17. Lens Performance

We want to calculate the intensity distribution at the image for an arbitrary object and optical system.

In linear system theory we have the concept of the impulse response of the system.

A linear system’s impulse response.

\(h(t)\) is the impulse response of the system. It is calculated by determining the output if the input is a \(\delta\) function.

We want to extend this to 2 dimensional optical systems.

17.1. Point Spread Function

2-dimensional response of an optical system.

In the above figure,

• \(f(x,y)\) is an arbitrary object.

• \(g(x,y)\) is the image of the object at the image plane.

Point Spread Function (PSF)

The 2D impulse respone of the optical system.

The PSF is used to characterize an optical system.

The object is a delta function:

\[ E_{obj} = \delta (x) \delta (y) \delta (z + d_0) \]

Optical delta object.

First, we’ll calculate the field of the object right before the lens.

\[\begin{split} \begin{align} E_1(x,y) &= \frac{e^{j k d_0}}{j \lambda d_0} \int \int _{-\infty} ^\infty \delta (x') \delta (y') \exp \left[ -\frac{j k}{2 d_0} \left( (x-x')^2 + (y-y')^2 \right) \right] dx' dy' \\ &= \frac{e^{j k d_0}}{j \lambda d_0} \exp \left[ -\frac{j k}{2 d_0} (x^2 + y^2) \right] \end{align} \end{split}\]

Now we’ll calculate the field immediately after the lens. Apply the quadratic phase shift:

\[\begin{split} \begin{align} E_2(x,y) &= E_1(x, y) \exp \left[ \frac{j k}{2 f} (x^2 + y^2) \right] \\ &= \frac{e^{j k d_0}}{j \lambda d_0} \exp \left[ -\frac{j k}{2 d_0} (x^2 + y^2) + \frac{j k}{2 f} (x^2 + y^2) \right] \\ &= \frac{e^{j k d_0}}{j \lambda d_0} \exp \left[ \frac{j k}{2} \left(\frac{1}{f} - \frac{1}{d_0} \right) (x^2 + y^2) \right] \end{align} \end{split}\]

Use a Fresnel diffraction integral to calculate the field at the image plane (\(z = d_i\)).

Since in the end we only care about relative magnitude, we’ll drop the common phase and amplitude terms.

\[ E_f(x,y) = \int \int _\sum \exp \left[ \frac{j k}{2} \left( \frac{1}{f} - \frac{1}{d_o} \right) (x'^2 + y'^2) \right] \exp \left[ - \frac{j k}{2 d_i} \left( (x-x')^2 + (y-y')^2 \right) \right] dx' dy' \]

Using the conjugate equation (4.5) from geometrical optics,

\[ \frac{1}{f} - \frac{1}{d_o} = \frac{1}{d_i} \]

\[\begin{split} \begin{align} E_f(x,y) &= \int \int _\sum \exp \left[ \frac{j k}{2 d_i} (x'^2 + y'^2) \right] \exp \left[ - \frac{j k}{2 d_i} \left( x'^2 + y'^2 \right) \right] \exp \left[ - \frac{j k}{2 d_i} (x^2 + y^2) \right] \exp \left[ \frac{j k}{d_i} (x x' + y y') \right] dx' dy' \\ &= \exp \left[ - \frac{j k}{2 d_i} (x^2 y^2) \right] \int \int _\sum \exp \left[ \frac{j k}{2} (x'^2 + y'^2) \left( \frac{1}{d_i} - \frac{1}{d_i} \right) \right] \exp \left[ \frac{j k}{d_i} (x x' + y y') \right] dx' dy' \\ &= \mathrm{PSF}(x,y) = \int \int _\sum \exp \left[ \frac{j k}{d_i} (x x' + y y') \right] dx' dy' \end{align} \end{split}\]

If the lens is perfect,

(17.1)\[ \mathrm{PSF}(x,y) = \mathfrak{F} \left. \left\{ \textrm{Aperture} \right\} \right|_{f_x = x / \lambda d_i, f_y = y / \lambda d_i} \]

\[ \mathrm{PSF}(x,y) = \mathfrak{F} \left. \left\{ \textrm{Aperture} \left( \text{difference from spherical wave} \right) \right\} \right|_{f_x = x / \lambda d_i, f_y = y / \lambda d_i} \]

Huygen’s principle:

(17.2)\[ E_i(x,y) = E_0(x, y) \otimes \mathrm{PSF}(x, y) \]

17.2. Modulation Transfer Function

Often what we want with an optical system is to determine the contrast of an imaging system.

See www.normankoren.com/Tutorials/MTF.html.

The contrast is

(17.3)\[ \gamma = \frac{E_{max} - E_{min}}{E_{max}+E_{min}} \]

It is more practical to use areance,

(17.4)\[ a = \frac{E_{max} - E_{min}}{2} + E_{min} \]

and variance of areance,

\[ b = E_{max} - a \]

The ratio is the modulation \(M = \frac{b}{a}\).

The modulation transfer function is

(17.5)\[ T = \frac{M_{image}}{M_{object}} \]

\(T\), as a function of spatial frequency, is

\[ \mathrm{Modulation} \mathrm{Transfer} \mathrm{Function} \quad \mathrm{MTF} = T(f) \]

Spatial frequency of an image

The number of lines or other detail within a given length, usually written in units of inverse mm.

17.2.1. Calculating MTF

MTF is the angular spectrum of the optical system.

Since most imaging systems use incoherent light, we use the power PSF

\[ \mathrm{PPSF} = |\mathrm{PSF}|^2 = \mathrm{PSF}(x,y) \mathrm{PSF}^*(x,y) \]

\[ \begin{equation} \mathrm{MTF} = | \int \int_{-\infty}^\infty \mathrm{PSF}(x,y) \mathrm{PSF}^*(x,y) \exp \left[ -j 2 \pi (f_x x + f_y y) \right] | \end{equation} \]

where

\[\begin{split} \begin{align} \mathrm{PSF}(x,y) &= \int \int_{-\infty}^\infty A(x_1, y_1) \exp \left[ j \frac{k}{f} (x_1 x + y_1 y) \right] dx_1 dy_1 \\ \mathrm{PSF}^*(x,y) &= \int \int_{-\infty}^\infty A^*(x_2, y_2) \exp \left[ -j \frac{k}{f} (x_2 x + y_2 y) \right] dx_2 dy_2 \end{align} \end{split}\]

\[ \mathrm{MTF} = | \int \int_{-\infty}^\infty A(x_1, y_1) A^*(x_2, y_2) \exp \left[ j \left( \frac{k x_1}{f} - \frac{k x_2}{f} - 2 \pi f_x \right) x + j \left( \frac{k y_1}{f} - \frac{k y_2}{f} - 2 \pi f_y \right) y \right] | \]

To eliminate the exponential, let

\[\begin{split} \begin{align} \frac{k x_1}{f} &= \frac{k x_2}{f} + 2 \pi f_x \\ x_1 &= x_2 + \frac{2 \pi f}{k} f_x \\ x_1 &= x_2 + \lambda f f_x \\ y_1 &= y_2 + \lambda f f_y \end{align} \end{split}\]

\[ \mathrm{MTF} = | \int \int_{-\infty}^\infty A(x_2 + \lambda f f_x, y_2 + \lambda f f_y) A^*(x_2, y_2) dx_2 dy_2 | \]

or, more simply

\[ \mathrm{MTF} = | \int \int_{-\infty}^\infty A(\hat{\alpha} + \alpha, \hat{\beta} + \beta) A^*(\hat{\alpha}, \hat{\beta}) d\hat{\alpha} d\hat{\beta} | _{\alpha = f_x \lambda f, \beta = f_y \lambda f} \]

Example: MTF of a perfect lens

Let’s calculate the MTF of a perfect lens with a circular aperture.

\[\begin{split} A(x,y) = \left\{ \begin{matrix} 1 & \sqrt{x^2 + y^2} \leq R \\ 0 & \mathrm{otherwise} \end{matrix} \right. \end{split}\]

The MTF is the overlap between two circles.

\[ \mathrm{MTF} = | \int \int_{-\infty}^\infty A(x+\alpha, y+\beta) A^*(x, y) dx, dy | \]

since the aperture function is purely real, \(A = A^*\).

The MTF is the overlap between two circles.

The overlapping area of the circles.

The shaded area is the area of \(A_1\) (the area of the sector of the circle) minus the area of \(A_2\), which is the area of the triangle.

Knowing that \(\Delta = \alpha / 2\), \(\theta = \cos^{-1} \Delta / R\), we get as the area of the sector

\[\begin{split} \begin{align} A_1 &= 2 \int_0^R \int_0^{\cos^{-1}(\Delta / R)} \rho d\rho d\phi \\ &= \rho^2 \theta |_0^R \\ &= R^2 \cos^{-1} \left( \frac{\Delta}{R} \right) \end{align} \end{split}\]

The area of the triangle is

\[\begin{split} \begin{align} A_2 &= (2) \left( \frac{1}{2} \right) (\Delta) \sqrt{R^2 - \Delta ^2} \\ &= R^2 \left[ \cos^{-1} \left( \frac{\Delta}{R} \right) - \frac{\Delta}{R} \sqrt{1 - \left( \frac{\Delta}{R} \right) ^2} \right] \end{align} \end{split}\]

\[ \mathrm{MTF} = 2A = 2 R^2 \left[ \cos^{-1} \left( \frac{\Delta}{R} \right) - \frac{\Delta}{R} \sqrt{1 - \left( \frac{\Delta}{R} \right) ^2} \right] \]

Given that \(\Delta = \alpha / 2\),

\[ \mathrm{MTF} = 2 R^2 \left[ \cos^{-1} \left( \frac{\alpha}{2 R} \right) - \frac{\alpha}{2 R} \sqrt{1 - \left( \frac{\alpha}{2 R} \right) ^2} \right] \]

where \(\alpha = f_r \lambda f\).

Normalizing yields

(17.6)\[ \mathrm{MTF} = \left( \frac{2}{\pi} \right) \left[ \cos^{-1} \left( \frac{f_r \lambda f}{2 R} \right) - \frac{f_r \lambda f}{2 R} \sqrt{1 - \left( \frac{f_r \lambda f}{2 R}^2 \right)} \right] \]

for \(0 \lt f_r \lt \frac{2 R}{\lambda f}\).

The maximum spatial frequency is \(f_{r, max} = \frac{2R}{f} \left( \frac{1}{\lambda} \right) = \frac{1}{\lambda f_\#}\).

Optical transfer functions with and without a Gaussian apodization.

Pupil amplitude transmittance and the corresponding OTF with and without a particular “inverse” apodization.