How far away does the image appear toJoe? Express your answer in meters, tothree

Question

How far away does the image appear toJoe? Express your answer in meters, tothree significant figures or as a fraction. How far away does the image appear toJoe? Express your answer in meters, tothree significant figures or as a fraction. Express your answer in meters, tothree significant figures or as a fraction. Joe is so startled by his image that hefalls forward. (Assume that his feet stay at the same position.) (Intro 2 figure) Part C Now what is the length (i.e., the distancefrom head to toe) of Joe's image? Express your answer in meters, tothree significant figures or as a fraction. Joe is so startled by his image that hefalls forward. (Assume that his feet stay at the same position.) (Intro 2 figure) Part C Now what is the length (i.e., the distancefrom head to toe) of Joe's image? Express your answer in meters, tothree significant figures or as a fraction.

Explanation / Answer

Object distance do = 5 m     Object heightho = 1.6 m (a) Use mirror equation   2 / R = 1 /do + 1 / di                                           di= (1 / f - 1 / do)-1                                              = (1 / 15 m - (1 / 5 m)-1
                                                           = (-2 / 15m )-1                                     = - 7.50 m is imagedistance
(b)     Use the ratio  hi / ho = - di / do                                       hi = - (- 5 / 7.5) * 1.6 m = 1.0656 m

This agrees with what we would expect for convex mirror (image is erect, smaller, virtual) (c)
Object distance (do) = 5 m -1.6 m                                       =   3.4 m         We have:              1 / f   = 1 / do + 1 / di          or 1 /di = 1 / f   - 1 / do                        = 1 / 15 - 1 / 3.4           Image Length (di)   = - 4.395 m    The length (i.e., the distance from head to toe)of Joe's image is also : - 7.50 - (-4.395) = - 3.1044 m
This agrees with what we would expect for convex mirror (image is erect, smaller, virtual) (c)
Object distance (do) = 5 m -1.6 m                                       =   3.4 m         We have:              1 / f   = 1 / do + 1 / di          or 1 /di = 1 / f   - 1 / do                        = 1 / 15 - 1 / 3.4           Image Length (di)   = - 4.395 m    The length (i.e., the distance from head to toe)of Joe's image is also : - 7.50 - (-4.395) = - 3.1044 m
Object distance (do) = 5 m -1.6 m                                       =   3.4 m         We have:              1 / f   = 1 / do + 1 / di          or 1 /di = 1 / f   - 1 / do                        = 1 / 15 - 1 / 3.4           Image Length (di)   = - 4.395 m    The length (i.e., the distance from head to toe)of Joe's image is also : - 7.50 - (-4.395) = - 3.1044 m

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