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Figure P35.20 (page 958) shows a curved surface separating a material with index of refraction n1 from a material with index n2. The surface forms an image I of object O. The ray shown in red passes through the surface along a radial line. Its angles of incidence and refraction are both zero, so its direction does not change at the surface. For the ray shown in blue, the direction changes according to Snell’s law, n1 sin θ1 = n2 sin θ2. For paraxial rays, we assume θ1, and θ2 are small, so we may write n1 tan θ1 = n2 tan θ2. The magnification is defined as M = h′/h. Prove that the magnification is given by M = −n1q/n2p.
Figure P35.20
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Chapter 36 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
- Two 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_forwardWhy is the following situation impossible? Consider the lensmirror combination shown in Figure P35.55. The lens has a focal length of fL = 0.200 m, and the mirror has a focal length of fM = 0.500 m. The lens and mirror are placed a distance d = 1.30 m apart, and an object is placed at p = 0.300 m from the lens. By moving a screen to various positions to the left of the lens, a student finds two different positions of the screen that produce a sharp image of the object. One of these positions corresponds to light leaving the object and traveling to the left through the lens. The other position corresponds to light traveling to the right from the object, reflecting from the mirror and then passing through the lens. Figure P35.55 Problem 55 and 57.arrow_forwardThe nickels image in Figure P23.36 has twice the diameter of the nickel when the lens is 2.84 cm from the nickel. Determine the focal length of the lens. Figure P23.36arrow_forward
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