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

Question

Part A How far away does the image appear toJoe? Express your answer in meters, tothree significant figures or as a fraction. Part A How far away does the image appear toJoe? Express your answer in meters, tothree significant figures or as a fraction. Part B Express your answer in meters, tothree significant figures or as a fraction. Intro 2 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

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

R=2f so f= 15m
u=-5m
v=?
using 1/v=1/f-1/u=1/15-1/-5=1/15+1/5=>
v=15/4m
magnification=-v/u=-15/4*5=-3/4
size of the image=size of theobject*magnification=-1.6*3/4=-1.2m
negative sign indicates that the image is virtual.
size of the image=1.2m

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