n 1.33 sin θ = water sin θ = sin 35◦ , air n water 1.00 air   i.e.

◦ θair = 49.7 .

Thus, the height above the horizon is

◦ ◦ θ = 90 θair = 40.3 . (4.7) − Because the sun is far away from the fisherman and the diver, the fisherman will see the sun at the same angle above the horizon.

4.4 Total Internal Reflection

Suppose that a light ray moves from a medium of refractive index n1 to one in which n1 > n2, e.g. glass-to-air, where n1 = 1.50 and n2 = 1.0003. ◦ If the angle of incidence θ1 = 10 , then by Snell’s law, •

n 1.50 θ = sin−1 1 sin θ = sin−1 sin 10◦ 2 n 1 1.0003  2    = sin−1 (0.2604) = 15.1◦ .

◦ If the angle of incidence is θ1 = 50 , then by Snell’s law, •

n 1.50 θ = sin−1 1 sin θ = sin−1 sin 50◦ 2 n 1 1.0003  2    = sin−1 (1.1487) =??? .

◦ So when θ1 = 50 , we have a problem:

Mathematically: the angle θ2 cannot be computed since sin θ2 > 1. •

142 Physically: the ray is unable to refract through the boundary. Instead, • 100% of the light reflects from the boundary back into the prism.

This process is known as total internal reflection (TIR).

Figure 8672 shows several rays leaving a point source in a medium with re- fractive index n1.

Figure 86: The refraction and reflection of light rays with increasing angle of incidence.

The medium on the other side of the boundary has n2 < n1. • Crossing the boundary from a medium of higher to lower refractive index • causes the ray to bend away from the normal.

72Knight, Figure 23.22, page 725

143 Two things happen as the angle θ1 increases: •

– The angle of refraction approaches 90◦.

– The fraction of light energy that is transmitted decreases, while that of the reflected ray increases.

So in the above case where n1 > n2, there must exist some critical angle,

θc, at which the refracted ray vanishes and the reflected ray becomes 100% for all θ1 θc. ≥

◦ This critical angle corresponds to when θ2 = 90 ; by Snell’s law,

n n θ = sin−1 2 sin 90◦ = sin−1 2 . (4.8) c n n  1   1 

Note: the existence of this critical angle is only true when n1 > n2. If n1 < n2, there is no critical angle, and hence no TIR

So for the above case of a glass/air boundary, if n1 = 1.50 and n2 = 1.0003, then from (4.8),

1.0003 θ = sin−1 = 42◦ . c 1.50  

◦ The fact that θc < 45 for the glass/air boundary has important applications. For example, in binoculars (figure 8773):

the lenses are much further apart than your eyes, so the light rays need • to be brought together before exiting the eyepieces.

Instead of using mirrors, binoculars use a pair of prisms on each side – • the light undergoes two TIR and emerges from the eyepiece.

73Knight, Figure 23.23, page 725

144

Figure 87: The use of total internal reflection in binoculars.

4.4.1 Application of TIR: Fibre Optics

The transmission of light through optical fibres provides arguably the most important application of TIR.

Figure 8874 shows a laser beam entering one end of a narrow-diameter glass tube. Light ray from laser beam passes easily from air into the glass fibre. • The ray hits the boundary – i.e. inside wall of the tube – at an angle of • ◦ incidence θ1 which approaches 90 .

74Knight, Figure 23.25(a), page 726

145 Figure 88: The use of total internal reflection in an optical fibre.

Since θ1 > θc, the light ray undergoes TIR and remains inside the tube. • The light ray undergoes a series of total internal reflections as it moves • down the fibre.

◦ By the time the ray reaches the end of the fibre, θ1 0 < θc, and the • ≈ ray refracts out of the fibre, and is easily detected.

Practical optical fibres (figure 8975) consist of a glass core surrounded by a layer of cladding.

The cladding protects the glass surface from scratches, etc., that can • alter the angle of incidence.

The cladding has a refractive index nclad < ncore, so the light ray under- • goes TIR at the boundary, and remains confined to the tube.

Note: realistically, there is always some absorption of light by the glass tube – no glass is perfectly transparent.

75Knight, Figure 23.25(b), page 726

146