Mirror Equations: Derivation of Newtonian Form

In summary: You're doing fine. Now just subtract sosi and sof from both sides, and you're done.In summary, the conversation was about solving an equation involving focal points and center of curvatures in physics. The relationship So = Do - f and Si = Di - f was given, and it was verified that 1/Do + 1/Di = 1/f through experiments. The final conclusion was to find a relationship between f, So, and Si that can be expressed analytically, graphically, and verbally. After some trial and error, the equation SoSi = f^2
  • #1
Zeros
2
0
Hallo,

First time I've experienced a real problem rearding equations in physics, so:

As conclusions to the lab on focal points and and center of curvatures (which will be explained at a later date), we were given the following relationships:

So = Do - f
Si = Di - f

And, we were ably to verify the Gaussian? form of the equation through a series of experiments:

1/Do + 1/Di = 1/f

And yadayada, now the final conclusion for the lab simply asks for a relationship between f, So, and Si that can be expressed analytically, graphically, and verbally.

After wrecking my brain on this question, I eventually checked the following site: http://www.sasked.gov.sk.ca/docs/physics/u3b32phy.html to find this relationship:

SoSi = f^2

However, I obviously cannot go that far, lol. Here is, as best as I can transcribe, how far I got:

So = do -f ----> do = So + f
Si = di -f ----> di = Si + f

then, I substituted those into the equation, getting this stuff:

1/(So +f) - 1/(Si + f) = 1/f

I know for sure (I think, well w/e) that that is the first correct step. After that, I tried a couple of different things, the most prominent being the LCM/LCD approach:

(Si + f) - (So + f)
all over or divided by
(So + f)(Si + f)

all equals = 1/f

then...some simplification:

Si - So
all over or divided by
(So + f)(Si + f)

all equals = 1/f

from there, a smart guy at school (whose done some college level physics...I thinK) told me I could flip the numerator/denominator in both sides of the equation...giving...

(Si + f)(So + f)
all over or divided by
Si - So

all equals = f

(I just put (si + f) in the front b/c it looks a bit better I think)
However, after this we were both rather clueless as to how further simplify the expression...I tried expanding the numerator:

SiSo + Sif + Sof + f^2
all over or divided by
Si - So

all equals = f

but, like, then what? Si, So, f...nothing is common in the numerator anyhow. Eventually I got kinda berserk and tried taking out a Si and an f from the numerator...not pretty:

(Si)(So + f) + (f)(So + f)
all over or divided by
Si - SO

all equals = f

yawn..now what?...mm then this:

(So + f)(Si + f)
all over or divided by
Si - So

all equals = f

and that last part, I'm not even sure about my algebra lol. so yeah, I'm sort of hopeless, and I can't find much besides the aformentioned site for the derviation of this formula, but I think I'm on the right track so if anyone has any ideas...go ahead.

--Zeros
 
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  • #2
Substituting the equations for the D's gives:
[tex]\frac{1}{S_0+f} + \frac{1}{S_i+f}=\frac{1}{f}[/tex]

You seem to have a minus sign wrong, which is probably the source of all your troubles. Try it now.

Since you want to get rid of the denominators, multiply both sides of the equation by [itex]f(S_0+f)(S_i+f)[/itex].
 
  • #3
Zeros said:
However, I obviously cannot go that far, lol. Here is, as best as I can transcribe, how far I got:

So = do -f ----> do = So + f
Si = di -f ----> di = Si + f

then, I substituted those into the equation, getting this stuff:

1/(So +f) - 1/(Si + f) = 1/f
First thing, correct that minus sign. :smile:

1/(So +f) + 1/(Si + f) = 1/f

Just multiply both sides by (So + f)(Si + f)f. Simplify. (Just get rid of the denominators.)

[Galileo beat me to it!]
 
  • #4
Eheh ^^;...that would explain a lot...

*seven minutes later*

O_O;;; Yes, where were we...:

1/(so + f) + 1/(si + f) = 1/f

now I'll take LCD/LCM:

(Si + f) + (So + f)
all over
(Si + f)(So + f)

all equals = 1/f

ok then I'll do what you guys said (needed to do LCD step, right? or else it gets...weird) (oh I'm not going to expand to see if stuff cancels, that's how you do it, right?)

f((si + f) + (so + f))

all equals

(So + f)(Si +f)

...er, now what? simplification/expansion gives the following:

f(si + so + 2f)

all equals

(sosi + sof + sif + f^2)

...yikes. erm, did I sort of screw up somewhere again? But I see, I think it's right, I just don't know what to do from here. But thanks for the help so far!

--Zeros
 
  • #5
Zeros said:
1/(so + f) + 1/(si + f) = 1/f

now I'll take LCD/LCM:

(Si + f) + (So + f)
all over
(Si + f)(So + f)

all equals = 1/f
That's overly complicated. Just multiply both sides by f(si + f)(so + f) right away. The reason is that you have an equation of the form:
1/a + 1/b = 1/c.
If you multiply both sides by abc, you get bc+ac=ab. You've immediately rid yourself of fractions.

Zeros said:
ok then I'll do what you guys said (needed to do LCD step, right? or else it gets...weird) (oh I'm not going to expand to see if stuff cancels, that's how you do it, right?)

f((si + f) + (so + f))

all equals

(So + f)(Si +f)

...er, now what?
You should expand to see if stuff cancels. How else are you going to simplify it?
 

1. What is the purpose of deriving the Newtonian form of mirror equations?

The Newtonian form of mirror equations is derived to mathematically describe the behavior of light as it reflects off of a mirror. This allows us to better understand and predict how an image will be formed by a mirror, which is important in various fields such as optics, astronomy, and imaging technology.

2. How is the Newtonian form of mirror equations derived?

The Newtonian form of mirror equations is derived using basic principles of geometric optics and the laws of reflection. By analyzing the angles and distances involved in a simple mirror setup, we can derive the mathematical equations that describe the relationship between the object, the mirror, and the resulting image.

3. What are the assumptions made when deriving the Newtonian form of mirror equations?

When deriving the Newtonian form of mirror equations, we assume that the mirror is perfectly flat and smooth, and that light travels in straight lines. We also assume that the object and the observer are relatively small compared to the distance between them and the mirror. These simplifying assumptions allow us to create a more straightforward and accurate mathematical model.

4. How are the mirror equations used in practical applications?

The mirror equations are used in various practical applications, such as designing and optimizing mirrors for telescopes, microscopes, and other optical instruments. They are also used in the field of computer graphics to create realistic reflections in images and animations. Additionally, the mirror equations are used in everyday situations, such as adjusting mirrors in a car for optimal visibility.

5. Are there any limitations to the Newtonian form of mirror equations?

While the Newtonian form of mirror equations is a useful tool in understanding the behavior of light in simple mirror setups, it does have some limitations. It does not take into account the effects of diffraction and interference, which are important in more complex optical systems. Additionally, it only applies to mirrors with a flat surface and cannot be used to describe the behavior of curved mirrors.

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