1) an object with a height of 36 cm is placed 2.1 m in front of a concave mirror with a focal length of 0.50 m. a) determine the location and size of the image using a ray diagram. b) is the image upright or inverted? c) is the image real or virtual? 2) using the mirror equations, find the: a. precise location of the image. b. magnification of the image. c. height of the image.

1) Solution of the problem using ray diagram (see attachment)

P is the point where the object is located, Q is the point where the image is located, F is the  focal length. Everything is in scale such that 1 small square = 0.10 m. As we can see:

(a) the location of the image (Q) is approximately at 7 small squares from the mirror, so the  distance of the image from the mirror will be between 0.65 m- 0.70m. Its size is a bit more than 1 small square, so it should be between 0.10 m- 0.15 m.

(b) the image is inverted

(c) the image is real, because it is on the same side of the object

2) Solution using mirror equation

(a) We have:

By using p=2.1 m and f=0.50 m, we find q=0.66 cm, which corresponds to what we found using ray diagrams.

(b) For the magnification, we can use the equation:

where  is the size of the image and  is the size of the object.
So, we find

So, the magnification is 0.31 (the negative sign means the image is inverted)

(c) The height of the image is given by the  same equation of step (b):

So, since M=-0.31 and h_o = 36 cm=0.36 m, we find

so, the size of the image is 11 cm, and the negative sign means again the image is inverted.