{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 23

# Chapter 23 - PHYS 132 WEEK#5 CHAPTER 23 23.7 Model Light...

This preview shows pages 1–4. Sign up to view the full content.

PHYS 132, WEEK #5, CHAPTER 23 23.7. Model: Light rays travel in straight lines and follow the law of reflection. Visualize: Solve: We are asked to obtain the distance h = x 1 + 5.0 cm. From the geometry of the diagram, tan q i = x 1 10 cm tan q r = x 2 15 cm x 1 + x 2 = 10 cm Because q r = q i , we have x 1 10 cm = x 2 15 cm = 10 cm - x 1 15 cm 15 cm ( ) x 1 = 100 cm 2 - 10 x 1 x 1 = 4.0 cm Thus, the ray strikes a distance 9.0 cm below the top edge of the mirror. 23.11. Model: Use the ray model of light and the law of reflection. Visualize: We only need one ray of light that leaves your toes and reflects in your eye. Solve: From the geometry of the diagram, the distance from your eye to the toe’s image is 2 d = (400 cm) 2 + (165 cm) 2 = 433 cm Assess: The light appears to come from your toes’ image. 23.12. Model: Use the ray model of light and Snell’s law. Visualize: Solve: According to Snell’s law for the air-water and water-glass boundaries, n air sin q air = n water sin q water n water sin q water = n glass sin q glass From these two equations, we have n air sin q air = n glass sin q glass sin q glass = n air n glass sin q air = 1.0 1.50 sin 60 ° q glass = sin - 1 sin 60 ° 1.5 = 35.3 °

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
23.16. Model: Use the ray model of light. For an angle of incidence greater than the critical angle, the ray of light undergoes total internal reflection. Visualize: Solve: The critical angle of incidence is given by Equation 23.9: q c = sin - 1 n cladding n core = sin - 1 1.48 1.60 = 67.7 ° Thus, the maximum angle a light ray can make with the wall of the core to remain inside the fiber is 90 ° - 67.7 ° = 22.3 ° . Assess: We can have total internal reflection because n core > n cladding . 23.22. Model: Use the ray model of light. Visualize: Solve: Using Snell’s law, n air sin 30 ° = n red sin q red q red = sin - 1 sin 30 ° 1.52 = 19.2 ° n air sin 30 ° = n violet sin q violet q violet = sin - 1 sin 30 ° 1.55 = 18.8 ° Thus the angular spread is D q = q red - q violet = 19.2 ° - 18.8 ° = 0.4 ° 23.26. Model: Use ray tracing to locate the image. Solve: The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. You can see from the diagram that the image is in the plane where the three special rays converge. The image is inverted and is located at s ¢ = 20.0 cm to the right of the converging lens.
23.28.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.