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# Dispersion Help

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## Dispersion

The index of refraction for a particular substance depends on the wavelength of the light passing through it. Glass, and virtually any other substance having a refractive index greater than 1, 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. 10-7).

Fig. 10-7. Light of different colors is refracted at different angles.

## Rainbow Spectra

The more a ray of light is slowed down by the glass, the more its path is deflected when it passes through a prism. This is why a prism casts a rainbow spectrum when white light passes through it. It is also responsible for the multi-colored glitter of jewelry, especially genuine diamonds, which have high indices of refraction and therefore spread out the colors to a considerable extent.

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. Second, dispersion degrades the quality of white-light images viewed through simple lenses. It is responsible for the “rainbow borders” often seen around objects when viewed through binoculars, telescopes, or microscopes with low-quality lenses.

## Dispersion Practice Problems

#### Practice 1

Suppose a ray of white light, shining horizontally, enters a prism whose cross-section is an equilateral triangle and whose base is horizontal (Fig. 10-8A). If the index of refraction of the prism glass is 1.52000 for red light and 1.53000 for blue light, what is the angle δ between rays of red and blue light as they emerge from the prism? Assume the index of refraction of the air is 1.00000 for light of all colors. Find the answer to the nearest hundredth of a degree.

#### Solution 1

There are several ways to approach this problem; all require several steps to complete. Let’s do it this way:

• Follow the ray of red light all the way through the prism and determine the angle at which it exits the glass
• Follow the ray of blue light in the same way
• Determine the difference in the two exit angles by subtracting one from the other

Refer to Fig. 10-8B. The ray of white light comes in horizontally, so the angle of incidence is 30° (consider this figure exact). The angle p 1 that the ray of red light subtends relative to N 1 , as it passes through the first surface into the glass, is found using the refraction formula:

Fig. 10-8 (A) Illustration for Practice 1.

Fig. 10-8 (B) Illustration for first part of Solution 1. (C) Illustration for second part of Solution 1.

Because the normal line N 1 to the first surface slants at 30° relative to the horizontal, the ray of red light inside the prism slants at 30° – 19.2049°, or 10.7951°, relative to the horizontal. The line normal to the second surface for the red ray (call it N 2r ) slants 30° with respect to the horizontal, but in the opposite direction from line N 1 (Fig. 10-8C). Thus, the angle of incidence p 2 , at which the ray of red light strikes the inside second surface of the prism, is equal to 30° + 10.7951°, or 40.7951°. Again we use the refraction formula, this time to find the angle p 3 , relative to the normal N 2r , at which the ray of red light exits the second surface of the prism:

Now we must repeat all this for the ray of blue light. Refer again to Fig. 10-8B. The ray of white light comes in horizontally, so the angle of incidence is 30°, as before. The angle β 1 that the ray of blue light subtends relative to N 1 , as it passes through the first surface into the glass, is:

Because the normal line N 1 to the first surface slants at 30° relative to the horizontal, the ray of blue light inside the prism slants at 30° - 19.0745°, or 10.9255°, relative to the horizontal. The line normal to the second surface for the blue ray (call it N 2b ) slants 30° with respect to the horizontal, but in the opposite direction from line N 1 (Fig. 10-8C). Thus, the angle of incidence β 2 , at which the ray of blue light strikes the inside second surface of the prism, is equal to 30° + 10.9255°, or 40.9255°. Again we use the refraction formula, this time to find the angle β 3 , relative to the normal N 2b , at which the ray of blue light exits the second surface of the prism:

The difference β 3p 3 is the angle δ we seek, the angle between the rays of blue and red light as they emerge from the glass. This, rounded off to the nearest hundredth of a degree, is:

#### Practice 2

Suppose you want to project a rainbow spectrum onto a screen, so that it measures exactly 10 centimeters (cm) from the red band to the blue band using the prism as configured in Problem 10-5. At what distance d from the screen should the prism be placed? Consider the position of the prism to be the intersection point of extensions of the red and blue rays emerging from the prism, as shown in Fig. 10-9. Consider d to be measured along the blue ray.

Fig. 10–9. Illustration for Practice 2.

#### Solution 2

This is a simple, straightforward right-triangle problem. The screen is placed so the blue ray is normal to it. We know that δ = 1.43° (accurate to three significant figures) from the solution to Problem 10-5. We are given that the length of the spectrum from red to blue, as it appears on the screen, is 10 cm, a figure that can be considered exact. We can solve for d as follows:

This should be rounded off to 401 cm, because we are given the value of δ to only three significant figures.

Practice problems for these concepts can be found at: Reflection and Refraction Practice Test

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