A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image. About the author Samantha
Given: Radius of curvature = 3.00 m Bus located from this mirror = 5.00 m Mirror formula: [tex] \frac{1}{v} + \frac{1}{u} = \frac{1}{f} [/tex] Formula of magnification of a mirror: [tex]m = – \frac{v}{u} [/tex] Here, v is the distance of the image and u is the distance of the object from the mirror. Solution: u = 5m r = 3m f = r/2 f = 3/2 f = 1.5m Now in the mirror formula: [tex] \frac{1}{v} = \frac{1}{f} – \frac{1}{u} \\ \frac{6.5}{7.5} [/tex] v = 1.15m Magnification: [tex]m = – \frac{v}{u} \\ m = – \frac{ – 1.15}{ – 5} \\ m = 0.23[/tex] Therefore, the size of the image on the convex mirror will be 0.23 times more than the bus (object). The image formed will be: Erect Diminished Reply
Refer the attachments….
Given:
Radius of curvature = 3.00 m
Bus located from this mirror = 5.00 m
Mirror formula:
[tex] \frac{1}{v} + \frac{1}{u} = \frac{1}{f} [/tex]
Formula of magnification of a mirror:
[tex]m = – \frac{v}{u} [/tex]
Here,
v is the distance of the image and u is the distance of the object from the mirror.
Solution:
u = 5m
r = 3m
f = r/2
f = 3/2
f = 1.5m
Now in the mirror formula:
[tex] \frac{1}{v} = \frac{1}{f} – \frac{1}{u} \\ \frac{6.5}{7.5} [/tex]
v = 1.15m
Magnification:
[tex]m = – \frac{v}{u} \\ m = – \frac{ – 1.15}{ – 5} \\ m = 0.23[/tex]
Therefore, the size of the image on the convex mirror will be 0.23 times more than the bus (object).
The image formed will be: