An orthodontist wishes to inspect a patient's tooth with a magnifying mirror. He places the mirror 0.750 cm behind the tooth. This results in an upright, virtual image of the tooth that is 15.0 cm behind the mirror. (a)What is the mirror's radius of curvature (in cm)? cm (b)What magnification describes the image described in this passage?

An object of height 5.50 cm is placed 33.0 cm to the left of a converging lens with a focal length of 10.5 cm. Determine the image location in cm, the magnification, and the image height in cm. HINT (a) the image location in cm cm (b) the magnification (c) the image height in cm cm (d) Is the image real or virtual? O real virtual (e) Is the image upright or inverted? O upright O inverted

A doctor wishes to inspect a patient's tooth with a magnifying mirror. He places the mirror 1.25 cm behind the tooth. This results in an upright, virtual image of the tooth that is 11.0 cm behind the mirror.   (a)What is the mirror's radius of curvature (in cm)?  ____ cm (b) What magnification describes the image described in this passage?

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A transparent photographic slide is placed in front of a converging lens with a focal length of 2.20 cm. An image of the slide is formed 14.5 cm from the slide. (a) How far is the lens from the slide if the image is real? (b) How far is the lens from the slide if the image is virtual? Step 1 (a) The real image case is shown in the ray diagram. or Object 1 Figure 1 + P 14.5 cm P Notice that here p + q = 14.5 cm, or q = 14.5 cm - p. The thin-lens equation, with focal length f = 2.20 cm, gives 1 - p² (14.5 cm)p+ p+q = F Description Image | cm² = 0. Using the quadratic formula to solve the above equation gives two solutions. The smaller solution is p= cm and the larger solution is p = cm cm. Both are valid solutions for the real image case.

A convex mirror has a focal length of -21 cm. Find the magnification produced by the mirror when the object distance is (a)12 cm and (b)22 cm. (a) m₁ = (b) m₂ = < < i M. Object Virtual image < <

An object is placed to the left of a lens, and a real image is formed to the right of the lens. The image is inverted relative to the object and is one-half the size of the object. The distance between the object and the image is 73.0 cm. (a) How far from the lens is the object? (b) What is the focal length of the lens? (a) do = (b) f=

A doctor wishes to inspect a patient's tooth with a magnifying mirror. He places the mirror 0.750 cm behind the tooth. This results in an upright, virtual image of the tooth that is 10.0 cm behind the mirror. (a) What is the mirror's radius of curvature (in cm)? cm (b) What magnification describes the image described in this passage?

You got a spherical concave mirror with focal length (f) of 15.0 cm. (a) Locate the image (q) and find the magnification for an object distance (p) of 30.0 cm in front of the mirror. Determine whether the image is real or virtual, inverted or upright, and larger or smaller. (b) Do the same for object distances of 15.0 cm Group of answer choices 1. (a) q = 30 cm, Image is virtual, upright, of same size, M = 1. (b) q = infinity, real, undefined, M = 2 2. (a) q = 30 cm, Image is real, inverted, of same size, M = 1. (b) q = infinity, virtual, undefined, M = undefined 3. (a) q = 15 cm, Image is real, upright, double of the object size, M = 2. (b) q = infinity, real, undefined, M = undefined 4. (a) q = 45 cm, Image is real, upright, of same size, M = 1. (b) q = infinity, virtual, upright, M = undefined