We are given the following situation:
Where:
[tex]\begin{gathered} h_0=height\text{ of the object} \\ h_i=\text{ height of the image} \\ d_i=\text{ distance to the image} \\ d_0=\text{ distance of the object} \end{gathered}[/tex]We will determine the height of the image by determining the height of the mirror. To do we use the area of the mirror. If the mirror is a square then its area is given by:
[tex]A=h^2[/tex]Substituting the value of the area:
[tex]35.27cm^2=h^2[/tex]Now, we take the square root to both sides:
[tex]\sqrt{35.27cm^2}=h[/tex]Solving the operations:
[tex]5.9cm=h[/tex]Therefore, the height of the image is 5.9cm.
Now, we determine the magnification using the following formula:
[tex]M=\frac{h_i}{h_0}[/tex]Substituting the values in the formula we get:
[tex]M=\frac{0.059m}{4.93m}=0.012[/tex]The magnification is also equal to:
[tex]M=-\frac{d_i}{d_o}[/tex]Now, we solve for the distance of the object. First, we multiply both sides by the distance of the object:
[tex]Md_o=-d_i[/tex]Now, we divide both sides by the magnification:
[tex]d_o=-\frac{d_i}{M}[/tex]Now, we plug in the values:
[tex]d_o=\frac{-60.02cm}{0.012}[/tex]Solving the operations:
[tex]d_o=-5015.23cm[/tex]Now, the distance from the observer to the pole is the difference between the distance of the object and the distance of the image:
[tex]d=d_0-d_i[/tex]Substituting we get:
[tex]d=5015.23cm-60.02cm=4955.2cm[/tex]Therefore, the pole is at 4955.2 cm.
