Question

6-10 A laser can be made by placing a plasma tube in an optical resornant cavity formed by two highly reflecting flat mirrors, which act like rigid walls for light waves (see figure). The purpose of the plasma tube is to produce light by exciting normal modes of the cavity. Plasma tube ㄈㄧ Vo frequency (a) What are the normal mode frequencies of the resonant cavity? (Express your answer in terms of the distance L between the mirrors and the speed of light c.) (b) Suppose that the plasma tube emits light centered at fre- quency vo = 5 × 1014 Hz with a spectral width ay, as shown in the sketch. The value of△v is such that all normal modes of the cavity whose frequency is within ±1.0 × 109 Hz of vo will be excited by the plasma tube. (1) How many modes will be excited if L = 1.5 m? (2) What is the largest value of L such that only one normal mode will be excited (so that the laser will have only one output frequency)? (c = 3 × 108 m/sec.)

This is the answer from the book. Please drive to this answer.

Answer 1

V_n = c/ lambda n = cn/2L (lowest mode has lambda = 2L) The mode number at V = V_o is n_o = 2L V_o/c = 2 times 1.5 times 5 times 10^14/3 times 10^8 = 5 times 10^6 and the frequency separation between mode is c/2L = 10^8 H_2 1 So there is a total of 21 modes with in plusminus 1 times 10^9 H_2 of V_o (2) A & L alpha 1/V_o L is largest. So V_o is smaller (V_o = 1 times 10^9 H 2 c/2L = 1 times 10^9 H2 L = 3 times 10^8/2 times 10^9 = 0.15m = 15 cm