1. Light entering a block of glass at an angle of incidence of 18.5° leaves the boundary between the air and the glass at an angle of 12.0°. What is the index of refraction of this type of glass?
2. Light is incident on a diamond at an angle of 10.0°. At what angle will it refract?
3. A beam of light is incident on a sheet of glass in a window at an angle of 30°. Describe exactly what path the light beam will take
(a) as it enters the glass.
(b) as it leaves the other side of the glass. Assume n = 1.500.
4. What optical device is designed specifically to separate the colours in “white light” using the fact that different wavelengths (colours) have different indices of refraction in glass? Which colour refracts most?
Using Snell’s Law to Calculate the Critical Angle
You can increase the angle of incidence of the light going from water into air until you can no longer observe refraction. The smallest angle at which you can observe no refraction and only reflection is called the critical angle (θic). At angles equal to or greater than the critical angle, the water-air boundary acts as a perfect mirror. At any angle equal to or greater than the critical angle, all light coming from the water will reflect back into the water. This is called total internal reflection. At incident angles less than the critical angle, refraction and weak reflection can both be observed.
In Figure 7.3.4, if a ray of light is coming from air into water, Snell’s law might be applied: incident ray reaches the water’s surface at the θic so it is reflected back.
For angles less than the critical angle, the path of light is reversible. For a ray of light coming from water out into air, what was the incident angle now becomes the angle of refraction and vice versa. For light going from water out into air:
sin θi sin θr = 1nw
Consider what happens as the incident angle approaches the critical angle, θic (Figure 7.3.4). The angle of refraction approaches 90°. Just as the critical angle is approached, we can write: sin θic
sin 90° = 1 nw
where nw is the index of refraction for light coming from air into water.
Solving for the critical angle, sin 90° = 1.0000 so:
sinθic = (1)(1.000)
nw = 1.000
1.33 = 0.750 and θic = 48.6°
The critical angle for water is therefore 48.6°.
The Rainbow — Dispersion by Raindrops
If you have ever looked at a rainbow carefully, you probably have noticed these conditions prevailed:
• The Sun was behind you.
• There were rain clouds in the sky in front of you.
• The rainbow forms an arc of a circle.
• In the primary bow (the brightest one), red is at the top and violet at the bottom.
• In the secondary bow, if it is visible, the colours are in reverse order.
Figure 7.3.5(a) illustrates what happens inside the raindrops. White light from the Sun refracts as it enters a drop. The white light is dispersed into component colours during the refraction. All colours of light experience internal reflection at the back of the drop. They are then refracted again and further dispersed as they leave the drop at the
“front.”
When you see a rainbow, you see the dispersive effects of millions of drops of water, not just one or two! You see red at the top of the primary bow with all the other rainbow colours below, ending in violet.
Quick Check
1. Calculate the critical angle for diamond, which has an index of refraction of 2.42.
2. What is the critical angle for a glass that has an index of refraction of 1.500?
3. A certain material has a critical angle of 52.0°. What is its index of refraction?
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Figure 7.3.5 (a) A primary rainbow: The red part of the rainbow is 42° from the “axis” of the bow; violet is 40° from the “axis.”
Sometimes a secondary, fainter rainbow may be seen above the primary rainbow.
A secondary rainbow has the colours in reverse order. Figure 7.3.5(b) shows the double internal reflection that happens inside droplets producing a secondary rainbow.
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Figure 7.3.5 (b) A secondary rainbow: The reversed colours of the secondary (higher) rainbow are due to double reflection of the Sun’s light within the droplets.
Wave Speed and Index of Refraction (Enrichment)
Proof That the Index of Refraction n = v1 v2
Consider a wavefront (labelled Aa in Figure 7.3.6) arriving at the boundary between two media. Medium 2 is one in which light slows down from speed v1 to speed v2. In the time t it takes for the wavefront at end A to reach the boundary B, the wavefront in medium 1 travels a distance AB, which equals v1t.
In the same time t, end a of the wavefront travels a distance ab inside medium 2.
The distance ab equals v2t.
In Figure 7.3.6, you will notice that sin θi = AB/aB and sin θr = ab/aB. Therefore the index of refraction, n, is equal to
n = sin θi
Therefore, the index of refraction is equal to the ratio of the speeds of light in the two media:
Figure 7.3.6 As a wavefront moves from one medium to the next, it changes speed.
When a beam of light of wavelength λ1 enters a medium in which its speed is reduced from v1 to v2, its wavelength is reduced to λ2 and the beam is refracted as in Figure 7.3.7. When the waves leave the “slow” medium, the speed is returned to the original v1 and the wavelength is again λ1. Refraction of a beam entering the second medium at an angle greater than 0° is caused by the change in speed of the waves.
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Figure 7.3.7 Wavelength changes as the waves move through the different media.
Recall that, for water, the index of refraction is 1.33. Snell’s law applied to light entering water from a vacuum (or air) says that sin θi
sin θr = 1.33.
The speed of light in air is 3.00 × 108 m/s.
The speed of light in water is 2.25 × 108 m/s.
The ratio of the speed of light in air to the speed of light in water is 3.00 × 108 m/s
2.25 × 108 m/s
Note that the ratio of the speeds is equal to the index of refraction for water. This relationship holds for other media as well. In general,
sin θi
sin θr = n = v1 v2
You will notice that the angle of incidence in Figure 7.3.7 is shown as the angle between the incident wavefront and the boundary between the two media. Also, the angle of refraction is shown as the angle between the refracted wavefront and the boundary between the media. When you do Investigation 7-3A, the angle of incidence will be taken as the angle between the incident ray and the normal. The angle of refraction will be defined as the angle between the refracted ray and the normal.
In Figure 7.3.8, lines have been drawn showing the paths taken by a single point on a wavefront as it moves into and out of the second medium. These lines, in effect, are identical with the incident ray and the refracted ray used in Investigation 7A. A little geometry will convince you that the incident angle (θi) in Figure 7.3.8 is equal to the incident angle i in Figure 7.3.7.
Likewise, the angle of refraction (θr) is the same in both diagrams.
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Figure 7.3.8 The lines show the paths taken by a single point on a wavefront.