15.3 Solutions
729
∴ sin r 1 =
n 1
n 2
sin 20
◦
=
1.1
1.6
× 0.342 = 0.2351
∴ r 1 = 13.6
◦
From the geometry of Fig. 15.16
r 2 = 40 − 13.6
◦
= 26.4
◦
(∵ r 1 + r 2 = A, the apex angle)
n 1 sin i 2 = n 2 sin r 2
sin i 2 =
n 2
n 1
sin r 2 =
1.6
1.1
sin 26.4
◦
= 0.6467
∴ i 2 = 40.3
◦
∴ φ = 90
◦
− 40.3
◦
= 49.7
◦
θ = 80 − 49.7
◦
= 30.3
◦
(∵ The exterior angle is equal to the sum of the interior angles).
Fig. 15.16 Triangular glass
prism
15.7 (a) sin C =
1
n g
=
1
1.45
= 0.6896
∴ C = 43.6
◦
(b) In applying Snell’s law instead of measuring angles with respect to the
normal, we will measure them with respect to the interface, for convenience. Then
1.2 cos 45
◦
= 1.0 cos θ
whence θ = 31.9 ◦ .
From Fig. 15.17 AC is parallel to EF and so B ˆ
AC = B ˆ
CA = 45 ◦ . Thus
in the triangle, ADC,
α = 180
◦
− 2(θ + 45
◦
) = 26.2
◦