Physics Hwk Problem on Convex Mirrors

[shawonna23]

Convex mirrors are being used to monitor the aisles in a store. The mirrors have a radius of curvature of 4 m. What is the image distance if a customer is 20 m in front of the mirror?

From using the mirror equation, i came up with an answer of 1.82m but when i submitted it for webassign, it was wrong. Is the answer supposed to be negative? Please Help!

 

[marlon]

I think this will provide your answer automatically.

Spherical Mirrors

May i give you a piece of advice here ?
When studying mirrors you should make a little list for yourself in which you make four columns and three row. Each row denotes the mirror type (plane, concave, convex) and the columns are like this:
1) mirror type
2) object location (inside or outside the focal point)
3) image (three sub-columns :location, type, orientation of the image with respect to the object)
4) the signs (four sub-columns where you write the signs of f,r,i,m for each row)

f : focal point
r: radius of spherical mirror
i : image distance
m: magnification
pbject distance

Nr 4 will eliminate the chance of you making any mistakes when filling in the parameters of the spherical mirror formula [tex]\frac{1}{p} + \frac{1}{i} = \frac{1}{f}[/tex]

You can do the same for lenses.
If you want a check just let me know and i will verify your tables :)

regards
marlon

 

[Doc Al]

From using the mirror equation, i came up with an answer of 1.82m but when i submitted it for webassign, it was wrong. Is the answer supposed to be negative? Please Help!

Just to add to marlon's post: Often the most confusing thing about using the mirror (and lens) equation is getting the sign convention straight. To add to the confusion, there are several different sign conventions.

With one standard sign convention, real images have positive image distances (meaning the image is on the same side of the mirror as the light is after reflection) and virtual images have negative image distances (meaning that the image appears to be on the side of the mirror surface where there really is no light). The image in this problem is virtual.

The mirror equation will tell you this, assuming you have the signs of object distance (positive for real objects, like a person) and focal length (negative for diverging mirrors, like convex mirrors). I suspect you made an error with one (or both) of these.