IOM 2020, Експеримент — Задача 15

• Q.15 a) $P_{3\omega} \sim P_{\omega}^{3}$
• Q.15 b) $P_{3\omega} \sim P_{2\omega}^{3/2}$
• Q.15 c) B = 2.38, 2 < B < 3
• Q.15 d) B = 1.2, 1 < B < 1.5
• Q.15 e) The filter transmits radiation of the fundamental and second harmonics.

Answer 15 a) $P_{3\omega} \sim P_{\omega}^{3}$ Cubic dependence

Answer 15 b) $P_{3\omega} \sim P_{2\omega}^{3/2}$ Degree «one and a half»

Answer 15 c)

for $I_{3\omega} = A \cdot I_{\omega}^{B}$ $A = 10^{-10}\ldots 10^{-9}$ $B = 2.38..$ $2 < B < 3$

The main thing is that B is less than 3, but more than 2

Answer 15 d)

The main thing is that B is less than 1.5, but more than 1

for $I_{3\omega} = A \cdot I_{2\omega}^{B}$ $A = 10^{-2}\ldots10^{-1}$ $B = 1.2..$ $1 < B < 1.5$

Построить зависимости в двойном логарифмическом масштабе. Провести линию (т.е. выявить наклон) и вычислить степень. Учитывая неточность построений. Величины коэффициентов A не так важны. Важен порядок величины. См. дополнительные листы.

Answer 15 e) The video showed that when we remove the crystal for 3w, then a signal remains on the photodetector 3. When we close 3 FP, then the signal is less, when we open, then more. This means that either w or 2w radiation passes through the filter, which should be absorbed.

The filter transmits radiation of the fundamental and second harmonics.

Additional forms 3

Third harmonic versus first. B = 2.38

Third harmonic versus second. B = 1.2