4
$\begingroup$

Can someone help me or guide me how the thin lens formula:

$$\frac{1}{s_1}+\frac{1}{s_0}=\frac{1}{f}$$

can be proven?

I was trying to prove it on my own using similar triangles, only to fail.

$\endgroup$
  • 3
    $\begingroup$ learnquebec.ca/en/content/curriculum/mst/opticks/chapter5/… $\endgroup$ – John Rennie Nov 7 '13 at 18:38
  • $\begingroup$ I found a full animated derivation here. $\endgroup$ – Keep these mind Nov 7 '13 at 19:02
  • $\begingroup$ Off the top of my head I believe that the typical proof only applies in the limit that the distances are much larger than both the radii of curvature of the focusing element and the heights of both image and object. When the limits are broken you get spherical aberration which is a real problem with spherical focusing elements. $\endgroup$ – dmckee Nov 7 '13 at 21:11
  • $\begingroup$ @hyg17 I noticed your user page. Have you read "Galois theory" by Emil Artin, or, best of all, seen the American Mathematical Monthly paper John Stillwell, "Galois Theory for Beginners", Jan 1994? I should think you'd be interested if you haven't already read them. $\endgroup$ – WetSavannaAnimal Nov 8 '13 at 9:55
  • $\begingroup$ Thank you so much for your help everyone. I was just swamped in work and did not have time to come see this because Physics is more of an extra curricular activity to me. $\endgroup$ – hyg17 Nov 12 '13 at 22:35
6
$\begingroup$

Further to the neat and standard similar triangles argument given by John Rennie's link http://www.learnquebec.ca/en/content/curriculum/mst/opticks/chapter5/5_eqlent.html there are two elegant approaches that are easy to recall.

Note that the thin lens formula is a paraxial formula: it only applies to light fields made up of plane waves skewed at small angles both to the nominal propagation direction, taken to be the $z$ direction in the following, and to any interface that the light field may meet. Under these conditions, $\sin\theta \to \theta$ for any angle between a ray and a surface unit normal Snell's law becomes $n_1\,\theta_1 = n_2\, \theta_2$


Method 1: Spherical Waves and their Interaction with Refracting Surfaces

Label $\mathbb{R}^3$ by Cartesian co-ordinates $x,y,z$ and consider a perfect point source at $(0,0,-R)$ a distance $R$ from the $x-y$ plane of constant axial co-ordinate $z=0$. The scalar field (i.e. any Cartesian component of the $\vec{E}$ or $\vec{H}$ fields or, in Lorentz gauge, potentials $\vec{A}$ or $\phi$) is a spherical wave by the time it reaches $z=0$ (strictly, one needs to assume $R\gg k\,s^2$ where $s$ is the diffraction limited spotsize of the point source and $k$ the wavenumber), but intuitively you can see that, far enough from the source, the source will be radiating spherical waves.

Therefore, the field's phase $\phi$ on the plane $z = 0$ varies like:

$$\phi(x,y) = k \,\sqrt{R^2 + x^2 + y^2}-k\,R\approx k\,\frac{x^2+y^2}{2\,R}\qquad(1)$$

the approximation being good if the field is paraxial, i.e. the radius of the support of $\phi$ in the plane is much less than $R$.

Now define a thin lens as an element that adds a curvature to its input field without significant diffractive effects. Its action is thus like a phase mask. A converging thin lens's transformation on the input field $\psi(x,y)$ is:

$$\psi(x,y)\to \exp\left(-i\,k\,\frac{x^2+y^2}{2\,f}\right)\psi(x,y)\qquad(2)$$

To understand the parameter $f$, we think of a plane wave (curvature of nought) with wavefront parallel to the $x-y$ plane (propagation exactly along the $z$ axis) as the input. The output field's phase on the $x-y$ plane is thus:

$$\tilde{\phi}(x,y) = -k\,\frac{x^2+y^2}{2\,f}\qquad(3)$$

By comparing (1) and (3) and noting the parameters $R$ and $f$ and their signs in the phase denominators, we see that the output field is that arising from a diffraction limited point field which is at $z = --f = +f$. So therefore $f$ is the thin lens's focal length, its sign setting whether we have a converging or diverging lens. So, if the field with phase (1) is input to the thin lens, when we impart transformation (2) on a field with phase (1) we get a field with phase:

$$\phi_{out}(x,y) = k\,\left(x^2+y^2\right)\left(\frac{1}{2\,R}-\frac{1}{2\,f}\right)\qquad(4)$$

again, comparing (4) with (1), we see that this is a field that arises from a point source at position $z=R^\prime$ where:

$$-\frac{1}{R^\prime} = \frac{1}{R}-\frac{1}{f}\qquad(5)$$

whence straight away follows the thin lens formula. QED


Method 2: Transformer Matrix Method

For an axisymmetric (invariant wrt rotation about the optic axis, here assumed to be the $z$ axis) system we can represent ray behavior through any plane through the optic axis. So any ray through a homogeneous medium has the linear equation $y = \theta z + \delta$ where $\theta$ (angle wrt $z$ axis) and $\delta$ are real constants characterizing the ray. Suppose a ray meets a refracting interface: it leaves the interface as another ray with a different direction, characterised by new $\theta^\prime$, $\delta^\prime$. Moreover, if the ray passes through any sequence of refracting surfaces or reflecting surfaces spaced by homogeneous mediums, the transformation wrought on the ray parameters $\theta$, $\delta$ is a homogeneous linear transformation because Snells law is paraxially linear in $\theta$ as noted in the introduction and the law of reflection is also linear in $\theta$. Note that this is true for a given ray, but in general the linear transformation depends on where (the value of $y$) the ray enters the transforming system. However, we either argue that, (1) to first order in $y$, the transformation does not depend on the entry point or (2) we design our refracting surfaces so that the transformation does not depend on the entry point.

So this means for these idealized, paraxial systems, a refracting/ reflecting system is characterized by a $2\times2$ real matrix that transforms the column vector $\left(\begin{array}{c}\theta\\\delta\end{array}\right)$ defining the input ray to $\left(\begin{array}{c}\theta^\prime\\\delta^\prime\end{array}\right)$ defining the output. It is customary to shift the $z$ axis origin to the input / output of each successive ray transformer so that the column vectors for the input assume that the input plane is $z=0$ and those for the output assume that $z=0$ is the output - we can define the transformation matrix accordingly.

So now we work out the matrices of the basic subsytems. A homogeneous medium does not transform the angle $\theta$ but it shifts a ray's height by $\theta\,\Delta z$; therefore the matrix for a medium of width $\Delta z$ is:

$$H(\Delta z) = \left(\begin{array}{cc}1&0\\\Delta z&1\end{array}\right)\qquad(6)$$

Now for the thin lens: $y$ does not shift, but an axially propagating ray ($\theta = 0$) is bent so that it passes through the optical axis a distance $f$ beyond the transformer's output plane; likewise a ray with $\theta = \delta/f$ (passing through the focal point a distance $f$ before the input) is mapped to an axially propagating ray, therefore:

$$\left(\begin{array}{c}0\\\delta\end{array}\right)\mapsto\left(\begin{array}{c}-\delta/f\\\delta\end{array}\right);\quad\left(\begin{array}{c}\delta/f\\\delta\end{array}\right)\mapsto\left(\begin{array}{c}0\\\delta\end{array}\right)\qquad(7)$$

and so the transformer matrix for the converging thin lens ("positive optical power" $1/f$) with focal length $f$ is:

$$T(f) = \left(\begin{array}{cc}1&-f^{-1}\\0&1\end{array}\right)\qquad(8)$$

So now we work out the transformer matrix for the system described by a thin lens: a homogenous medium of width $s_0$ followed by thin lens of focal length $f$ followed by homogenous medium of width $s_1$; we compose the ray transformations by multiplying the relevant matrices:

$$H(s_1)\, T\left(\left(s_0^{-1}+s_1^{-1}\right)^{-1}\right)\, H(s_0) = \left( \begin{array}{cc}-\frac{s_0}{s_1} & -\frac{s_0+s_1}{s_0\,s_1} \\0 & -\frac{s_1}{s_0} \\\end{array}\right)\qquad(9)$$

whereunder the family of rays through the optical axis characterized $\theta = \theta_0\,\delta = 0$ is mapped to another family of ray through the optical axis $\theta = -s_0\,\theta_0/s_1\,\delta = 0$ (i.e. a focal point is mapped to a focal pojnt) and moreover the system magnification is the reciprocal $-s_1/s_0$ of the angular scaling at the matrix element $(1,1)$. QED.

What's interesting about this method is that (6) and (8) are unimodular, so therefore so are all their products, so any system of thin lenses spaced by homogenous mediums also has a unimodular transformation matrix: if some of the spacing mediums are not vacuums, we represent the transition into them by Snell's law, which in this notation means a flat interface orthogonal to the optical axis transforms by:

$$R(n_1,n_2) = \left(\begin{array}{cc}n_2/n_1&0\\0&1\end{array}\right)\qquad(10)$$

which is not unimodular, but as long as we end up with the inputs and outputs in the same medium, the matrix product includes $R(n_1,n_2), R(n_2,n_3), R(n_3,n_4)\cdots R(n_N,n_1)$, whose determinants all multiply to yield 1. So our system still has a unimodular matrix. With a little more work, you can show that there is a paraxial system to realize any member of the unimodular group $SL(2,\mathbb{R})$. This is most readily done by understanding that the "infinitesimal" versions of (Lie algebra members in $\mathfrak{sl}(2,\mathbb{R})$ corresponding to) (6) and (8) are:

$$h = \left(\begin{array}{cc}0&0\\1&0\end{array}\right);\quad t=\left(\begin{array}{cc}0&1\\0&0\end{array}\right);\quad i.e.\;H(\Delta z) = \exp(\Delta z \, h);\quad T(f) = \exp\left(-\frac{1}{f}\,t\right);\qquad(11)$$

and that:

$$[t,h] = s = \left(\begin{array}{cc}1&0\\0&-1\end{array}\right);\quad [s,t] = 2\,t;\quad [s,h] = -2\, h\qquad(12)$$

so that $s, t$ and $h$ span the whole Lie algebra $\mathfrak{sl}(2,\mathbb{R})$, therefore there is a finite product of matrices of the form (6) and (8) to realize any member in the identity-connected component of $SL(2,\mathbb{R})$, which is simply the whole unimodular group. If we add a reflecting system $\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)$ (flips an image and has determinant -1) to the mix as well scale factors to the determinant through an "unbalanced" $R(n_{in}, n_{out})$ arising from a different input and output medium, we can indeed realize any member of the whole group $GL(2,\mathbb{R})$ of invertible $2\times2$ matrices as a paraxial optical ray transformer.


Afternotes to Method 1

Note that you can also get the system magnification and the lateral positions of images relative to sources if you do method 1 again assuming an offset phase mask, i.e. by replacing (2) by:

$$\psi(x,y)\to \exp\left(-i\,k\,\frac{(x-x_0)^2+(y-y_0)^2}{2\,f}\right)\psi(x,y)\qquad(13)$$

and interpreting the linear terms of the output phase appropriately as lateral co-ordinates of the image point.

This method assumes perfectly unaberrated waves. However, witness that any lens whose surface sag (height) near its vertex can be described by an analytic function of distance $r$ from the optical axis (axis of symmetry) will impart a phase mask well described by the above in the paraxial limit, i.e. by limiting the numerical aperture (maximum skew angle) in any field such that the support of any field in the transverse plane $z=0$ is small enough. As long as the width of the support needed to validate the analysis above stays big compared with a wavelength, the analysis above will be valid in the paraxial limit. This means it works in the paraxial limit for all lenses of practical curvatures in the neighbourhood of the optical axis.

You can also dig deeper into the scalar wave theory behind the thin lens equation with the theory of Gaussian beams. You begin with the Helmholtz equation in a homogeneous medium $(\nabla^2 + k^2)\psi = 0$. If the field comprises only plane waves in the positive $z$ direction then we can represent the diffraction of any scalar field on any transverse (of the form $z=c$) plane by:

$$\begin{array}{lcl}\psi(x,y,z) &=& \frac{1}{2\pi}\int_{\mathbb{R}^2} \left[\exp\left(i \left(k_x x + k_y y\right)\right) \exp\left(i \left(k-\sqrt{k^2 - k_x^2-k_y^2}\right) z\right)\,\Psi(k_x,k_y)\right]{\rm d} k_x {\rm d} k_y\\ \Psi(k_x,k_y)&=&\frac{1}{2\pi}\int_{\mathbb{R}^2} \exp\left(-i \left(k_x u + k_y v\right)\right)\,\psi(x,y,0)\,{\rm d} u\, {\rm d} v\end{array}\qquad(14)$$

In words:

  1. Take the Fourier transform of the scalar field over a transverse plane to express it as a superposition of scalar plane waves $\psi_{k_x,k_y}(x,y,0) = \exp\left(i \left(k_x x + k_y y\right)\right)$ with superposition weights $\Psi(k_x,k_y)$;
  2. Note that plane waves propagating in the $+z$ direction fulfilling the Helmholtz equation vary as $\psi_{k_x,k_y}(x,y,z) = \exp\left(i \left(k_x x + k_y y\right)\right) \exp\left(i \left(k-\sqrt{k^2 - k_x^2-k_y^2}\right) z\right)$;
  3. Propagate each such plane wave from the $z=0$ plane to the general $z$ plane using the plane wave solution noted in step 2;
  4. Inverse Fourier transform the propagated waves to reassemble the field at the general $z$ plane.

Now we make the paraxial approximation to the propagation relationship in step 2 above, i.e. we assume that the plane waves aren't skewed at too steep angles relative to the $z$ axis so that $k_x^2+k_y^2 \ll k^2$. Then our two propagation equations above become the Fresnel propagation integral:

$$\begin{array}{lcl}\psi(x,y,z) &=& \frac{1}{2\pi}\int_{\mathbb{R}^2} \left[\exp\left(i \left(k_x x + k_y y\right)\right) \exp\left(i \frac{k_z^2+k_y^2}{2\,k} z\right)\,\Psi(k_x,k_y)\right]{\rm d} k_x {\rm d} k_y\\ \Psi(k_x,k_y)&=&\frac{1}{2\pi}\int_{\mathbb{R}^2} \exp\left(-i \left(k_x u + k_y v\right)\right)\,\psi(x,y,0)\,{\rm d} u\, {\rm d} v\end{array}\qquad(15)$$

Now witness that a beam $\psi(x,y,0)$ with Gaussian dependence on $x$ and $y$ becomes, under the Fresnel diffraction integral, $\psi(x,y,z)$ with Gaussian dependence on $x$ and $y$. The Fourier transform of a Gaussian is a Gaussian, a fact which does not change when we multiply by the $\exp\left(i \frac{k_z^2+k_y^2}{2\,k} z\right)$ kernel in the Fresnel diffraction integral, and of course a Gaussian is recovered by the inverse Fourier transform. Moreover, the transformation of the phase mask in (2) and (13) maps a Gaussian beam into a Gaussian beam. Best of all, for Gaussian beams the integrals split up into a product of separate $x$ and $y$ dependences and these separate dependences are also operated on independently by the phase masks (2) and (13), so propagation analysis can be done as a product of two decoupled one-transverse dimensional diffraction problems. Therefore the propagation through a system comprising homogeneous, diffractive mediums and thin lenses characterized by (2) and (13) can be done wholly in closed form expressions (use Mathematica though!) and you end up with something slightly more general than the thin lens formula that becomes the thin lens formula when the axial distances involved become longer than the Rayleigh diffraction length (of the order of wavelengths).

$\endgroup$
2
$\begingroup$

I know this formula can be derived in 2 ways, and not sure of the best one for you, so I will show both of them. Summed up I would call them "Thin lens method" and "two surface method" or "two index method"

The thin lens method is most simple, so I will start with that...

Assume the thin lens (magically) turns the light at a changing slope that increases the further one enters away from the center of the lens, if you cannot do this, move on to the "two surface method". In the real world the rate of changing slope as a ray enters further away from the center of the lens depends on a lot of things, but in this case let's just say this is linear. So let's use "$r$" to represent the distance from the center of the lens that the light ray enters and "$m$" to represent the slope of entry relative to the normal of the lens. I use "$P$" to represent the constant that "$r$" should be multiplied by to get the value of the slope change. So if we have an object at "$S_1$" from the lens and "$R$" from the center axis, we know one ray passing through the center of the lens will have no slope change, let's call this the "chief ray". We can know the slope of a different ray with this $$ m_{in} = { R - r \over S_1} $$ Now this slope is changed with "$r$" and "$P$"... $$ m_{out} = m_{in} + r * P = { R - r \over S_1} + r * P $$ Remember $$ m_{chief}={R \over S_1} $$ So given the new slope, we will observe an intersection with the "chief ray" at: $$ S_0 = {r \over \left( m_{out}-m_{chief} \right)} $$ $$ {1 \over S_0}={m_{out} \over r}-{m_{chief} \over r} ={{ R - r \over S_1} + r * P \over r}-{R \over r * S_1}={{ R - r \over r * S_1} + P }-{R \over r * S_1} $$ Now we can know the relation between "$P$", "$S_1$", and "$S_2$" is: $$ {1 \over S_0}+{R \over r * S_1}-{{ R - r \over r * S_1} } = P ={1 \over S_1} + {1 \over S_0} $$ Thus: $$ P={1 \over f} $$ $$ \space $$

The "two surface method" uses the index of refraction properties, so unless Snell's Law has been covered, this method probably cannot be. Also keep in mind an ideal lens is made with parabolic curves, though most real lenses use spherical surfaces. Mostly for ease of manufacturing and compatibility with other lenses(also spherical), this is why we get spherical aberrations. I am not going to try and explain the math behind spherical lenses, as they do not work like ideal lenses. So we start with Snell's Law and move toward defining one over the focal length.

If we are using a parabolic lens than we know the surface has a slope relative to the line perpendicular to the lens of: $$ m_{s1} = {1 \over r} \space \space \space \space \space \space and \space on \space the \space other \space side \space \space \space \space \space \space m_{s0} = {- 1 \over r} $$ Using the same variables as before: $$ m_{in} = { R - r \over S_1} $$ $$ m_{chief}={R \over S_1} $$ Only this time a difference between the distance from the object to the actual lens surface and the distance "$S_1$" exists. Just to note that fact. We will say that the plane, of which "$S_1$" is the distance to from the object, is the plane an ideal lens would be placed at to simulate the same result as this real system with an index "$n$". We will also assume air has an index of one. For the "chief ray" we know the ray should be directly perpendicular to the surface. Thus this ray will exit at a point on the lens that also has a surface perpendicular to the ray.

So using Snell's Law we can find the new slope of the ray as it is inside the lens: $$ {\sin\left(\theta_{going-in}\right) \over \sin\left(\theta_{inside}\right)} = {1 \over n} \space \space \space \space \space \space \rightarrow \space \space \space \space \space \space {{\sin\left(\arctan\left(m_{s1}\right)-\arctan\left(m_{in}\right)\right) \over \sin\left(\arctan\left(m_{inside}\right)-\arctan\left({-1 \over m_{s1}}\right)\right)}} = {1 \over n} $$ OK, so let's make that a bit more solvable, by assuming this ray is directly perpendicular to what would be the thin lens as it enters this lens (basically "$R=r$" and "$m_{in}=0$"): $$ \sin\left(\arctan\left(m_{inside}\right) - \arctan\left({-1 \over m_{s1}}\right)\right) = \sin\left(\arctan\left(m_{s1}\right)\right) * n $$ Now to find the exit angle with the same Snell's Law: $$ {\sin\left(\theta_{inside}\right) \over \sin\left(\theta_{going-out}\right)} = {n \over 1} \space \space \space \space \space \space \rightarrow \space \space \space \space \space \space n={{\sin\left(\arctan\left(m_{inside}\right)\right) \over \sin\left(\arctan\left({{r+m_{chief}*S_0} \over S_0}\right)\right)}} $$ Now we can start getting something from: $$ n*\sin\left(\arctan\left({{r+m_{chief}*S_0} \over S_0}\right)\right)=\sin\left(\arctan\left(m_{inside}\right)\right) $$ using $$ \arctan\left(m_{inside}\right) = \arcsin\left( \sin\left(\arctan\left(m_{s1}\right) * n\right) + \arctan\left({-1 \over m_{s1}}\right)\right) $$ $$ n*\sin\left(\arctan\left({{r+m_{chief}*S_0} \over S_0}\right)\right)=\sin\left(\arcsin\left( \sin\left(\arctan\left(m_{s1}\right)\right) * n\right) + \arctan\left({-1 \over m_{s1}}\right) \right) $$ $$ n*\sin\left(\arctan\left({R \over S_0}+{R \over S_1}\right)\right)=\sin\left(\arcsin\left( \sin\left(\arctan\left({1 \over R}\right)\right) * n\right) + \arctan\left({-R}\right) \right) $$ $$ \arcsin\left(n*\sin\left(\arctan\left({R \over S_0}+{R \over S_1}\right)\right)\right)=\arcsin\left( \sin\left(\arctan\left({1 \over R}\right)\right) * n\right) + \arctan\left({-R}\right) $$ $$ \sin\left(\arctan\left({R \over S_0}+{R \over S_1}\right)\right)= \sin\left(\arctan\left({1 \over R}\right)\right) + {\sin\left(\arctan\left(-R\right) \right) \over n} $$ OK so that is a pain, but I think you can now see the "${1 \over S_0}+{1 \over S_1}$" forming with a trigonometric relation to "$n$". I would recommend using Jone's Vectors, but I suppose this is meant to be a proof, Good Luck.

$\endgroup$
  • $\begingroup$ +1 Luke the actual shape doesn't invalidate your proof and doesn't matter for the paraxial approximation, you can assume spherical and then argue that we need to keep the numerical aperture small enough that the lens differs negligibly from the spherical one over the transverse support of the field. Moreover, standard lens grinding technology with "random" rotation of the grinding cup about two independent axes gives you spherics. See the last paragraph of my answer. $\endgroup$ – WetSavannaAnimal Nov 8 '13 at 10:12
  • $\begingroup$ George, I think you may have been right. Like Rod Vance has said the change in slope of the lens over "r" makes no difference to the proof by much. I was trying to make it simple, but by bringing up the topic I probably made it more complex. $\endgroup$ – Luke Burgess Nov 8 '13 at 14:28
1
$\begingroup$

A lot depends on how much you want to learn, just to teach elementary (geometrical) optics. There are many good books, but often they cover too much that you aren't interested in.

By far the best textbook on actual geometrical optics, relating to lens design, including aberrations etc, is one that as a mathematician, you would enjoy for yourself.

Applied Optics & Optical design, By A. E. Conrady written in 1926 , taught all of the world's good optical designers of WW-II, on all sides of the conflict. It's a two volume Dover Press paper back, that you can buy at Amazon, or B&N for peanuts. And it is fully mathematical, but high school level.

High school, algebra, geometry, and trigonometry required. NO calculus needed, and it will show you lens formulae that are much more useful, than the one you give above.

I should add, that all of the computations are done with logarithm tables, which was the best in those days, but now you simply use a calculator.

I managed to convince a good number of employers for over 50 years, that I knew a little bit of optics; solely because of Conrady.

His daughter, Hilda Conrady Kingslake,, was the wife of chief Kodak optics guru, Rudolph Kingslake, and she was for years, historian for the Optical Society of America. Sadly, both are now doing optics in the clouds.

$\endgroup$
1
$\begingroup$

Well I was hoping somebody else would present this high school solution, since I don't have any way of posting drawings or graphs etc. But you should be able to follow this.

In my sign convention (here) all quantities are positive.

......oh|<----x---->|<---f--->|<---f'-->|<--y-->|ih'.....

.............<---------l---------> <------l'------->|........

So my object (o) of height (h) is distant (x) from the front focal point, which is distant (f) from the thin lens. And we have x + f = l

From the thin lens to the back focal point is (f') where f and f' are equal. My image (i) of height h' is distant (y) from the back focal point, and y + f' = l'

Now I need another line which you will have to imagine. It goes from the top of the object (oh) at height (h) , through the front focal point, and continues to the lens, at a height (down) h.f/x = h.f/(l-f)

Now this ray passed through the focal point, and therefore it refracts parallel to the axis, and must pass through the top of the image, at height h' (down)

Ergo h' = h.f/x = h.f/(l-f)

By symmetry, if I reverse everything, I can write:-

 h = h'.f'/y = h'f'/(l'-f')..... = h.f/x .f'/y = h.f/(l-f) .f'/(l'-f')  

By substituting for h and h' in the two sets of equations.

Then one can see that: h = h.f/x.f'/y or x.y = ff' = f^2

This is Newton's thin lens formula. Then from the other equations I have:

 h = h.f/(l-f).f'/(l'-f') , so f.f' = (l-f).(l'-f') = l.l'- f(l + l') + f.f' ;(f'=f)

So finally l.l' = f(l + l') or 1/f = l/l.l' + l'/l.l' = 1/l + 1/l'

QED

Sorry to the mathematicians for dragging out what you can do in your head. OP should take note of Newton's formula, xy = f^2 x and y measured from the focus.

Notice that x = y = f gives l + l' = 4f which is the minimum object to image distance.

Sorry about the bush algebra; I can write it with a #2 pencil in 1/10th the time.

$\endgroup$
  • $\begingroup$ Dear everyone who has posted an answer. If I could, I would give every one of you points for helping me out, but I am unfortunately not as strong as any of you in physics as I am a mathematics major, so it's completely another language to me. However, I was able to go through the basic geometric proof and it made sense and that was all I needed so far. I can at least see that there is much more into it than I have ever imagined, and I have a much greater respect towards this subject now. Thank you very much. $\endgroup$ – hyg17 Nov 12 '13 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.