From the given figure, the interface is horizontal and the normal is along the y-axis.
The incident ray is along the vector AO=2i^−3j^.
The angle of incidence α is the angle between the incident ray and the normal.
sinα=22+(−3)22=132
The refracted ray is along the vector OB=Ci^−4j^.
The angle of refraction β is the angle between the refracted ray and the normal.
sinβ=C2+(−4)2C=C2+16C
Applying Snell's law at the interface:
μ1sinα=μ2sinβ
1×132=1.5×C2+16C
132=23C2+16C
Squaring both sides:
134=49(C2+16C2)
16(C2+16)=117C2
16C2+256=117C2
101C2=256
C2=101256
C=10116≈1.592
Approximating 101≈10, we get C≈1.6.
Answer: 1.6