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986_PartUniversity Physics Solution

# 986_PartUniversity Physics Solution - 34-24 Chapter 34...

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34-24 Chapter 34 34.84. I DENTIFY and S ET U P : Treating each of the goblet surfaces as spherical surfaces, we have to pass, from left to right, through four interfaces. Apply a b b a n n n n s s R + = to each surface. The image formed by one surface serves as the object for the next surface. E XECUTE : (a) For the empty goblet: 1 1 1 1.50 0.50 12 cm 4.00 cm a b b a n n n n s s s R s + = + = = . 2 2 2 1.50 1 0.50 0.60 cm 12 cm 11.4 cm 64.6 cm. 11.4 cm 3.40 cm s s s = = − + = = − 3 3 3 1 1.50 0.50 64.6 cm 6.80 cm 71.4 cm 9.31cm. 71.4 cm 3.40 cm s s s = + = + = = − 4 4 4 1.50 1 0.50 9.31cm 0.60 cm 9.91cm 37.9 cm. 9.91cm 4.00 cm s s s = + = + = = − The final image is 37.9 cm 2(4.0 cm) 29.9 cm = to the left of the goblet. (b) For the wine-filled goblet: 1 1 1 1.50 0.50 12 cm 4.00 cm a b b a n n n n s s s R s + = + = = . 2 2 2 1.50 1.37 0.13 0.60 cm 12 cm 11.4 cm 14.7 cm. 11.4 cm 3.40 cm s s s = = − + = = 3 3 3 1.37 1.50 0.13 6.80 cm 14.7 cm 7.9 cm 11.1cm. 7.9 cm 3.40 cm s s s = = − + = = 4 4 4 1.50 1 0.50 0.60 cm 11.1cm 10.5 cm 3.73 cm 10.5 cm 4.00 cm s s s = = − + = = . The final image is 3.73 cm to the right of the goblet. E VALUATE : If the object for a surface is on the outgoing side of the light, then the object is virtual and the object distance is negative. 34.85. I DENTIFY : The image formed by refraction at the surface of the eye is located by a b b a n n n n s s R + = . S ET U P : 1.00 a n = , 1.35 b n = . 0 R > . For a distant object, s ≈ ∞ and 1 0 s . E XECUTE : (a) s ≈ ∞ and 2.5 cm s ′ = : 1.35 1.35 1.00 2.5 cm R = and 0.648 cm 6.48 mm R = = . (b) 0.648 cm R = and 25 cm s = : 1.00 1.35 1.35 1.00 25 cm 0.648 s + = . 1.35 0.500 s = and 2.70 cm 27.0 mm s ′ = = . The image is formed behind the retina. (c) Calculate s for s ≈ ∞ and 0.50 cm R = : 1.35 1.35 1.00 0.50 cm s = . 1.93 cm 19.3 mm s ′ = = . The image is formed in front of the retina. E VALUATE : The cornea alone cannot achieve focus of both close and distant objects. 34.86. I DENTIFY : Apply a b b a n n n n s s R + = and a b n s m n s = − to each surface. The overall magnification is 1 2 m m m = . The image formed by the first surface is the object for the second surface. S ET U P : For the first surface, 1.00 a n = , 1.60 b n = and 15.0 cm R = + . For the second surface, 1.60 a n = , 1.00 b n = and R =→ ∞ . E XECUTE : (a) 1 1.60 0.60 36.9 cm. 12.0 cm 15.0 cm a b b a n n n n s s s R s + = + = = − The object distance for the far end of the rod is 50.0 cm ( 36.9 cm) 86.9 cm. − − = The final image is 4.3 cm to the left of the vertex of the hemispherical surface. 1.60 1 0 54.3 cm. 86.9 cm a b b a n n n n s s s R s + = + = = − (b) The magnification is the product of the two magnifications: 1 2 1 2 36.9 1.92, 1.00 1.92.

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