Answers:
In the convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object
a) Position of the crayon (the object)
The Mirror Formula is:
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex] (1)
Where:
[tex]f[/tex] is the focal distance
[tex]u[/tex] is the distance between the object and the mirror
[tex]v[/tex] is the distance between the image and the mirror
We already know the values of [tex]f[/tex] and [tex]v[/tex], let's find [tex]u[/tex] from (1):
[tex]u=\frac{v.f}{v-f}[/tex] (2)
Taking into account the explanation at the beginign of this asnswer:
[tex]f=-49.1cm[/tex] and [tex]v=-21.9cm[/tex]
The negative signs indicate the focal distance and the distance between the image and the mirror are virtual
Then:
[tex]u=\frac{(-21.9cm)(-49.1cm)}{-21.9cm-(-49.1cm)}[/tex]
[tex]u=39.532cm[/tex] (3) >>>>Position of the crayon
b) magnification of the image
The magnification [tex]m[/tex] of the image is given by:
[tex]m=-\frac{v}{u}[/tex] (4)
[tex]m=-\frac{(-21.9cm)}{39.532cm}[/tex]
[tex]m=0.553[/tex] (5)>>>the image is 0.553 smaller than the object
The fact that this value is positive means the image is upright
c) Describe the image
According to the explanations and results obtained in the prior answers, the correct option is 4:
The image is virtual, upright, smaller than the object
d) Height of the crayon (object)
Another way to find the magnification is by the following formula:
[tex]m=\frac{h_{i}}{h_{o}}[/tex] (6)
Where:
[tex]h_{i}[/tex] is the image height
[tex]h_{o}[/tex] is the object height
We already know the values of [tex]m[/tex] and [tex]h_{i}[/tex], let's find [tex]h_{o}[/tex]:
[tex]h_{o}=\frac{h_{i}}{m}[/tex] (7)
[tex]h_{o}=\frac{2.93cm}{0.553}[/tex]
[tex]h_{o}=5.29cm[/tex] (8) >>>height of the crayon
