# Phase delay in a thin lens?

Let the maximum thickness of a lens (on its axis) be $\Delta_0$, and let the thickness at coordinates $(x,y)$ be $\Delta(x,y)$.

I have read that phase delay suffered by the wave at coordinates $(x,y)$ in passing through a thin lens may be written as $$\phi(x,y) = kn \Delta(x,y) + k(\Delta_0 - \Delta(x,y)),$$ where $n$ is the refractive index of the lens material, $kn\Delta(x,y)$ is the phase delay introduced by the lens, and $k(\Delta_0 - \Delta(x,y)$ is the phase delay introduced by the remaining region of free space between the two planes. Equivalently the lens may be represented by a multiplicative phase transformation of the form $$t_l = e^{jk\Delta_0} e^{jk(n-1)\Delta(x,y)}.$$

No explanation is provided for where these expression comes from. So why is it that the first expression represents the phase delay in thin lens, and what can the phase transformation function $t_l$ be used for in applications?

The situation is that there is an incoming light wave that is transmitted through a lens. The thickness of the lens is chosen in the z-direction and the cross section of the lens is in the x,y-plane. A plane is defined at one edge of the lens at $z = z_1$ and another plane is defined at the other edge of the lens at $z = z_2$. We want to know the phase difference of the light between the moments that the light enters the plane at $z=z_1$ and the moment the light leaves the plane at $z = z_2$. In general the light will travel partly in the vacuum and partly in the medium of the lens. If the light would be incident exactly at $\Delta_0$ there would only be travel in the lens medium, but in all other situations part of the light path will be in the vacuum and the other part will be in the lens medium. The total difference in phase is equal to the contribution caused by the part in the lens plus the contribution caused by the part in the vacuum.

Let's assume the incoming light can be written as an amplitude function times an exponential, so: $A(r)e^{ikr}$. The phase has to do with the complex exponential, namely $kr$.

Let's make the discussion a bit general and let some light travel a distance between two planes. The electric and magnetic fields of the light oscillate with a period $T$ and a angular frequency $\omega$. The light has a certain wave length $\lambda$. If you look a the light at position $x = x_0$ and a position $x = x_0 + \lambda$ you don't see a difference due to the periodicity of the light. So, if the light travels a certain distance $r$ you need to subtract the wave length $\lambda$ as many times as you can and the difference that is left at the end is interesting for the difference in phase.

Let the light enter the first plane at position $r_1$ with a light phase $\phi_1$ and let the light hit the second plane at position $r_2$ with a light phase $\phi_2$. Then $mod(|r_2 - r_1|, \lambda)$ is the fraction of the period we need to shift. Also $\phi$ is just a number in the interval $\left[0, 2 \pi \right)$. The phase at position 2 is $\phi_2 = \phi_1 + mod(|r_2 - r_1|, \lambda) \cdot 2 \pi$. Thus: $\phi_2 - \phi_1 = mod(|r_2 - r_1|, \lambda) \cdot 2 \pi$. Let's put the phase in a complex exponential, because that is the formula for the light. $e^{i(\phi_2 - \phi_1)} = e^{i( mod(|r_2 - r_1|, \lambda) \cdot 2 \pi)}$. The right hand side can be rewritten using the periodicity of the complex exponential. $e^{i(\phi_2 - \phi_1)} = e^{i( (|r_2 - r_1|/ \lambda )\cdot 2 \pi)}$.

When light enters a medium with a different optical density, such as a lens, the wave length changes but not the frequency. $\lambda_{medium} = \lambda_{vacuum}/n$. The $\lambda$ used in the previous paragraph is $\lambda_{medium}$, so you can replace that $\lambda$ in the previous paragraph with $\lambda_{vacuum}/n$. In the formula's $k$ is the wave vector in vacuum.

The definition of the wave vector: $k = 2 \pi / \lambda_{vacuum}$ leads to $e^{i (\phi_2-\phi_1)} = e^{i k n |r_2 - r_1|}$. What we know now is that when you let some light of the form $A(r)e^{ikr}$ travel a distance $|r_2 - r_1|$ you can multiply with $e^{i k n|r_2 - r_1|}$ to get the correct phase. Let's call this factor $t$. So $t = e^{i k n|r_2 - r_1|}$.

Back to our set-up. The total distance between the two planes $z_2-z_1 = \Delta_0$. Distance travelled through the lens is $\Delta(x,y)$ and the distance travelled through the vacuum is $\Delta_0 - \Delta(x,y)$. $t$ is the factor we need to multiply our light formula with to get the correct phase after the light has travelled through the two planes.

In the vacuum $n=1$. $t$-factor due to vacuum part: $e^{i k \cdot 1 \cdot distance} = e^{i k \cdot (\Delta_0 - \Delta(x,y))}$.

$t$-factor due to lens part: $e^{i k n \cdot distance} = e^{i k n \cdot \Delta(x,y)}$

Total t-factor is: t = $e^{i k n \cdot \Delta(x,y)} \cdot e^{i k \cdot (\Delta_0 - \Delta(x,y))}$

$$t = e^{i k \Delta_0} e^{i k (n-1) \Delta(x,y)}$$

The resulting formula neglects the fact that the light actually bents due to the lens. We assumed that light travelled in a straight line when we said that the distance the light travels in the vacuum is equal to $\Delta_0 - \Delta(x,y)$ and the distance the light travels in the lens is equal to $\Delta(x,y)$. Because the light rays actually bent, the distances are an underestimation and the phase shift is in reality larger. The approximation becomes exact in the limit of an infinitely thin lens.

An application I can think of is that you can make a computer program to calculate interference patterns. For example define a lens with a certain $\Delta(x,y)$. Then calculate the phase difference for light rays incident at different coordinates $(x, y, z_1)$. Based on the phase differences you can work out what the interference pattern looks like for a lens with a certain $\Delta(x,y)$ function.

In practice, I think the phase of the light is only important in applications where interference plays a role. So you might want to use the formula to make an estimation of a phase shift and compare it with a different phase.

• I appreciate the detailed answer! – eurocoder Sep 21 '17 at 5:57
• "I think the phase of the light is only important in applications where interference plays a role" It depends what you mean by interference in that this can be used as the starting point of a method to define the focal point/plane, image position etc for a thin lens? – Farcher Sep 21 '17 at 9:14

The phase delay that you have found is $\phi = kn \Delta(x,y) + k(\Delta_0 - \Delta(x,y)) = k\Delta_0 + k(n-1)\Delta(x,y)$

Suppose you started with a wave of the form $\cos(\omega t)$ after passing through the lens it becomes $\cos(\omega t + \phi)$.

Using complex numbers such a transformation is made easier.

Assume that the incoming wave is the real part of $e^{j\omega t}$ then the outgoing wave is the real part of $e^{j(\omega t+\phi)} = e^{j\phi} e^{j\omega t}$ and what is $e^{j\phi}$?

It is $e^{j(k\Delta_0 + k(n-1)\Delta(x,y))} = e^{jk\Delta_0} e^{k(n-1)\Delta(x,y)}$ your multiplicative phase transformation.

Basically, for a simple monochromatic wave, the phase is given by

$\phi= \omega t - k x + \varphi_0$. Let's ignore the last term by just start counting when it is 0. So the phase is $\omega t - k x$, (or its opposite).

For two waves like that, $\omega$ is the same and the time elapsed is the same. Consequently, the value of $\omega t$ is the same in both waves and that causes no difference. What makes both waves different? The other term: $k x$.

In other words, the delay is $\Delta \phi = \Delta(\omega t - kx) = \stackrel{=0}{\overbrace{\Delta (\omega t)}}-\Delta (kx)\rightarrow\Delta (kx)$

But k is $2\pi/\lambda$, and $\lambda$ varies in different media. But we can use $k_0$ as the $k$ in vacuum (with wavelenght in vaccum). The relation is simple:

$k=n\cdot k_0$

because $k=\omega/v = \omega / (c/n) = \omega/c \cdot n = k_0\cdot n$.

Then, we can take $k_0$ as common factor outside (in your book, it seems that they use $k$ as $k_0$, but be careful that it refers to the one in vaccuum).

In sum, only the Optical path length $(n\cdot x)$ causes the delay. In your book it is $n\cdot \Delta$. Check that, for air, $n=1$ and it doesn't appear, but it is there! haha.

The expression is using the paraxial approximation, i.e. that rays are all propagating within a small angular range. In that case, the distance any ray propagates through the lens is, to a good approximation, simply the thickness $\Delta(x,\,y)$ of the lens, measured along the $z$ direction. The true thickness varies roughly like $\Delta(x,\,y)\,\cos\theta$, where $\theta$ is the angle between the ray and the $z$ axis - the geometry is a bit more complicated in general than this, but the $\cos\theta$ variation is correct up to second order. And, of course, $\cos\theta\approx 1$ to first order.

Given the above paraxial approximations, the calculation runs as follows. The path followed by a ray at transverse position $(x,\,y)$ in reaching the plane $z=\Delta_0$ comprises a thickness $\Delta(x,\,y)$ of lens material, and the complement is a thickness $\Delta_0-\Delta(x,\,y)$ of air. The total phase delay is thus $k\,(\Delta_0-\Delta(x,\,y))$ for the path through air and $k\,n\,(\Delta_0-\Delta(x,\,y))$ for the path through the lens. The sum is the expression you are given. As in the other answers, phase delay $\phi$ corresponds, in phasor notation, to scaling by the unity magnitude quantity $e^{i\,\phi}$, which will yield the second expression.

## protected by Qmechanic♦Aug 17 '17 at 12:01

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).