Q2. Find the critical angle for total internal reflection oflight at interface b

Question

Q2. Find the critical angle for total internal reflection oflight at interface between

1) SiO2 (n=1.44) andSiO2:Ti (n=1.46).          

2) air-diamond (n=2.42)  

Q3. Consider a planar dielectricoptical waveguide made of SiO2 (n=1.44) andSiO2:Ti (n=1.46). The light isguided inside the core SiO2:Ti by total internalreflection. Find the amplitude of the evanescent wave in thecladding at distance 40 m from the interface core/cladding ifthe angle of incidence is 85 degrees. Hint: It’s enoughto find |E(40µm)/E t,0|2 in dB.

Explanation / Answer

Given index of refraction of SiO2 is n = 1.44

Index of refraction of SiO2:Ti is n ‘ = 1.46

From snell’s law sin I / sin r = n / n ‘

For critical angle I = C and r = 90 degrees

Plug the values we get   sinC / 1 = 1.44 / 1.46

                                               C = 80.5 degrees

(2). For air diamond surface :

n = 1

n’=2.42

So, from snell’s law   sin C / sin 90 = 1 /2.42

                                                    C = 24.4 degrees

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