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BIO People with normal vision cannot focus their eyes underwater if they aren’t wearing a face mask or goggles and there is water in contact with their eyes (see Discussion Question Q34.23). (a) Why not? (b) With the simplified model of the eye described in Exercise 34.50, what corrective lens (specified by focal length as measured in air) would be needed to enable a person underwater to focus an infinitely distant object? (Be careful―the focal length of a lens underwater is not the same as in air! See Problem 34.92. Assume that the corrective lens has a refractive index of 1.62 and that the lens is used in eyeglasses, not goggles, so there is water on both sides of the lens. Assume that the eye-glasses are 2.00 cm in front of the eye.)
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University Physics (14th Edition)
- Figure P26.72 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F1 and F2 of the lens are 5.00 cm from the center of the lens. (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the final image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.arrow_forwardTwo converging lenses having focal lengths of f1 = 10.0 cm and f2 = 20.0 cm are placed a distance d = 50.0 cm apart as shown in Figure P35.48. The image due to light passing through both lenses is to be located between the lenses at the position x = 31.0 cm indicated. (a) At what value of p should the object be positioned to the left of the first lens? (b) What is the magnification of the final image? (c) Is the final image upright or inverted? (d) Is the final image real or virtual?arrow_forwardThe left face of a biconvex lens has a radius of curvature of magnitude 12.0 cm, and the right face has a radius of curvature of magnitude 18.0 cm. The index of refraction of the glass is 1.44. (a) Calculate the focal length of the lens for light incident from the left. (b) What If? After the lens is turned around to interchange the radii of curvature of the two faces, calculate the focal length of the lens for light incident from the left.arrow_forward
- A converging lens made of crown glass has a focal length of 15.0 cm when used in air. If the lens is immersed in water, what is its focal length? (a) negative (b) less than 15.0 cm (c) equal to 15.0 cm (d) greater than 15.0 cm (e) none of those answersarrow_forwardFigure P38.43 shows a concave meniscus lens. If |r1| = 8.50 cm and |r2| = 6.50 cm, find the focal length and determine whether the lens is converging or diverging. The lens is made of glass with index of refraction n = 1.55. CHECK and THINK: How do your answers change if the object is placed on the right side of the lens? FIGURE P38.43arrow_forwardIn Figure P26.38, a thin converging lens of focal length 14.0 cm forms an image of the square abcd, which is hc = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a, b, c, and d represent the respective corners of the image. Let qa represent the image distance for points a and b, qd represent the image distance for points c and d, hb represent the distance from point b to the axis, and hc represent the height of c. (a) Find qa, qd, hb, and hc. (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a and d, for which the object distance is p. Let h represent the distance from the axis to the point at the edge of the image between b and c at image distance q. Demonstrate that h=10.0q(114.01q) where h and q are in centimeters. (d) Explain why the geometric area of the image is given by qaqdhdq (e) Carry out the integration to find the area of the image. Figure P26.38arrow_forward
- An object is placed a distance of 10.0 cm to the left of a thin converging lens of focal length f = 8.00 cm, and a concave spherical mirror with radius of curvature +18.0 cm is placed a distance of 45.0 cm to the right of the lens (Fig. P38.129). a. What is the location of the final image formed by the lensmirror combination as seen by an observer positioned to the left of the object? b. What is the magnification of the final image as seen by an observer positioned to the left of the object? c. Is the final image formed by the lensmirror combination upright or inverted? FIGURE P38.129arrow_forwardA thin plastic lens with index of refraction n = 1.67 has radii of curvature given by R1 = 12 0 cm and R2 = 40.0 cm. Determine (a) the focal length of the lens, (b) whether the lens Ls converging or diverging and the image distances for object distances of (c) infinity, (d) 8,00 cm, and (e) 50.0 cm.arrow_forwardA man inside a spherical diving bell watches a fish through a window in the bell, as in Figure P23.26. If the diving bell has radius R = 1.75 m and the fish is a distance p = 1 00 m from the window, calculate (a) the image distance and (b) the magnification. Neglect the thickness of the window. Figure P23.26arrow_forward
- A thin plastic lens with index of refraction n = 1.67 has radii of curvature given by R1 = 12 0 cm and R2 = 40.0 cm. Determine (a) the focal length of the lens, (b) whether the lens Ls converging or diverging and the image distances for object distances of (c) infinity, (d) 8,00 cm, and (e) 50.0 cm.arrow_forwardA diverging lens has a focal length of magnitude 20.0 cm (a) Locate the images for object distances of (i) 40.0 cm. (ii) 20.0 cm, and (iii) 10.0 cm. For each case, state whether the image is (b) real or virtual and (c) upright or inverted. (d) For each case, Find the magnification.arrow_forwardIn Figure P35.30, a thin converging lens of focal length 14.0 cm forms an image of the square abed, which is he = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a, b, c. and d represent the respective corners of the image. Let qa represent the image distance for points a and b, qd represent the image distance for points c and d, hb, represent the distance from point b to the axis, and hc represent the height of c. (a) Find qa, qd, hb, and hc. (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a and d, for which the object distance is p. Let h represent the distance from the axis to the point at the edge of the image between b and c at image distance q. Demonstrate that h=10.0q(114.01q) where h and q are in centimeters. (d) Explain why the geometric area of the image is given by qaqdhdq (e) Carry out the integration to find the area of the image. Figure P35.30arrow_forward
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