Refractive index

The light passes through the interface between air and glass with a refractive index of 1.5. Find:
(a) the angle of refraction if light strikes the interface from the air at an angle of 40°.
(b) the angle of refraction when light strikes the glass interface at an angle of 40°.
(c) the angle of incidence if the light is refracted at an angle of 20° on impact from the glass.
(d) cut-off angle (justify for which way the light passes through the interface)

Correct result:

A =  25.374 °
B =  74.6186 °
C =  30.8659 °
D =  41.8103 °

Solution:

n=1.5 α=40   sinα:sinA=n  A1=arcsin(sinα/n)=arcsin(sin40 /n)=arcsin(sin40 /1.5)=arcsin(0.642788/1.5)=0.44286 A=A1  =A1 180π   =0.44285973862 180π   =25.374  =25.374=252226"
 β=40   sinB:sinβ=n  B1=arcsin(n sinβ)=arcsin(n sin40 )=arcsin(1.5 sin40 )=arcsin(1.5 0.642788)=1.30234 B=B1  =B1 180π   =1.3023397001 180π   =74.619  =74.6186=74377"
 γ=20   sinC:sinγ=n  C1=arcsin(n sinγ)=arcsin(n sin20 )=arcsin(1.5 sin20 )=arcsin(1.5 0.34202)=0.53871 C=C1  =C1 180π   =0.538711275689 180π   =30.866  =30.8659=305157"
δ=90   sinδ:sinD=n  D1=arcsin(sinδ/n)=arcsin(sin90 /n)=arcsin(sin90 /1.5)=arcsin(1/1.5)=0.72973 D=D1  =D1 180π   =0.729727656227 180π   =41.81  =41.8103=414837"



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