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# Snell's Law Help (page 2)

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McGraw-Hill Professional
Updated on Sep 12, 2011

## 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 )

#### Problem 1

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?

#### Solution 1

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°

#### Problem 2

What is the critical angle for light rays shining upward from beneath a freshwater pond?

#### Solution 2

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.

## Dispersion

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.

Practice problems of these concepts can be found at: Optics Practice Test

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