A 15.0 cm tall object is 75.0 cm to the left of a diverging lens (with focal length of magnitude 40.0 cm). A second diverging lens (with a focal length of magnitude 100.0 cm) is 57.2 cm to the right of the first lens. The object and both lenses are on the same optic axis. What is the height of the final image that is produced by light passing through both lenses?
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A 15.0 cm tall object is 75.0 cm to the left of a diverging lens (with focal length of magnitude 40.0 cm). A second diverging lens (with a focal length of magnitude 100.0 cm) is 57.2 cm to the right of the first lens. The object and both lenses are on the same optic axis. What is the height of the final image that is produced by light passing through both lenses?

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- On one side of a diverging lens of focal length 59.0 cm, you position an object of height 3.92 cm somewhere along the principal axis. The resultant image has a height of 2.74 cm. How far from the lens is the object located? 17.8 cm 25.4 cm 43.2 cm 33.0 cmA converging lens is placed 36.0 cm to the right of a diverging lens of focal length 7.0 cm. A beam of parallel light enters the diverging lens from the left, and the beam is again parallel when it emerges from the converging lens. Calculate the focal length of the converging lens. f = _____ cmTwo lenses are separated by 50 cm. Lens 1 is convex and has a radius of curvature of magnitude 30 cm. Lens 2 is concave and has a radius of curvature of magnitude 40 cm and is located to the right of Lens 1. An object is located 20 cm to left of lens 1. Find the location of the final image of the object. What is the magnification of the final image?
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- A 3.0 cm tall object is positioned 15.5 cm from a converging lens. The focal length of the lens is 10.5 cm. What is the distance to the image? What is the magnification of the image? What is the height of the image? Is the image real or virtual? Is the image inverted or upright?A student's uncorrected near and far points are at 9.0 cm and infinity, respectively. She looks through a meniscus-shaped lens as shown in the figure. The absolute value of the radius of curvature is 2.4 cm on the side of the lens near her eye, and 2.6 cm on the side away from her eye. The index of refraction of the lens is 1.4. What is the closest an object can get to the lens and still be clearly visible to the student? [You may assume that the lens is very close to her eye.] eye 3.5 cm 78.0 cm 10.2 cm 8.1 cm 2.3 cm