 cm laser rod, index of refraction = 1.76, has its ends coated for reflectivity of 99.9 % and 90 %. If there is no excited state absorption (  calculate a) the fractional loss per pass and b) the decay time of the cavity

Answer 1

Answer:

the fractional loss per pass =0.053

the decay time of the cavity =11 ns

Explanation:

laser is an acronym for Light Amplification by Stimulated Emission of Radiation and it has lot of military applications : basically ranging from target acquisition, fire control, and training. It is also being used  in LIDAR communications, radar systems , laser guided landing systems, laser pointers, holography, photography, guidance systems, scanning machines, metal working,  and medicine.

the fractional loss per pass= $\alpha l - ln(R_1R_2)^ \frac {1}{2}$

but no excited state absorption, therefore $\alpha$ is 0

$= - ln(R_1R_2)^ \frac {1}{2}\\\\=- ln(0.999* 0.9)^ \frac {1}{2}$

=0.053

b) the decay time of the cavity:

$T_c = \frac{nd}{cl_f}=\frac{1,76 * 10 *10^-2}{3 *10^8 *0.53}$

=1.1 *10^-8 seconds

=11 ns