A coin is placed in front of a concave mirror that curvature R = 0.50m. The imag

Question

A coin is placed in front of a concave mirror that curvature R = 0.50m. The image of the coin is inverted and five times the size of the object. Determine the distance between the coin and the mirror. A) 0.10 m B) 0.15 m C) 0.20 m D) 0.30 m E) 0.45 m A concave mirror (R = 54 cm) is used to project a transparent slide onto a wall. The slide is located at 46.0 cm from the mirror, and light bulb shines light through the slide and onto the mirror. The setup is shown below. (a) How far from the wall should the mirror be located? (b) The height of the object on the slide is 0.80 cm. What is the height of the image? (c) How should the slide be oriented so that the picture on the wall looks normal?

Explanation / Answer

4

image height y' = 5*y

magnification m = y'/y = s'/s

5 = s'/s

s' = 5*s

from mirror equation

1/s + 1/s' = 2/R

1/s + 1/(5s) = 2/0.5

s = 0.3 m <<<<<---answer

OPTION (D)

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object distance s = 46 cm

1/s + 1/s' = 2/R

1/46 + 1/s' = 2/54

s' = 65.4 cm <<<<<-----answer

(b)

height of the image/height of the object = -s'/s

height of the image / 0.8 = 65/46

height of the image = 1.13 cm <<<<<-----answer

(c)

slide is inverted (upside down)

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