The characteristics of convex mirrors

Question

The problem says:
An object is placed 25 cm in front of a convex mirror with a radius of curvature of 15 cm. The height of the object is 4 cm. Draw a ray diagram and find the location and the height of the image. Repeat this again when the object is placed 5 cm in front of the mirror. 
(b) Does the magnitude of the image height become larger or smaller as the object moves closer to the mirror?
(c) What is the ratio of the image height when the object distance is 5 cm to its height when the object distance is 25 cm?

For part b I drew ray diagrams and got that the image becomes larger. However when I found the image height in both situations I got 25cm away to be 2.5cm high and I got 5cm away to be 1cm high. So I am confused on whether my diagram was incorrect or if my math was incorrect.

For part c my teachers answer is 2.6 but I am unsure on how to get this ratio. I thought it was hi when at 5cm over hi when at 25cm but I just got .4

Please help me, I would be so grateful!

Answer 1

Hi Olivia!

 

Three overriding principles that are always true for convex mirrors are:

 

1.  they can never create a real image

2.  they can never create an image larger than the object; and,

3.  the image gets larger as the object gets closer

 

The image starts at infinitely small height. and located at the focal point, hen the object is at "infinity."  Then, as the object approaches the surface of the mirror, the image does, too; and grows in size until -- when the object is right in front of the mirror -- the image is basically the same size as the object and "just inside" the surface of the mirror.  This is because, when the object is that close, the convex mirror looks a lot like a plane mirror, which would just produce an equal-sized virtual image the same distance on the other side of the mirror surface from the object.  The same thing is true for an object right next to a concave mirror.  That also produces an equal-sized virtual image right on the other side of the mirror, because, that close, the concave mirror looks basically like a plane mirror, as well.

 

So, for part b, I would say the ray diagram must be incorrect somehow, because of overriding principle 2.

 

When locating an image with a ray diagram, I usually rely on two chief rays:

 

1.  A ray that comes in parallel to the optical axis, which then reflects off the mirror so that it looks like it come from the principal focal point, which is inside a convex mirror

2.  A ray that reflects at the point where the optical axis intersects the mirror surface.  That ray just reflects around the axis as it would off a plane mirror (so that its incident angle with respect to the axis is the same as its reflected angle with respect to the axis).

 

I also think there was a calculation error, since you calculate a smaller image size when the object was closer, when the image size should get larger as the object approaches the convex mirror.

 

So I also checked the answers you got with calculation, using the spherical mirror (and thin lens) equation:

 

(1/o)+(1/i) = (1/f)

 

where 

 

o = object distance

i = image distance

f = focal length

 

I got different answers than you did for the image height in each case.  I think I know what happened for the 25 cm object, but I am not sure about the 5 cm object.  So let me take you through what I did, so you can compare.

 

The main rules for using this equation for convex mirrors are that:

 

1.  object distance is always positive

2.  for convex mirrors, focal length is negative

 

However we are not given focal length.  We are given radius of curvature C = 15 cm.  Fortunately, for spherical mirrors, there is a direct relationship between the focal length and radius of curvature, in that f = C/2.  So we can calculate the focal length right away to be (15/2) cm (=7.5 cm) [note: I think, at least for the first case, you may gave used to the radius of curvature as the focal length].  And this value is defined to be negative for convex mirrors.

 

So, when the object distance is 25 cm, the equation becomes:

 

(1/25)+(1/i) = -1/(15/2) = -2/15

 

Then 1/i = -2/15 - 1/25 =  -10/75 - 3/75 = -13/75 --> so i = -75/13 ~ -5.77 cm   (negative, as we expect, since the image must be virtual for a convex mirror, and less than the focal length, which it must be for convex mirror...so that is a nice "warm fuzzy" that suggests the calculation was okay)

 

So m = -i/o = (75/13)/25 = 75/13 * 1/25 = 3/13 

 

then  h_i = mh_o = (3/13)(4) = 12/13 cm

 

Doing this same procedure for an object distance of 5 cm, I obtained h_i = 12/5 = 2.4 cm (larger than the image was when the object was at 25 cm, as I expect)

 

Taking the ratio of these two image heights gives your teacher's answer for Part (c).  Try working through that and see if you can find your errors.  If you still have problems, just let me know!