Snell's Law Help (page 2)
Light Rays At A Boundary
A qualitative example of refraction is shown in Fig. 19-2, where the refractive index of the first (lower) medium is higher than that of the second (upper) medium. A ray striking the boundary at a right angle (that is, angle of incidence equal to 0°) passes through without changing direction. However, a ray that hits at some other angle is bent; the greater the angle of incidence, the sharper is the turn the beam takes. When the angle of incidence reaches a certain critical angle , then the light ray is not refracted at the boundary but instead is reflected back into the first medium. This is known as total internal reflection.
Fig. 19-2 . Rays of light are bent more or less as they cross a boundary between media having different properties.
A ray originating in the second (upper) medium and striking the boundary at a grazing angle is bent downward. This causes distortion of landscape images when viewed from underwater. You have seen this effect if you are a scuba diver. The sky, trees, hills, buildings, people, and everything else can be seen within a circle of light that distorts the scene like a wide-angle lens.
If the refracting boundary is not flat, the principle shown by Fig. 19-2 still applies for each ray of light striking the boundary at any specific point. The refraction is considered with respect to a flat plane passing through the point and tangent to the boundary at that point. When many parallel rays of light strike a curved or irregular refractive boundary at many different points, each ray obeys the same principle individually.
When a ray of light encounters a boundary between two substances having different indices (or indexes) of refraction, the extent to which the ray is bent can be determined according to an equation called Snell’s law .
Look at Fig. 19-3. Suppose that B is a flat boundary between two media M r and M s , whose indices of refraction are r and s , respectively. Imagine a ray of light crossing the boundary as shown. The ray is bent at the boundary whenever the ray does not strike at a right angle, assuming that the indices of refraction r and s are different.
Fig. 19-3 . A ray passing from a medium with a relatively lower refractive index to a medium with a relatively higher refractive index.
Suppose that r < s ; that is, the light passes from a medium having a relatively lower refractive index to a medium having a relatively higher refractive index. Let N be a line passing through some point P on B such that N is normal to B at P . Suppose that R is a ray of light traveling through M r that strikes B at P . Let x be the angle that R subtends relative to N at P . Let S be the ray of light that emerges from P into M s . Let y be the angle that S subtends relative to N at P . Then line N , ray R , and ray S all lie in the same plane, and y < x . The two angles x and y are equal if, but only if, ray R strikes the boundary at an angle of incidence of 0°. The following equation holds for angles x and y in this situation:
sin y /sin x = r / s
This equation also can be expressed like this:
s sin y = r sin x
Now look at Fig. 19-4. Again, let B be a flat boundary between two media M r and M s whose absolute indices of refraction are r and s , respectively. In this case, imagine that r > s ; that is, the ray passes from a medium having a relatively higher refractive index to a medium having a relatively lower refractive index. Let N, B, P, R, S, x , and y be defined as in the preceding example. As before, x = y if, but only if, ray R strikes B at an angle of incidence of 0°. Then line N , ray R , and ray S all lie in the same plane, and x < y . Snell’s law holds in this case, just as in the situation described previously:
sin y /sin x = r/s and s sin y = r sin x
Determining The Critical Angle
Refer again to Fig. 19-4. The light passes from a medium having a relatively higher index of refraction r into a medium having a relatively lower index s . Therefore, r > s . As angle x increases, angle y approaches 90°, and ray S gets closer to the boundary plane B . When x , the angle of incidence, gets large enough (somewhere between 0° and 90°), angle y reaches 90°, and ray S lies exactly in plane B . If angle x increases even more, ray R undergoes total internal reflection at the boundary plane B . Then the boundary acts like a mirror.
The critical angle is the largest angle of incidence that ray R can subtend relative to the normal N without being reflected internally. Let’s call this angle x c . The measure of the critical angle is the arcsine of the ratio of the indices of refraction:
x c = sin −1 ( s/r )
Determining The Critical Angle Practice Problems
Suppose that a laser is placed beneath the surface of a freshwater pond. The refractive index of fresh water is approximately 1.33, whereas that of air is 1.00. Imagine that the surface is perfectly smooth. If the laser is directed upward so that it strikes the surface at an angle of 30.0° relative to the normal (perpendicular), at what angle, also relative to the normal, will the beam emerge from the surface into the air?
Envision the situation in Fig. 19-4 “upside down.” Then M r is the water and M s is the air. The indices of refraction are r = 1.33 and s = 1.00. The measure of angle x is 30.0°. The unknown is the measure of angle y . Use the equation for Snell’s law, plug in the numbers, and solve for y . You’ll need a calculator. Here’s how it goes:
sin y/sin x = r/s
sin y /(sin 30.0°) = 1.33/1.00
sin y /0.500 = 1.33
sin y = 1.33 × 0.500 = 0.665
y = sin −1 0.665 = 41.7°
What is the critical angle for light rays shining upward from beneath a freshwater pond?
Use the formula for critical angle, and envision the scenario of Problem 19-3, where the laser angle of incidence x can be varied. Plug in the numbers to the equation for critical angle x c :
x c = sin −1 ( s/r )
x c = sin −1 (1.00/1.33)
x c = sin −1 0.752
x c = 48.8°
Remember that the angles in all these situations are defined with respect to the normal to the surface, not with respect to the plane of the surface.
The index of refraction for a particular substance depends on the wavelength of the light passing through it. Glass slows down light the most at the shortest wavelengths (blue and violet) and the least at the longest wavelengths (red and orange). This variation of the refractive index with wavelength is known as dispersion . It is the principle by which a prism works (Fig. 19-5). The more the light is slowed down by the glass, the more its path is deflected when it passes through the prism. This is why prisms cast rainbows when white light is shone through them.
Fig. 19-5 . Dispersion is responsible for the fact that a glass prism “splits” white light into its constituent colors.
Dispersion is important in optics for two reasons. First, a prism can be used to make a spectrometer , which is a device for examining the intensity of visible light at specific wavelengths. (Fine gratings are also used for this.) Second, dispersion degrades the quality of white-light images viewed through simple lenses.
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