A convex mirror used as a rear-view mirror in a car has a radius of curvature of 3m .if a bus is located at a distance of 5m from this mirror , find the position of image . what is nature of the image
Answers
Answered by
21
R = + 3.0 m
u = - 5.0 m
mirror formula: 1/v + 1/u = 2/ R
1/v = 2/R - 1/U
= 2/(3.0) - 1/(-5.0)
= 2/3 + 1/5 = 10+3 /15
= 13/15
V = 15/13 m = +1.16 m
+ve sign of image means that the image is virtual and formed behind the mirror.
Now for magnification: m = - v/u
= -(1.16) / (-5.0)
= +0.23.
The image is diminished in size.
u = - 5.0 m
mirror formula: 1/v + 1/u = 2/ R
1/v = 2/R - 1/U
= 2/(3.0) - 1/(-5.0)
= 2/3 + 1/5 = 10+3 /15
= 13/15
V = 15/13 m = +1.16 m
+ve sign of image means that the image is virtual and formed behind the mirror.
Now for magnification: m = - v/u
= -(1.16) / (-5.0)
= +0.23.
The image is diminished in size.
Answered by
25
given :
R=3m
f=R/2=3/2=1.5m
u=5m
v=?
Mirror formula:
1/f=1/u+1/v
by sign conventions:
1/1.5=1/-5+1/v
1/v=1/1.5+1/5⇒6.5/7.5
v=7.5/6.5
=1.15m
∴The image is virtual seen at the back of mirror.
m=hi/ho=-v/u
-1.15/-5=0.23
∴Image is erect and diminished in size by a factor of 0.23
R=3m
f=R/2=3/2=1.5m
u=5m
v=?
Mirror formula:
1/f=1/u+1/v
by sign conventions:
1/1.5=1/-5+1/v
1/v=1/1.5+1/5⇒6.5/7.5
v=7.5/6.5
=1.15m
∴The image is virtual seen at the back of mirror.
m=hi/ho=-v/u
-1.15/-5=0.23
∴Image is erect and diminished in size by a factor of 0.23
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