---
title: "Binocular Prism Lab Interactive Lab | Sandbox Physics"
description: "How do binocular prisms turn the image upright, and why can a roof prism blur it?"
canonical_url: "https://sandboxphysics.com/labs/binocular-prism-lab/"
language: "en"
page_type: "lab"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/labs/binocular-prism-lab/"
last_modified: "2026-10-27"
---

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.

> Interactive simulation: open this page in a web browser to change the parameters and run the model.

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

**Background** A 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 matters** How 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 misconception** If both roof faces reflect all the light, phase correction cannot improve resolution.

**Better mental model** Power 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

1. 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.

2. 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.

3. 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.

Physical conclusion

Check where light really meets the prism faces, treat image orientation and wave image quality as separate questions, and compute the polarization instead of setting an arbitrary phase.

Waves & optics · Starter path

## From one ray to interference measurement

Start with refraction, find where the ray picture stops working, then use interference to measure tiny distances.

[Start here → **Optical Boundary Explorer** Turn the incident angle and watch reflected and transmitted power trade off. Then switch to the critical-angle scene. Observe · 11 min](https://sandboxphysics.com/labs/optical-boundary-explorer/)

1. [1 Optical Boundary Explorer](https://sandboxphysics.com/labs/optical-boundary-explorer/)
2. [2 Rainbow Caustic Observatory](https://sandboxphysics.com/labs/rainbow-laboratory/)
3. [3 Camera Obscura](https://sandboxphysics.com/labs/camera-obscura/)
4. [4 Coherent Wave Composer](https://sandboxphysics.com/labs/coherent-wave-composer/)
5. [5 Young Double-Slit Bench](https://sandboxphysics.com/labs/young-double-slit-bench/)
6. [6 Michelson Precision Lab](https://sandboxphysics.com/labs/michelson-precision-lab/)

[View the full path →](https://sandboxphysics.com/topics/waves-and-optics/#starter-path)