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SCIENCE · INVESTIGATE AND EXPLAIN

Locate a virtual image without sending light behind a mirror

A plane-mirror image can be located by tracing physical reflected rays backward as straight-line extensions. The reflected light stays on the object side, but an observer interprets the diverging rays as coming from a point behind the mirror. That apparent point is a virtual image. Dashed extensions are a geometric construction, not rays passing through the reflecting surface.

Learning outcome: Retain object distances 20 and 40 cm at the same height. Follow each physical segment and its dashed backward extension. Compare image distance, height and the object-to-image separation.

Starting recap

Define normal angles and signed object/image distances. Distinguish actual ray intersections from backward extensions, and keep the focal/critical boundaries explicit rather than forcing a finite answer.

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Try

An object 30 cm in front of the mirror has i=−30 cm, height unchanged and M=+1. Object-to-image separation is 60 cm. If the object moves 10 cm farther away, the separation becomes 80 cm.

Open the experiment

Predict

If a virtual image is visible, must physical reflected rays pass through its apparent location?

Experiment

Retain object distances 20 and 40 cm at the same height. Follow each physical segment and its dashed backward extension. Compare image distance, height and the object-to-image separation.

Locate a plane-mirror image by backward extensions

Worked teaching model · synthetic values · Vertical plane mirror x=0; object on left, rays travel toward mirror then reflect left. Image behind mirror is virtual, equal height and equal distance. Dashed lines are backward extensions, not light traveling through the mirror.

Reset keeps your prediction, captures and practice history.

Object tipVirtual image tipObject/image ray geometry in cm; +x right, +y up

Physical ray direction follows its listed first coordinate to last coordinate. Dashed lines are not extra propagating rays. Display scale changes between settings; within the diagram both axes use the same scale.

Ray segments and construction lines — coordinates in diagram units
SegmentCoordinates in listed directionType
Incoming ray at mirror height 0(-30, 6) → (0, 0)Physical ray
Reflected ray at mirror height 0(0, 0) → (-30, -6)Physical ray
Backward extension at mirror height 0(0, 0) → (30, 6)Extension / construction
Incoming ray at mirror height 3(-30, 6) → (0, 3)Physical ray
Reflected ray at mirror height 3(0, 3) → (-30, 0)Physical ray
Backward extension at mirror height 3(0, 3) → (30, 6)Extension / construction
Plane mirror(0, -6) → (0, 12)Extension / construction
Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Signed image distance-30cm
Image height6cm
Magnification1ratio
Accessible plot coordinates
Primary plotted points

Backward extensions meet behind the mirror, while reflected light remains in front.

Positive object distance is in front of the mirror; signed image distance is negative for this virtual image. A screen behind an opaque mirror does not capture this image. The virtual image can still be seen by an observer receiving reflected rays.

Backward extensions meet behind the mirror, while reflected light remains in front.

Positive object distance is in front of the mirror; signed image distance is negative for this virtual image. A screen behind an opaque mirror does not capture this image. The virtual image can still be seen by an observer receiving reflected rays.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Establish the model and evidence

A plane-mirror image can be located by tracing physical reflected rays backward as straight-line extensions. The reflected light stays on the object side, but an observer interprets the diverging rays as coming from a point behind the mirror. That apparent point is a virtual image. Dashed extensions are a geometric construction, not rays passing through the reflecting surface.

A plane mirror has no finite focal point: parallel incident rays remain parallel after reflection, corresponding to infinite focal distance. Predict 1/o+1/i for i=−o and check that it is zero, consistent with 1/f=0.

For a plane mirror the image is the same perpendicular distance behind the mirror as the object is in front, with equal height and upright orientation. Under the real-is-positive mirror convention, object distance is positive and image distance negative. Magnification M=−i/o is therefore +1. The sign describes orientation, not whether someone can see the image.

A virtual image is visible when suitable rays enter an observer’s eye, yet it cannot be caught directly by placing a screen at the apparent image point behind an opaque mirror. Failure to obtain a sharp screen image is therefore not automatically an equipment failure. Real-image formation requires actual rays from one object point to meet at an image point.

Moving the object away from the mirror increases both object-to-mirror and mirror-to-image distances, so object-to-image separation changes by twice the object displacement if the mirror stays fixed. The mirror reverses the coordinate perpendicular to its plane, rather than applying an unexplained universal left-right swap. Viewing orientation and comparison conventions explain familiar apparent reversals.

Worked evidence

An object 30 cm in front of the mirror has i=−30 cm, height unchanged and M=+1. Object-to-image separation is 60 cm. If the object moves 10 cm farther away, the separation becomes 80 cm.

Predict and investigate

If a virtual image is visible, must physical reflected rays pass through its apparent location?

Retain object distances 20 and 40 cm at the same height. Follow each physical segment and its dashed backward extension. Compare image distance, height and the object-to-image separation.

Transfer to changed conditions

An object of height 4 cm stands 15 cm in front of a plane mirror. Find image height, signed magnification and object-to-image separation.

Compare the worked transfer

The image height is +4 cm, magnification +1 and separation 30 cm. It is upright and virtual; the negative image distance −15 cm is a location convention, not a negative physical length.

Explain the boundary

Explain why no sharp image on a screen behind the mirror can be the correct prediction rather than a measurement failure.

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. Original text, rays and computed/synthetic records, not completed optical measurements. All investigations are virtual: no lasers, intense-light viewing, sunlight focusing, flames or eye tests are requested. Any real practical work needs qualified supervision and independent safety review.

Clause-level source/action reconciliation checked 31 October 2026 against the linked current framework and Fall 2026 clarification. Added packets are original virtual comparisons; no new practical or learner-completion claim.

Apply

A real object 20 cm in front of a plane mirror produces which signed image distance under this convention?

Public practice checks a specific claim. It does not establish independent mastery.

Explain why no sharp image on a screen behind the mirror can be the correct prediction rather than a measurement failure.

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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