# Question about the derivation of linear magnification

In summary, the lecturer is trying to derive Snell's law by using a diagram. He introduces the right hand term when ##r=infinity##. Later, he introduces the magnification formula. However, he is still confused about the image height and object height.
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?

#### Attachments

• 15638045496385286667207122650364.jpg
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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.

kuruman said:
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).

#### Attachments

• Screenshot_20190723_065733.jpg
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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.

#### Attachments

• Screenshot_20190723_202236.jpg
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Last edited:
Tom.G

## Question 1: What is linear magnification?

Linear magnification is a measurement of the size change of an object when viewed through a lens or optical instrument. It is the ratio of the size of the image to the size of the object.

## Question 2: How is linear magnification calculated?

Linear magnification is calculated by dividing the height of the image by the height of the object. The formula is M = hi/ho, where M is the magnification, hi is the height of the image, and ho is the height of the object.

## Question 3: What is the difference between linear and angular magnification?

Linear magnification measures the size change of an object, while angular magnification measures the change in angle of the object as seen through a lens. Linear magnification is a ratio of lengths, while angular magnification is a ratio of angles.

## Question 4: How does the focal length affect linear magnification?

The focal length of a lens determines the amount of magnification that will occur. A shorter focal length will result in a larger magnification, while a longer focal length will result in a smaller magnification.

## Question 5: Can linear magnification be negative?

Yes, linear magnification can be negative. This occurs when the image is inverted compared to the object. In this case, the magnification will have a negative value, indicating a decrease in size.

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