Deriving linear magnification from refraction angles

[Baal Hadad]

Homework Statement

Given the diagram below, I would like to derive the expression for linear magnification:

Relevant Equations

$$m=\mod{\frac{n_1v }{n_2u}}$$

I tried assumed $\theta \approx sin \theta \approx tan \theta$.
By Snell's law(after approximation),
$$n_1 \tan( i_1)= n_2 \tan( i_2)$$
If $\tan (i_1)=\frac {h_o}{ u}$ and $\tan (i_2)=\frac {h_i}{ v}$,then
$$m=\frac {h_i}{h_o}=\mod{\frac {v n_1}{un_2}}$$
Which is the expected expression.
I don't consider orientation of image here.
The point that I confused is why$\tan (i_1)=\frac {h_o}{ u}$ and $\tan (i_2)=\frac {h_i}{ u}$.
I even can't find the height of image and object from this diagram,and how the height is related with the angle $i_1$ and $i_2$.
Can anyone help me?
Thanks in advance.

 

[kuruman]

The point that I confused is why$\tan (i_1)=\frac {h_o}{ u}$ and $\tan (i_2)=\frac {h_i}{ u}$.

I am confused about the same point. I see neither $h_o$ nor $h_i$ in the figure. Can you post a revised figure? If you do, please make sure that it's at the appropriate orientation.

 

[Baal Hadad]

I am confused about the same point. I see neither $h_o$ nor $h_i$ in the figure. Can you post a revised figure? If you do, please make sure that it's at the appropriate orientation.

Here is the situation:
The diagram above is used to derive the refraction formula by my lecturer.After deriving,he directly introduce (without new diagrams ) when $r=infinity$ the right hand term becomes zero and then below the magnification formula.Do these two are related?
Anyway,so I assume the magnification formula is derived from the same diagram as no new diagram is given.
For now,I only found that Pedrotti has the relevant topics but still the diagram is as below attached.But still don't have the image height and object height.
Sorry that I can't find relevant diagrams (or I don't understand which diagram can be used).

 

[Baal Hadad]

The point that I confused is why$\tan (i_1)=\frac {h_o}{ u}$ and $\tan (i_2)=\frac {h_i}{ u}$.

Oh,and sorry for typos.It should be $\tan (i_2)=\frac {h_i}{ v}$.

Well,I found the relevant diagram just next page of the posted page of Pedrotti,sorry for bringing everyone much trouble.
Thank you.