Visualize first concentrate on the optic axis and the

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Visualize: First concentrate on the optic axis and the ray parallel to it. Geometry says if parallel lines are both cut by a diagonal (in this case the line through the center of curvature and normal to the mirror at the point of incidence) the interior angles are equal; so i . φ θ = The law of reflection says that i r , θ θ = so we conclude r . φ θ = Now concentrate on the triangle whose sides are R , a , and b . Because two of the angles are equal then it is isosceles; therefore . b a = Apply the law of cosines to this triangle. Solve: 2 2 2 2 cos b a R aR φ = + Because , a b = they drop out. 2 2 cos R aR φ = 2 cos R a φ = We want to know how big a is in terms of , R so solve for . a 2cos R a φ = If 1 φ ± then cos 1, φ so in the limit of small , 2, a R φ = and then since f R a = it must also be that 2 R f = Assess: Many textbooks forget to stress that 2 f R = only in the limit of small , φ i.e. , for paraxial rays.
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23.79. Model: Use the ray model of light and assume the lens is a thin lens. Visualize: Please refer to Figure 23.47. Solve: Let n 1 be the refractive index of the fluid and n 2 the refractive index of the lens. The lens consists of two spherical surfaces having radii of curvature R 1 and R 2 and the lens thickness t 0. For the refraction from the surface with radius R 1 , we use Equation 23.21: 1 2 2 1 1 1 1 n n n n s s R + = For the refraction from surface with radius R 2 , 2 1 1 2 1 2 2 n n n n s s R + = A negative sign is used with 1 s because the image formed by the first surface of the lens is a virtual image. This virtual image is the object for the second surface. Adding the two equations, 1 1 2 1 1 2 1 2 1 1 n n n n s s R R + = 2 1 1 2 1 1 2 1 1 1 ( ) 1 1 n n s s f n R R + = = (b) In air, R 1 = + 40 cm (convex toward the object), R 2 = 40 cm (concave toward the object), n 1 = 1.0, and n 2 = 1.50. So, 1 1.50 1.0 1 1 1.0 40 cm 40 cm f ⎞⎛ = ⎟⎜ ⎠⎝ f = 40 cm In water, n 1 = 1.33 and n 2 = 1.50. So, 1 1.50 1.33 1 1 1.33 40 cm 40 cm f ⎞⎛ = ⎟⎜ ⎠⎝ f = 156 cm
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23.80. Model: Use the ray model of light. Solve: (a) The time ( t ) is the time to travel from A to the interface ( t 1 ) and from the interface to B ( t 2 ). That is, ( ) 2 2 2 2 1 2 1 2 1 1 2 2 1 2 1 2 1 2 1 2 d d d d n d n d n n t t t x a w x b v v c n c n c c c c = + = + = + = + = + + + (b) Because t depends on x and there is only one value of x for which the light travels from A to B in the least possible amount of time, we have ( ) ( ) 2 1 2 2 2 2 0 n w x dt n x dx c x a c w x b = = + + The solution (hard to do! ) would give x min . (c) From the geometry of the figure, 1 2 2 1 sin x x d x a θ = = + ( ) 2 2 2 2 sin w x w x d w x b θ = = + Thus, the condition of part (b) becomes 1 2 1 2 sin sin 0 n n c c θ θ = 1 1 2 2 sin sin n n θ θ =
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23.81. Model: Assume the ray model of light. The ball is not a thin lens. However, the image due to refraction from the first surface is the object for the second surface.
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