Visualize solve a from the graph in figure 2329 we

This preview shows pages 23–32. Sign up to view the full content.

Visualize: Solve: (a) From the graph in Figure 23.29, we estimate the index of refraction for the red light (656 nm) to be n red = 1.572 and for the blue light (456 nm) to be n blue = 1.587. (b) The angle of incidence onto the rear of the prism is 35 . ° Using these values for the refractive index and Snell’s law, red air red sin35 sin n n θ ° = 1 red 1.572sin35 sin 64.4 1.0 θ ° = = ° blue air blue sin35 sin n n θ ° = 1 blue 1.587sin35 sin 65.5 1.0 θ ° = = ° blue red 1.1 θ θ θ Δ = = °

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

View Full Document
23.24. Model: Use the ray model of light and the phenomenon of dispersion. Visualize: Solve: Using Snell’s law for the red light, air air red red sin sin n n θ θ = air 1.0sin 1.45sin26.3 θ = ° ⇒ ( ) 1 air sin 1.45sin26.3 40.0 θ = ° = ° Now using Snell’s law for the violet light, air air violet violet sin sin n n θ θ = violet 1.0sin40.0 sin25.7 n ° = ° ⇒ n violet = 1.48 Assess: As expected, n violet is slightly larger than n red .
23.25. Model: The intensity of scattered light is inversely proportional to the fourth power of the wavelength. Solve: We want to find the wavelength of infrared light such that IR 500 0.01 I I = . Because ( ) 4 500 500 nm I and 4 IR , I λ we have 4 500 IR 100 500 nm I I λ = = 1580 nm λ =

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

View Full Document
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.
23.27. 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 located at s = 15 cm to the right of the converging lens, and is inverted.

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

View Full Document
23.28. 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 that the rays after refraction do not converge at a point on the refraction side of the lens. On the other hand, the three special rays, when extrapolated backward toward the incidence side of the lens, meet at P , which is 15 cm from the lens. That is, s = 15 cm. The image is upright.
23.29. Model: Use ray tracing to locate the image. Solve: The figure shows the ray-tracing diagram using the steps of Tactics Box 23.3. The three rays after refraction do not converge at a point, but they appear to come from P . P is 6 cm from the diverging lens, so s = 6 cm. The image is upright.

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

View Full Document
23.30. Model: Assume the biconvex lens is a thin lens. Solve: If the object is on the left, then the first surface has R 1 = + 40 cm (convex toward the object) and the second surface has R 2 = 40 cm (concave toward the object). The index of refraction of glass is n = 1.50, so the lensmaker’s equation is ( ) ( ) 1 2 1 1 1 1 1 1 1.50 1 40 cm 40 cm n f R R = = f = 40 cm
23.31.

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.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern