HW 5b solution

HW 5b solution - Problem 1 1 . The phase 4R 2 1+ sin (1 R)...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 1 (1) Since there is no absorption, T=1-R and 2 sin ) 1 ( 4 1 1 2 2 0 δ R R I I t - + = . The phase difference is λ θ π t t n cos 4 2 = and the coefficient of finesse is 360 ) 1 ( 4 2 = - = R R F and 2 n =1. So t t t I I cos 2 sin 360 1 1 2 0 × + = 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 incident angle, deg Normalized transmittance (2) Maxima: t m m 2 cos 2 = = , for =632.8nm and o o 5 . 1 5 . 0 < < the o 07 . 1 max = and m=3160. Minima: t m m 2 ) 2 1 ( cos 2 ± = ± = , the o 48 . 1 min = , m=3160 or m=3159 (3) o A t FSR 0 . 2 2 2 = = Δ , pm t R R 7 . 6 2 1 2 min = - = Δ (Note: the angle interval in the question is too small, if you plot the transmission in this
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online