A virtual image is formed 23.0 cm from a concave mirror having a radius of curvature of 32.0 cm. (a) Find the position of the object. (b) What is the magnification of the mirror?
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A virtual image is formed 23.0 cm from a concave mirror having a radius of curvature of 32.0 cm.
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- The distance between the eyepiece and the objective lens (L) in a certain compound microscope must be adjusted. The focal length of the eyepiece is 3.00 cm and that of the objective is 0.500 cm. The overall required magnification of the microscope is – 400. Specify L in cm.A concave mirror (R = 68.2 cm) is used to project a transparent slide onto a wall. The slide is located at adistance of 43.5 cm from the mirror, and a small flashlight shines light through the slide and onto the mirror.(a) How far from the wall should the mirror be located? (b) The height of the object on the slide is 1.78 cm. What is the height of the image?Question in attachment .
- An object of 1 cm tall is placed in front of a converging lens of focal length of 2 cm. The object is 2.5 cm away from the mirror. (a 5') Use the ray tracing to find the image. (b 5') Compute the image distance and height using the lens equation.pls answer a-b plsA convex mirror has a radius of curvature of 46.7 cm. A 11.5-cm-tall object is placed in front of the mirror and produces an image that is 5.13-cm tall. (a) What is the magnification? (b) How far is the object from the mirror?In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the bending occurs at the cornea, whose vertex is 2.75cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 60.0cm from the cornea’s vertex is focused on the retina?
- Ch. 35A doctor wishes to inspect a patient's tooth with a magnifying mirror. She 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)? (b) What magnification describes the image described in this passage?The objective lens in a microscope with a 15.0 cm long tube has a magnification of -50.0 and the eyepiece has a magnification of 21.0. (a) What is the focal length of the objective? cm (b) What is the focal length of the eyepiece? cm (c) What is the overall magnification of the microscope?
- 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 1.50 cm high is held 2.95 cm from a person's cornea, and its reflected image is measured to be 0.166 cm high. (a) What is the magnification? ✕ (b) Where is the image (in cm)? cm (from the corneal "mirror") (c) Find the radius of curvature (in cm) of the convex mirror formed by the cornea. (Note that this technique is used by optometrists to measure the curvature of the cornea for contact lens fitting. The instrument used is called a keratometer, or curve measurer.) cmA concave spherical mirror has a radius ofcurvature of 35 cm . The object distance is51.9 cm . Scale: 10 cm Find the magnitude of the image distance.Answer in units of cm.