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Unformatted text preview: Physics 2760 Recitation 11/18/2008 34.17 A small tropical fish is at the center of a water‐filled spherical fish bowl 28 cm in diameter. Find the apparent position of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. Find the magnification of the fish to an observer outside the bowl. A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl? 34.26 A converging lens with a focal length of 90 cm forms an image of a 3.20‐cm‐tall real object that is to the left of the lens. The image is 4.5 cm tall and inverted. Where are the image and the object located in relation to the lens? Is the image real or virtual? 34.29 A double‐convex thin lens has surfaces with equal radii of curvature of magnitude 3.00 cm. Looking through this lens, you observe that it forms an image of a very distant tree at a distance of 1.50 cm from the lens. What is the index of refraction of the lens? 34.89 Two thin lenses with focal lengths of magnitude 15 cm, the first diverging and the second converging, are placed 12 cm apart. An object 4 mm tall is placed 5 cm to the left of the first diverging lens. A) Where is the image formed? B) How far from the object is the final image formed? C) Is the final image real or virtual? D) What is the height of the final image? Is the final image upright or inverted? 34.82 A transparent rod 30.0 cm long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 cm at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 cm from the flat end and 15.0 cm from the vertex of the curved end. When viewed from the flat end of the rod, the apparent depth of the object is 9.50 cm from the flat end. What is its apparent depth when viewed from the curved end? 34.103 You have a camera with a 35.0 mm focal length lens and 36.0 mm‐wide film. You wish to take a picture of a 12.0 m long sailboat but find that the image of the boat fills only ¼ of the width of the film. A) How far are you from the boat? B) How much closer must the boat be to you for its image to fill the width of the film? 34.17. IDENTIFY: na nb nb − na n s′ += . m = − a . Light comes from the fish to the person’s eye. s s′ R nb s SET UP: R = −14.0 cm . s = +14.0 cm . na = 1.333 (water). nb = 1.00 (air). Figure 34.17 shows the object and the refracting surface. 1.333 1.00 1.00 − 1.333 (1.333)( −14.0 cm) + = EXECUTE: (a) . s′ = −14.0 cm . m = − = +1.33 . −14.0 cm 14.0 cm s′ (1.00)(14.0 cm) The fish’s image is 14.0 cm to the left of the bowl surface so is at the center of the bowl and the magnification is 1.33. n n − na (b) The focal point is at the image location when s → ∞ . b = b . na = 1.00 . nb = 1.333 . R = +14.0 cm . s′ R 1.333 1.333 − 1.00 = . s′ = +56.0 cm . s′ is greater than the diameter of the bowl, so the surface facing the sunlight does s′ 14.0 cm not focus the sunlight to a point inside the bowl. The focal point is outside the bowl and there is no danger to the fish. EVALUATE: In part (b) the rays refract when they exit the bowl back into the air so the image we calculated is not the final image. Figure 34.17 34.26. IDENTIFY: Apply m =
y′ s′ 111 = − to relate s′ and s and then use + = . s s′ f y s SET UP: Since the image is inverted, y′ < 0 and m < 0 . EXECUTE: m = s′ 1 1 1 y′ −4.50 cm 111 = and = = −1.406 . m = − gives s′ = +1.406 s . + = gives + s s 1.406s 90.0 cm y 3.20 cm s s′ f s = 154 cm . s′ = (1.406)(154 cm) = 217 cm . The object is 154 cm to the left of the lens. The image is 217 cm to the right of the lens and is real. EVALUATE: For a single lens an inverted image is always real. 34.29. IDENTIFY: Apply ⎛1 1 1⎞ = ( n − 1) ⎜ − ⎟ . f R1 R2 ⎠ ⎝ SET UP: For a distant object the image is at the focal point of the lens. Therefore, f = 1.5 cm . For the double‐convex lens, R1 = + R and R2 = − R , where R = 3.0 cm . 1 R 3.0 cm ⎛ 1 1 ⎞ 2(n − 1) = (n − 1) ⎜ − . n = +1 = + 1 = 2 . ⎟= f R −R ⎠ R 2f 2(1.5 cm) ⎝ EVALUATE: f > 0 and the lens is converging. A double‐convex lens is always converging. EXECUTE: 34.89. IDENTIFY: SET UP: EXECUTE: The first lens forms an image which then acts as the object for the second lens. s′ 111 The thinlens equation is + = and the magnification is m = − . s s′ f s (a) For the first lens: 111 1 1 1 +=⇒ += ⇒ s′ = −3.75 cm , to the left of the lens (virtual ′f ′ −15.0 cm ss 5.00 cm s image). (b) For the second lens, s = 12.0 cm + 3.75 cm = 15.75 cm. 111 1 1 1 +=⇒ += ⇒ s′ = 315 cm , or 332 cm from the object. s s′ f 15.75 cm s′ 15.0 cm (c) The final image is real. s′ (d) m = − , m1 = 0.750, m2 = −20.0, mtotal = −15.0 ⇒ y′ = −6.00 cm, inverted. s EVALUATE: Note that the total magnification is the product of the individual magnifications. 34.82. na nb nb − na += . Use the image distance when viewed from the flat end to determine the refractive s s′ R index n of the rod. SET UP: When viewing from the flat end, na = n , nb = 1.00 and R → ∞ . When viewing from the curved end, na = n , IDENTIFY: Apply nb = 1.00 and R = −10.0 cm . EXECUTE: na nb n 1 15.0 + =0⇒ + =0⇒n= = 1.58. When viewed from the curved end of the rod s s′ 15.0 cm −9.50 cm 9.50 na nb nb − na n 1 1− n 1.58 1 −0.58 += ⇒+= ⇒ += , and s′ = −21.1 cm . The image is 21.1 cm within the rod s s′ R s s′ R 15.0 cm s′ −10.0 cm from the curved end. 34.103. IDENTIFY: The camera lens can be modeled as a thin lens that forms an image on the film. s′ 111 SET UP: The thin‐lens equation is + = , and the magnification of the lens is m = − . ′f ss s EXECUTE: (a) m = − s′ y′ 1 (0.0360 m) == ⇒ s′ = (7.50 × 10−4 ) s , s y 4 (12.0 m)
111 1 1⎛ 1 1 ⎞1 +=+ = ⎜1 + ⇒ s = 46.7 m . ⎟= = s s′ s (7.50 × 10−4 ) s s ⎝ 7.50 × 10−4 ⎠ f 0.0350 m (b) To just fill the frame, the magnification must be 3.00 × 10−3 so: 1⎛ 1 1 ⎞1 ⇒ s = 11.7 m . ⎜1 + ⎟= = s ⎝ 3.00 × 10−3 ⎠ f 0.0350 m Since the boat is originally 46.7 m away, the distance you must move closer to the boat is 46.7 m – 11.7 m = 35.0 m. EVALUATE: This result seems to imply that if you are 4 times as far, the image is ¼ as large on the film. However this result is only an approximation, and would not be true for very close distances. It is a better approximation for large distances. ...
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This note was uploaded on 06/21/2010 for the course PHYS 2760 taught by Professor Kozstin during the Spring '08 term at Missouri (Mizzou).
 Spring '08
 kozstin
 Physics

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