Question

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.

Answer 1

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:
$\frac{1}{f} = \frac{1}{p}+ \frac{1}{q}$
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:
$M= \frac{h_i}{h_o} = - \frac{q}{p}$
where $h_i$ is the size of the image and $h_o = 36 cm =0.36 m$ is the size of the object.
So, we find
$M = -\frac{q}{p}= -\frac{0.66 m}{2.1 m}=-0.31$
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):
$M= \frac{h_i}{h_o}$
So, since M=-0.31 and h_o = 36 cm=0.36 m, we find
$h_i = -M h_o = (--0.31)(0.36 m)=-0.11 m=-11 cm$
so, the size of the image is 11 cm, and the negative sign means again the image is inverted.