Optics 032 · Imaging, instruments, and visual systems

Binocular Prism Lab

How do binocular prisms turn the image upright, and why can a roof prism blur it? In a pair of Porro prisms the light reflects four times, which flips the image both top to bottom and left to right, so it comes out upright. A roof prism splits the beam into two halves that reflect off the two roof faces. The halves pick up different polarization phases, so unless that is compensated they interfere imperfectly and blur the image, even though no light is lost.

Open up two orthogonal Porro prisms and trace finite Snell–Fresnel paths and the pupil they deliver. Then compare the polarization-resolved diffraction of a bare roof pupil with that of an ideally compensated one.

Porro rotation / bare roof retardance

\Phi_P,\Delta_{sp}

$0.50$

Smallest total-reflection margin

\min(\theta_i)-\theta_c

$50\%$

Delivered power / roof on-axis peak

T,S_0

$0.00\pi$

Model regime $\text{valid model regime}$

The Porro part is a finite geometric module with a square entrance port, and the field cards are orientation diagrams drawn at equal scale. The roof part is a separate on-axis module that turns light 90 degrees inside the glass, with an oversized circular pupil and ideal retardance compensation. It is not a complete binocular or a fabricated coating design.

Physics tutorial

Audit the physics of Binocular Prism Lab

BackgroundA positive telescope objective forms an upside-down image. The reflections turn it upright, and each one is described by the operator $H_{\hat{\mathbf n}}=I-2\hat{\mathbf n}\hat{\mathbf n}^{\mathsf T}$ . Each Porro prism has two reflecting faces at right angles. Two Porros set at right angles to each other flip both transverse coordinates and send the light forward again.

Why it mattersHow do binocular prisms turn the image upright, and why can a roof prism blur it?

Start with the essentials

Focus question: How do binocular prisms turn the image upright, and why can a roof prism blur it?
One-sentence intuition: A roof prism divides the pupil into two halves that meet its two faces in opposite order, and the halves do not form two separate images. For unpolarized light, the image intensity is $I(\mathbf q)=\frac12\left\|J_+A_+(\mathbf q)+J_-A_-(\mathbf q)\right\|_F^2$ . The two complex vector fields interfere first; only then is the result averaged over the two independent input polarizations. Reflecting all the power does not guarantee a diffraction-limited point image.

Core mathematical model

Four reflections, two coordinate flips

H_4H_3H_2H_1=\operatorname{diag}(1,-1,-1)

The first coordinate is the forward optical axis; the other two are the transverse world coordinates. The image inverted by the objective is therefore upright after the second prism. All field cards use the same symbolic scale. They are symbols, not photographs, and they do not trace a full image relay.

Critical angle and finite prism faces

\theta_c=\arcsin(1/n_g),\qquad \hat{\mathbf k}_r=H_{\hat{\mathbf n}}\hat{\mathbf k}_i

Angles are measured from the face normal. A traced ray reflects only if it actually hits the finite face. Below the critical angle, part of the light refracts out into the air, and the reflected part follows the Fresnel amplitudes. The square entrance stop and the blackened prism ends are modelled as absorbing hardware.

Lossless reflection still changes phase

r_s=\frac{\cos\theta-i\gamma}{\cos\theta+i\gamma},\quad r_p=\frac{\eta^2\cos\theta-i\gamma}{\eta^2\cos\theta+i\gamma},\quad \gamma=\sqrt{\sin^2\theta-\eta^2},\quad \eta=1/n_g

In total internal reflection both amplitudes have magnitude one, but their phases differ. The two roof faces have different planes of incidence, and the two pupil halves meet them in opposite order. The model transforms the full electric-field vector at each face.

Jones diffraction from the two pupil halves

A_\pm(\mathbf q)=\frac1\pi\iint_{P_\pm}e^{-i\mathbf q\cdot\boldsymbol\rho}\,\mathrm d^2\boldsymbol\rho,\qquad S_0=I(\mathbf 0)

The roof model uses a uniform circular pupil at 550 nm. The point image comes from numerical integration over each half-disk and the analytic transform of the full disk. All point images share the same 100 arcsecond field, are scaled to an ideal peak of one and show brightness as the square root of intensity. Full correction means an ideal lossless element that equalizes the retardance; no particular multilayer or thickness is specified.

Common difficulties

Does total reflection guarantee a sharp image?

Typical misconceptionIf both roof faces reflect all the light, phase correction cannot improve resolution.

Better mental modelPower and complex field are different things. Because the two halves reflect in opposite order, the pupil field changes differently for each polarization, and energy in the image is redistributed. The roof reference leaves out the losses where light enters and leaves the glass; the direct Porro throughput includes them. Neither number is the full transmission of a binocular.

Run the experiment

01
Scene 1: Double Porro erection
Open Double Porro. Follow the four reflection points and compare the asymmetric field before, between and after the prisms. Orbit the view to see that the two folds are perpendicular.
What to observe: In the double Porro, light leaves parallel to the way it came in, but shifted sideways, and the field is rotated by 180 degrees.
02
Scene 2: Finite pupil & TIR leakage
Open Finite pupil. Raise the field angle or lower the index, and watch the pink leakage and the margin to the critical angle. Enlarge the pupil or narrow the Porro to see light clipped by the stop and the end faces.
What to observe: Fresnel losses and geometric clipping are separate effects, so delivered power and accepted pupil area are reported separately.
03
Scene 3: Roof pupil diffraction
Open Roof. Move ideal correction from zero to one. Compare the uncorrected, current and ideal circular-pupil point images at the same brightness. The index sets the Fresnel phase of the bare faces; the pupil diameter sets the angular scale of the diffraction.
What to observe: The roof module turns the light through 90 degrees. It isolates the phase problem and is not a complete straight-through Schmidt–Pechan instrument.