1 Question - Convex Mirror Help

1 question - Convex mirror help?

For a convex mirror, with f = radius of curvature, focal length, etc, do = distance of object from mirror, and di = distance of image from mirror. The equation to use is:
1/f = 1/do + 1/di
1/(2.0 m) = 1/(3.0 m) + 1/di
1/di = 3/(6.0 m) - 2/(6.0 m) = 1/(6.0 m)
di = 6.0 m

PHYSICS Convex mirror question?

The basic lens equation, which applies equally to spherical concave amd convex mirrors, is
1/u + 1/v = 1/f...therefore 1/u = 1/f - 1/v
where u = object distance...v = image distance...f = focal length
v and f are negative for a convex mirror.
1/u =.. (-1/f ) - (-1/v)
1/u =. . -1/46 - (-1/23)
1/u = . -1/46 -(- 2/46)
1/u . . -1/46 + 2/46
1/u = . . 1/46
u = 46cm

Magnification = -( image distance) / (object distance) . . = -(-23) /46 = 23 /46 = 0.5

Image size is 1.7cm which is 0.5 times the object size
Object size (the crayon) = 2 x 1.7cm = 3.4 cm

The image is upright

Physics question about convex mirrors?

For part A, since we are dealing with mirrors, we should take the thin lens equation which also applies to mirrors:

(1 / f) = (1 / s) + (1 / s')

where f is the focal length (f = - 8.0 cm) and is negative because it is a convex mirror, s is the object distance, and s' is the image distance which will also be negative since it is virtual. So since the image distance is one third the magnitude of the object distance, then:

|s'| = |(1 / 3)*s|
- s' = (1 / 3)*s

So if we substitute this in to the thin lens equation, we get:

(1 / - 8) = (1 / s) - (1 / (1 / 3)*s)
(1 / - 8) = (-2 / s)
s = 16 cm {ANSWER}

For part B, we can now substitute our answer for part A back into the relation equation, and get magnitude. So:

|s'| = |(1 / 3)*s|
|s'| = |(1 / 3)*(16)|
|s'| = |5.33| = 5.3 cm {ANSWER}

Keeping in mind that the previous answer is just the magnitude (really s' = - 5.3 cm), we can now use the magnification formula to see whether the image is upright or inverted. So:

m = - (s' / s)
m = - (- 5.33 / 16)
m = 0.333

Since m > 0, then:

image is upright {ANSWER}

CONVEX MIRROR HELP!!!?

this is my physics homework and i really need help!
Shiny lawn spheres placed on pedestals are convex mirrors. One such sphere has a diameter of 48.0 cm. A 12 cm robin sits in a tree 1.3 m from the sphere. Where is the image of the robin?

_______ cm from the mirror

thanks so much!

Physics problem please help.....convex mirror question?

A convex mirror has a focal length of magnitude 3.25 cm.

(a) If the image is virtual, what is the object location for which the magnitude of the image distance is one third the magnitude of the object distance? (Enter distance in cm from the front of the mirror. If the image is real, enter NONE.)
____________ cm

(b) Find the magnification of the image.


State whether it is upright or inverted.


Confused on how to find q and p when i only have the focal length. If you could help me with how to calculate (a), I'm pretty sure I will be able to do the rest thanks

Reduced image by a convex mirror question.?

An object is placed in front of a convex mirror with a 41cm radius of curvature. A virtual image half the size of the of the object is formed. At what distance is the object from the mirror?

I know that the equation of a mirror is 1/p + 1/q = 1/f. I also know that f=-R/2 for a convex mirror so f=-0.5. I have m=-q/p and solved for q to get q=-0.5p to plug into the mirror equation so I can then solve for p so that gives me 1/p + 1/(-0.5p) = 1/f. I guess what I'm getting at is trying to solve for p and the algebra to go with it... I do it and it gives me the wrong answer so I know I'm doing something wrong. Any suggestions of solving this problem another way would be very helpful or even showing me the algebra part for this method would be wonderful (math isn't my strongest subject, haha). Thanks so much!