---
title: "Ghost Image, Pepper’s Ghost & HUD Interactive Lab | Sandbox Physics"
description: "Where does a half-mirror image appear, and why can a bright background or the mirror’s edge hide it?"
canonical_url: "https://sandboxphysics.com/labs/ghost-image-peppers-ghost-hud/"
language: "en"
page_type: "lab"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/labs/ghost-image-peppers-ghost-hud/"
last_modified: "2026-10-27"
---

Optics 039 · Imaging, instruments, and visual systems

# Ghost Image, Pepper’s Ghost & HUD

**Where does a half-mirror image appear, and why can a bright background or the mirror’s edge hide it?** Behind the glass: as far behind the half-mirror as the display is in front of it, and at the same size. The mirror only adds the display's reflected light on top of whatever you see through it, so a bright background swamps it. Wherever your line of sight to the image passes outside the mirror's edge, that part of the image is gone.

On an exact half-mirror bench, a flat display forms a virtual image of the same size. A computed view at fixed exposure and a map of the viewpoints the finite sheet admits show how the light adds, how a bright background washes out the image and how moving the viewpoint clips it.

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

The combiner is ideal, lossless and of zero thickness; this is not a volumetric hologram or a complete automotive HUD. The virtual image is a geometric construction and emits no light, and the reference contrast applies only where the geometry admits the view. Ghosts from the rear surface of the plate and calibrated luminance are not modeled.

Physics tutorial

## Audit the physics of Ghost Image, Pepper’s Ghost & HUD

**Background** A tilted half-mirror reflects a flat display into a virtual image. For a mirror through the origin with a unit normal, the image position is $\mathbf V=\mathbf S-2\mathbf n(\mathbf n\cdot\mathbf S)$. This reflection preserves every distance, so the image keeps its size. Real light turns at the sheet and enters the observer; only its backward extensions reach the virtual plane.

**Why it matters** Where does a half-mirror image appear, and why can a bright background or the mirror’s edge hide it?

Start with the essentials

- **Focus question**: Where does a half-mirror image appear, and why can a bright background or the mirror’s edge hide it?
- **One-sentence intuition**: For a ray that gets through the sheet, the display light adds to the transmitted real background: $L_{\mathrm{out}}=R L_d+(1-R)L_s$. Black display pixels add no light, but they do not remove any scene light either, which is why a bright background washes out the overlay. The virtual glyph is flat, not a volumetric hologram. A real automotive HUD also needs image-forming optics, and often collimating optics, which are not included here.

### Core mathematical model

#### Reflection preserves distances, so it cannot resize the image

$$
H=I-2\mathbf n\mathbf n^{\mathsf T},\quad H^{\mathsf T}H=I,\quad \|H(\mathbf S_1-\mathbf S_2)\|=\|\mathbf S_1-\mathbf S_2\|
$$

The real emitter and the virtual plane have the same 240 by 180 mm bounds. With the sheet tilted forty-five degrees, a horizontal display becomes an upright virtual plane. Its apparent size changes with perspective, which is not optical magnification. The overview can show the virtual construction from any orbit angle; only the computed card shows what the selected observer sees.

#### Each drawn reflection meets the actual sheet

$$
\mathbf M=\mathbf E+t(\mathbf V-\mathbf E),\quad t=\frac{\mathbf n\cdot\mathbf E}{\mathbf n\cdot(\mathbf E-\mathbf V)},\quad \mathbf d_r=\mathbf d_i-2\mathbf n(\mathbf n\cdot\mathbf d_i)
$$

Intersect the line from the observer to the virtual point with the actual tilted sheet. The ray from the source to that intersection reflects specularly into the observer. The dashed segment from the intersection to the image is a backward construction, never a light ray passing through an opaque mirror. At full reflectance the transmitted background ray disappears.

#### Finite sheet, finite viewing window

$$
|M_x|\le a,\quad |\mathbf M\cdot\mathbf t|\le b,\quad f_{\mathrm{area}}=\frac{\operatorname{area}(P_{\mathrm{admitted}})}{240\cdot180}
$$

The four aperture inequalities clip a polygon in source coordinates exactly. From a green observer position the whole source gets through, from amber only part of it, and from black none. The map is a fixed slice at 700 mm observer depth, spanning horizontal positions from minus 700 to 700 mm and vertical positions from minus 100 to 500 mm. Its white ring follows the selected point. This geometric source-area measure does not depend on brightness, and it is not the power through a finite pupil or a three-dimensional eye-box.

#### The observer view and the aperture map measure different things

$$
\mathbf d=(q_x,q_y,-1),\quad |q_x|\le0.30,\quad |q_y|\le0.225
$$

Each observer pixel sends this backward ray from the selected point. At the sheet the ray splits into a reflected and a transmitted branch; otherwise the opaque border blocks it, or it passes through open space outside the finite sheet. The nearest-hit test includes the one-sided source. The angular window and the exposure never zoom to fit a moving image. A virtual point that the aperture admits may still fall outside this view, because the area map measures what the aperture admits, not how the camera frames it.

#### Radiance mixing and an explicitly defined contrast

$$
T=1-R,\quad L_{\mathrm{on}}=R L_d+T L_s,\quad L_{\mathrm{off}}=T L_s,\quad C_M=\frac{L_{\mathrm{on}}-L_{\mathrm{off}}}{L_{\mathrm{on}}+L_{\mathrm{off}}}=\frac{R L_d}{R L_d+2T L_s}
$$

This is a reference Michelson contrast between white and black display patches over a uniform background patch of unit radiance, for a ray that gets through. It is not a visibility score for objects in the patterned city image, and it does not predict human perception. With no light, the contrast is undefined. Free propagation preserves radiance, so no inverse-square factor is applied to a resolved surface pixel. All relative linear-radiance images use a fixed Reinhard exposure and sRGB encoding, without normalizing each image.

### Common difficulties

#### Can a black virtual pixel hide the real scene?

**Typical misconception** A dark pixel on the display should erase whatever is behind the ghost.

**Better mental model** A passive combiner adds radiance. A black source pixel adds nothing, and the background is still transmitted. Full reflectance can block the background across the whole physical sheet, but it cannot hide the scene pixel by pixel. Real occlusion needs extra hardware.

### Run the experiment

1. 01

   #### Scene 1: Dark-stage phantom

   Start with the dark-stage phantom. Change the real display distance and the sheet tilt, and compare how the hardware moves with the same-size virtual plane and the observer image.

   **What to observe:** The overview is there to explain the construction. The observer pixels are traced backward independently through the same finite aperture. If the source itself falls inside the fixed view, it is seen directly. The sheet border is opaque, while the space outside the finite sheet is open.

2. 02

   #### Scene 2: Bright-background washout

   Choose bright-background washout. Raise the background radiance, then the display radiance. Finally set the reflectance to zero and then to one: the reflected and transmitted branches each disappear at one of these limits.

   **What to observe:** The reference contrast can stay high even when the ghost is clipped away entirely. Read the geometric admission separately; more brightness cannot fix a blocked path.

3. 03

   #### Scene 3: Finite viewing window

   Choose finite viewing window. Move the observer through the green, amber and black regions, then enlarge the sheet to bring the ghost back. Brightness does not change the geometric admission.

   **What to observe:** The classical tilted half-mirror is not the holographic mirror described in the cited research chapter. The model makes no claim about plate thickness, double-surface ghosts or the collimation in commercial HUDs.

Physical conclusion

A convincing phantom needs the correct virtual geometry, enough radiance contrast and a ray path that actually reaches the viewer. A transparent 3D sprite is not enough.

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)