Model assume the planoconvex lens is a thin lens

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Model: Assume the planoconvex lens is a thin lens. Solve: If the object is on the left, then the first surface has R 1 = and the second surface has R 2 = 40 cm (concave toward the object). The index of refraction of polystyrene plastic is 1.59, so the lensmaker’s equation is ( ) ( ) 1 2 1 1 1 1 1 1 1.59 1 40 cm n f R R = = 1 0.59 40 cm f = f = 68 cm
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23.32. Model: Assume the biconcave lens is a thin lens. Solve: If the object is on the left, then the first surface has R 1 = 40 cm (concave toward the object) and the second surface has R 2 = + 40 cm (convex toward the object). The index of refraction of glass is 1.50, so the lensmaker’s equation is ( ) ( ) ( ) 1 2 1 1 1 1 1 1 1 1.50 1 0.50 40 cm 40 cm 20 cm n f R R = = = + f = 40 cm
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23.33. Model: Assume the meniscus lens is a thin lens. Solve: If the object is on the left, then the first surface has R 1 = 30 cm (convex toward the object) and the second surface has R 2 = 40 cm (convex toward the object). The index of refraction of polystyrene plastic is 1.59, so the lensmaker’s equation is ( ) ( ) 1 2 1 1 1 1 1 1 1.59 1 30 cm 40 cm n f R R = = f = 203 cm 200 cm
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23.34. Model: The water is a spherical refracting surface. Consider the paraxial rays that refract from the air into the water. Solve: If the cat’s face is 20 cm from the edge of the bowl, then s = + 20 cm. The spherical fish bowl surface has R = + 25 cm, because it is the convex surface that is toward the object. Also n 1 = 1 (air) and n 2 = 1.33 (water). Using Equation 23.21, 1 2 2 1 n n n n s s R + = 1 1 1.33 1.33 1 0.33 0.0132 cm 20 cm 25 cm 25 cm s + = = = ( ) 1 1.33 0.0132 0.050 cm s = s = 36 cm This is a virtual image located 36 cm outside the fishbowl. The fish, inside the bowl, sees the virtual image. That is, the fish sees the cat’s face 36 cm from the bowl.
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23.35. Model: Model the bubble as a point source and consider the paraxial rays that refract from the plastic into the air. The edge of the plastic is a spherical refracting surface. Visualize: Solve: The bubble is at P, a distance of 2.0 cm from the surface. So, s = 2.0 cm. A ray from P after refracting from the plastic-air boundary bends away from the normal axis and enters the eye. This ray appears to come from P , so the image of P is at P and it is a virtual image. Because P faces the concave side of the refracting surface, R = 4.0 cm. Furthermore, n 1 = 1.59 and n 2 = 1.0. Using Equation 23.21, 1 2 2 1 n n n n s s R + = 1 1.59 1.0 1.0 1.59 0.59 0.1475 cm 2.0 cm 4.0 cm 4.0 cm s + = = + = 1 1 1 0.1475 cm 0.795 cm s = s = 1.54 cm That is, the bubble appears 1.54 cm 1.5 cm beneath the surface.
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23.36. Model: Assume the lens is thin. Visualize: 1 1 1 fs s s s f s f + = = Solve: (20 cm)(60 cm) 30 cm 60 cm 20 cm fs s s f ′ = = = The magnification is 30 cm 60 cm 1 2. m s s = − = − = − This means the image is inverted and has a height of 0.50 cm.
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