EE140 Mid-Term 2 Dr. Ray Kwok P = ? Student Name: (Last) ____________________ (First) ______________________________ 1. [35pts] Glass isosceles triangular prisms shown in Figure are used in optical instruments. Assuming εr = 4 for glass, calculate the normally incident light power reflected back (in dB) by the prism (as shown); if the light is unpolarized. 312121242121−=+−=+−=Γ==nnnnnglassoccoaircglassnn3021sin90sinsin===θθθAt the first water-glass interface Inside of the prism: total reflection for both polarizations 3112122=+−=ΓAt the last glass-water interface ()()dBP 02.1%7979.09111122221−===−=Γ−Γ−=Total power coming back at the bottom is: Note: The polarization here doesn’t do anything since both polarizations have the same reflection / transmission properties.EE140 Mid-Term 2 Dr. Ray Kwok incidence transmission reflection θθθθi θθθθr θt n1 n2 Hr ki Er Ht kt Et Hi kr Ei z y O 2. [35 pts] A plane wave in air with mA/m, is incident upon the planar surface of a dielectric material with n = 1.5, occupying the half-space z > 0. Write the full expression for the reflected E field and the transmitted H field. )43sin(2ˆzytxH −+=ωrotttiiinnckkyzk6.23sin)5.1(53)1(sinsin43tan105.1543ˆ3ˆ42191221====⋅===+=−=θθθθθωr( ) ( )[ ]( ) ( ) ( )[ ])43105.1sin()08.0ˆ06.0ˆ()5/4()5/3(5sin002.0)(377)(134.0))5/4(ˆ5/3ˆ(cossinsin)cosˆsinˆ(134.0)5/4)(5.1(6.23cos)1()5/4)(5.1(6.23cos)1(coscoscoscos91//2121//zytyzEzytyzEzyktHyzEnnnnrrrroiorrrooitit++⋅−=−−−−−=−−−Γ−=−=+−=+−=ΓrrrωθθωηθθθθθθV/m [ ]( )[ ]( )zytxHzytxHzytHxHnkknnntootttoiotooiti87.63105.1sin27.2ˆ)6.23cos6.23sin(5.7sin)2)(5.1(76.0ˆ)cossin(5.7sinˆ5.7)5.1(576.0)5/4)(5.1(6.23cos)1()5/4)(1(2coscoscos292//22211//−+⋅=+−−=+−−=====+=+=rrrωθθωηητθθθτmA/mEE140 Mid-Term 2 Dr. Ray Kwok y z 3. [30 pts] It is extremely difficult to recover an underwater Black Box by air. Assuming the Box is emitting a 100W, RHCP, 10-kHz signal from the bottom of the ocean (8 km deep), estimate the power intensity reaching a helicopter 100 m above the surface. Use εr = 81, µr = 1, σ = 4 S/m for seawater. The following steps would guide you through your estimation: (a) What is the max power intensity reaching the surface (from below) if water is lossless (assume isotropic radiation). (b) With the attenuation of sea water, what is the power intensity right underneath the surface. (c) What is the critical angle? (Remember this is a lossy medium). (d) Estimate the total power reaching the surface within this critical “cone”. The cone is defined by rotating the critical angle about the normal. (e) What is the fractional power being reflected at normal incident? (f) Assume the reflection coefficient is uniform (does not depends on incident angle), how much (total) power is being transmitted. (g) What is the max power intensity reaching a helicopter flying by at 100 m above water? (a) S = (100W)/(4π·80002)=1.24 x 10-7 W/m2 (b) ( )( )( )( )( )( )( )( )22787278072780)8000)(40.0(2274494101.4103.31024.127800103.340.0241041022109.836/10811024tanmWSdBeer−−−−−−−−⋅=⋅⋅=−=⋅===⋅==⋅===αππωµσαππωεσδGood conductor ( )radianskkckkoCoC44221148421103.503.090sin101.2sin4.0sinsin101.210310240.0−−−⋅==⋅==⋅=⋅=====θθθθπωαβ(d) For this small angle, one can estimate the power within the critical angle is: (c) (f) 0.11% transmitted = 2.5·10-2788 W (e) ( ) ( ) ( )( )( )%89.999989.003.09995.01.013771.0137745141.01.0144.011212122==Γ−∠=+++−=+−=Γ∠=+=+=+=oojjjjjηηηησαη()WSRSPC2785242103.2)103.58000(−−⋅=⋅⋅=⋅=θπ(g) Power intensity = (2.5·10-2788 W)/(4π·1002) = 2·10-2793
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