Dispersion Help

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 ₃ – p 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:

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.