Question

• Estimate the transmitted power and reflected one If the light of 1mW hint an interface of: • 1) GaAs (n=3.5) • 2) SizN4 (n=2.1) • 3)Water (n=1.33)

Answer 1

7 When an . When the angle of incident is not mentioned in the problems it is supposed that the light incidents normally to the surface of interface. for normal incident, Coefficient of transmision of 4n2 power, Tp = [12, +42,32 To +22)² & - Coefficient of reflection of power, Rp = (22-02) for air to another medium, (021=1) ło = 42 p To - I and Rp = Where, n = n2. = M2 (since, ni Also, Incident power, P=.mw..

(3) for Gads 2= 3.5 So, .. ap - un T1+2² 4x3.5 - 0.691 (1+3.5)2 = 0.691 02-1 and 'Rp = (12-2) and . Rp = n +1 = 0.309 Therefore, 1. Transmitted power =.Tp P = 0.691 mW and Reflected power = Rp P = 0.309 mW . (since, P=1mW) (2) for Sig Ny , 2 = 2.1 To = h 21= 0.874 and Rp = (2+) = (2+0) 0.126

The refore, transmitted power, Pq = Top = 0.874x2 mW = 0.874 mW and reflected power PR = RpP = 0.126 x 1 mW = 0.126 mw (3) for n=2.33 , (Water) - = 0.489 2X1-33 (1.3371)2 Rp = (1.33-1) = 0.02 Therefore, transmitted power =0.489 x 1 mW = 0.489 mw and reflected power = 0.02. xlmW = 0.02 mW