Question

4) Look again at Equation 4.79. Assume light passes through a gas that has no free electrons and for low frequency light has only 1 important resonance frequency. Only one term in Eq. 4.79 will then be important (not the free electron term). Using this model, consider the index of refract of an ideal gas, where each molecule has a single resonance state. Assume the only resonance frequency is hw-leV.For gas at standard temperature and pressure and for wave length of light of 600 nm, the index of refraction is found to be 1.000300. a. Determine the wave length of light corresponding the ho leV in nanometers. b. What will the index be for the same conditions if the wave length of light is 800 nm? What will the index be for 600 nm light if the gas pressure is doubled with temperature held constant? c.

Answer 1

h w = lev x = ? h w = lev = h 2 pi v = lev = h/2 pi times 2 pi v = lev E_ph = h v = lev alpha = hc/E_ph = 6.63 times 10^-34 times 3 times 10^8/1 times 1.6 times 10^-19 = 1243 nm = 1.24 mm n = alpha v/alpha m = CK/W K = 2 pi/alpha = wave no W = 2 pi v n = c 2 pi v/alpha 2 pi v alpha = 800 mm \r\n h w = lev w = 2 pi v = l ev/6.58 times 10^-16 = 1.52 times 10^15 n = 2 times 3.14 times 3 times 10^8/800 times 10^-9 times 1.52 times 10^15 = 1.549 alpha = 600 nm n = c k/w = c 2 pi/alpha = 3 times 10^8 times 2 times pi/w times 600 nm = 2.065