Visualize solve a from the graph in figure 2329 we

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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 = = °
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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 .
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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 =
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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.
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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.
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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.
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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.
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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
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23.31.
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