Fundamental of Optical Engineering Lecture 9

 The amount of light reflected when a beam moves from one media to another can be reduced by placing a thin coating layer between them.

 A 12 A 23 > 0 and we want R min.  cos  = -1.

 n 1 = 1.5, n 3 = 1.7. What should be n 2 for antireflection film?

 Find the thinnest film to be coated to prevent the reflected light give n 1 = 1 and n 3 = 3.6 if λ=0.83μm.

 Consider the case of non normal incidence as shown in the previous figure.  The emerging beam travels with an optical- path difference between them as

 By Snell’s law, and, this yields  Then we have

 So that, an optical-path difference is  As EB = tcos  t, finally, we have

 Therefore, a round trip phase shift in this case equals to  Therefore,

 Consider a film of thickness t and refractive index 1.6 sandwiched between two media of refractive index 1.5. ◦ (a) determine all values of t for which the reflectance will be a maximum at normal incidence for λ = 1 μm and calculate the reflectance.

◦ (b) For an angle of incidence of 20 relative to the normal, calculate the wavelength at which the reflectance will maximum. Use the smallest value of t determined in (a).

◦ (c) Calculate the reflectance for both s- and p-polarization for the case considered in (b).

 These are instruments which utilize coherent summation of wave amplitudes.  Two beam interferometer:

 In general,  BSx =  +  BSz  Assume they are lossless beam splitters.  For 50:50 beam splitter.

 Suppose in a MZ interferometer for λ = μm, P Ax = 0 and P Az = P in. Then, a microscope slide 2 mm thick with a reflective index of 1.55 is placed in one arm of the interferometer. What are the new values of P ax and P az.

 For a Michelson interferometer in air with λ = 1.06 μm, P out = 0.5 P in. One of the mirrors is displaced by increasing L 1 continuously and Pout increases continuously to a final value of 0.65 P in. How large is the displacement?

 After one round trip  After 2 round trips

 After n round trips  Steady state (N   )

 Therefore,

 If  = 0 (lossless resonator), e -  = 1

Light from a laser of wavelength λ is transmitted through a lossless Fabry-Perot interferometer in air. The mirror reflectances are equal to R. As the mirror separation is increased from an initial value, the transmitted power increases to a maximum of 21 mW for a mirror separation D. As the mirror separation is further increased D+0.25 μ m, the transmitted power decreases to a minimum of 0.3 mW. (a) What is λ in μ m? (b) What is R? (c) What is the transmitted power when the mirror separation is D μ m?

 Sol n