# Model use the ray model of light solve a using snells

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Model: Use the ray model of light. Solve: (a) Using Snell’s law at the air-glass boundary, with φ being the angle of refraction inside the prism, air sin sin n n β φ = sin sin n β φ = Considering the triangle made by the apex angle and the refracted ray, ( ) ( ) 1 2 90 90 180 φ φ α φ α °− + °− + = ° ⇒ = Thus ( ) ( ) 1 1 1 2 2 sin sin sin sin( ) n n β α β α = = (b) Using the above expression, we obtain 1 2 sin sin52.2 1.58 sin( ) sin30 n β α ° = = = °

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23.59. Model: The bubble is a point source of light. The surface is a spherical refracting surface. Solve: The bubble is in zircon, so n 1 = 1.96 and n 2 = 1.00. The surface is concave (object facing into a “cave”) as seen from the bubble (the object), so R = 3.0 cm. Equation 23.21 is 1 2 2 1 n n n n s s R + = 1.96 1.00 1.00 1.96 3.0 cm 3.0 cm 3.0 cm s s + = = − Thus, seen from outside, the bubble appears to be 3.0 cm beneath the surface. That is, a bubble at the center actually appears to be at the center.
23.60. Model: Use the ray model of light. The surface is a spherically refracting surface. Visualize: Solve: Because the rays are parallel, s = . The rays come to focus on the rear surface of the sphere, so s = 2 R , where R is the radius of curvature of the sphere. Using Equation 23.21, 1 2 2 1 1 1 2.00 2 n n n n n n n s s R R R + = + = =

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23.61. Model: Assume that the converging lens is a thin lens. Use ray tracing to locate the image. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three rays after refraction converge to give an image at s = 40 cm. The height of the image is h = 2 cm. (b) Using the thin-lens formula, 1 1 1 1 1 1 40 cm 20 cm s s f s + = + = 1 1 40 cm s = 40 cm s ′ = The image height is obtained from 40 cm 1 40 cm s M s = − = − = − The image is inverted and as tall as the object, that is, h = 2.0 cm. The values for h and s obtained in parts (a) and (b) agree.
23.62. Model: Use ray tracing to locate the image. Assume that the converging lens is a thin lens. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three special rays that experience refraction do not converge at a point. Instead they appear to come from a point that is 15 cm on the same side as the object itself. Thus s = 15 cm. The image is upright and has a height of h = 1.5 cm. (b) Using the thin-lens formula, 1 1 1 1 1 1 10 cm 30 cm s s f s + = + = 1 1 15 m s = − 15 cm s ′ = − The image height is obtained from 15 cm 1.5 10 cm s M s = − = − = + The image is upright and 1.5 times the object, that is, 1.5 cm high. These values agree with those obtained in part (a).

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23.63. Model: Use ray tracing to locate the image. Assume that the converging lens is a thin lens. Solve: (a) The figure shows the ray-tracing diagram using the steps of Tactics Box 23.2. The three special rays after refracting do not converge. Instead the rays appear to come from a point that is 60 cm on the same side of the lens as the object, so 60 cm. s ′ = − The image is upright and has a height of 8.0 cm.
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• Winter '10
• E.Salik
• Light, refractive index, Total internal reflection, Geometrical optics

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