Answer:
Angle of incidence that entered material b= 63.1°
Angle of incidence between a and b = 55.9°
Explanation: Using the formular:
n1sintheta1= n2sintheta2
The light ray which enters material B will be
1.4Sin72.8° = 1.5Sin theta
1.3373= 1.5Sintheta
sintheta = 1.3373/1.5
Sin^-1 0.8916 = Theta
63.1 = theta
When the ray hits interface with material a
1.5Sin63.1 = 1.3 Sin theta
1.3374 = 1.3Sin theta
Sintheta= 1.3374/1.3
Sin theta = 1.0877
There will be total reflection off the boundary b c because sin theta exceeded 1 in value.
The equation should be
1.4sin63.1 = 1.4 sin theta
Sin theta=72.8°
When the ray hits air-c boundary:
1.4sin72.8=1.00sin theta
Sin theta=1.3374/1 =1.3374
There is total reflection.
In material a,the ray will:
1.3sin72.8° = 1.00sin theta
There will be total reflection when the ray hits a-b boundary.
1.3sin72.8= 1.5sintheta
Sin theta= 1.2419/ 1.5
Sin theta =0.8279
Theta= Sin^-10.8279= 55.88°
When ray hits c-air boundary
1.4sin63.1= 1.00sintheta
1.2485= sin theta = Toal reflection.
Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.
