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
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.
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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