Why a Mirror’s Image Appears Behind the Mirror

Stand 0.80 m in front of a plane mirror. Your reflection appears 0.80 m behind it. Yet if you place a screen where the image seems to be, the screen stays dark.

How can an image exist somewhere that light never reaches?

The answer is that reflected light travels from the mirror to your eyes along real paths. If you extend those paths backwards in straight lines, they meet at a point behind the mirror. Your eyes interpret the light as coming from that apparent position. This is called a virtual image.

For an ideal plane mirror, the image forms the same perpendicular distance behind the mirror as the object is in front.

01Why the image appears behind the mirror

Imagine a small illuminated object in front of a mirror. Light travels from the object towards the mirror, reflects from its surface, and then travels towards your eye.

The important detail is what happens after reflection.

The reflected rays spread out as they travel towards your eye. They do not actually converge behind the mirror. However, if you extend their paths backwards, they appear to meet at a common point behind the reflecting surface.

That point is the virtual image.

Your visual system interprets incoming light using the direction from which it arrives. Because light normally travels in straight lines, your brain effectively locates the apparent source along the incoming rays.

It therefore appears that the light originated behind the mirror.

There is an important distinction:

• Actual reflected rays travel from the mirror towards your eye.
• Backward extensions are imaginary lines used to locate the apparent source.
• The actual rays never meet behind the mirror.

This is why the image is virtual rather than real.

A real image forms where actual light rays converge. A virtual image forms where the backward extensions of diverging rays meet.

Why the image is the same distance behind the mirror

A plane mirror produces an image that is symmetrical with the object across the reflecting surface.

For example:

The perpendicular distance from the object to the mirror equals the perpendicular distance from the image to the mirror.

If the object moves further away, the image moves further behind the mirror. If the object moves closer, the image moves closer too.

The image does not depend on where you stand. Moving your head changes which reflected rays enter your eye, not the actual position of the image.

02Draw the normal and measure the angles correctly

To understand why the reflected rays have the directions they do, you need the law of reflection.

Start by drawing a normal.

A normal is an imaginary line drawn perpendicular to the mirror surface at the point where the incident ray strikes it.

Two angles are then defined:

• Angle of incidence, \(i\): the angle between the incident ray and the normal.
• Angle of reflection, \(r\): the angle between the reflected ray and the normal.

The law of reflection states:

\[ \boxed{i=r} \]

Both angles are measured from the normal, not from the mirror surface.

This convention is important in HSC Physics questions. Measuring from the mirror surface instead gives the complementary angle.

For example, if an incident ray makes an angle of \(30^\circ\) with the normal, it makes an angle of \(60^\circ\) with the mirror surface.

The reflected ray also makes \(30^\circ\) with the normal.

Which media are involved?

There is another distinction worth making.

In the idealised plane-mirror model used for basic HSC ray diagrams, the incident and reflected rays both travel through air. The light remains on the same side of the reflecting surface.

This is different from refraction.

For reflection, use:

\[ i=r \]

For refraction between two transparent media, use Snell’s law:

\[ n_1\sin\theta_1=n_2\sin\theta_2 \]

Do not apply Snell’s law to the reflected ray in an ideal plane-mirror question.

A real rear-silvered mirror has a glass layer in front of its reflective coating. A more detailed model would account for refraction as light enters and exits that glass. Unless a question explicitly includes this construction, use the ideal reflecting-surface model.

For a broader explanation of reflection and plane-mirror image properties, see Reflection and Image Formation in Mirrors for HSC Physics.

03How the experiment distinguishes a real image from a virtual image

Suppose you want to establish whether the apparent image behind a mirror is a real location where light converges.

A tempting first experiment is to place a white screen behind the mirror.

The first attempt: use a screen

Set up an illuminated object in front of an upright plane mirror.

Place a white screen at the position where you expect the image to form.

If actual light rays converge at that location, the screen should intercept them and display an image.

But the screen stays dark.

Does this mean the image does not exist?

No. The problem is that an opaque mirror blocks the light from reaching the screen. This experiment cannot directly locate an apparent source behind the mirror.

It only establishes that the image is not formed by actual light convergence at that position.

The revised approach: trace the reflected rays

Instead of trying to catch the image, trace the directions of the light that reaches your eye.

A standard pin-sighting investigation can do this.

Arrange the apparatus in the following order:

• Stand a plane mirror upright along a marked line on a sheet of paper.
• Place a small illuminated object, such as an LED, in front of the mirror.
• Look at its reflection from a first viewing position.
• Place two sighting pins between your eye and the mirror, aligning their tips with the reflected image.
• Move your eye to a second position and repeat with another pair of pins.

The pins mark the directions of the reflected rays on the observer’s side of the mirror.

Now transfer the pin positions and mirror position onto your paper.

Draw straight lines through each pair of pins and extend those lines backwards behind the mirror.

The extensions intersect at a common point.

That intersection identifies the apparent image position.

Notice what the experiment has achieved: you have located the image without placing a screen where the image appears to be.

You can also draw the incident rays from the object to their respective reflection points. At each point, draw a normal and compare the angles of incidence and reflection.

Illustrative experimental readings

The following values are illustrative, not measurements from an actual experiment.

The small differences are consistent with ordinary drawing and protractor-reading uncertainty.

The important observation is that the two angles are approximately equal in every trial.

The experiment supports the law of reflection and provides a way to construct the virtual image using backward extensions.

A simple scale check makes the prediction clear.

Suppose the object is 0.60 m in front of the mirror.

The ideal plane-mirror model predicts:

\[ d_i=d_o=0.60\ \mathrm{m} \]

The object-to-image separation is therefore:

\[ 0.60+0.60=1.20\ \mathrm{m} \]

This is a geometrical prediction. A screen cannot directly measure the virtual image at that location.

04When a virtual image is not visible

There is one more detail that can catch students out.

A virtual image can exist behind a mirror without being visible to a particular observer.

Why?

Because a real mirror has a finite size.

Imagine a small mirror that only extends from 0.2 m to 1.2 m above the floor. An object might produce a virtual image at a height of 1.6 m behind the mirror.

You might initially predict that the image cannot be seen because it is above the mirror’s top edge.

That prediction is not necessarily correct.

The image’s position and the point where light strikes the mirror are different things.

To determine whether the image is visible, draw a straight line from the observer’s eye to the virtual image. The line must intersect the actual reflecting surface.

If it intersects within the mirror, the image is visible.

If it misses the mirror, no reflected ray from that image point reaches the observer.

Finding where the ray hits a finite mirror

We can calculate the height at which the line from the observer to the virtual image intersects the mirror.

Assume:

• The mirror is vertical at \(x=0\).
• The object is a perpendicular distance \(a\) in front of the mirror.
• The observer is a perpendicular distance \(d\) in front of the mirror.
• The object height is \(y_O\).
• The observer’s eye height is \(y_E\).
• The ray intersects the mirror at height \(y_M\).

The virtual image is at the same height as the object, on the other side of the mirror.

Using straight-line geometry:

\[ y_M=y_E+\frac{d}{a+d}(y_O-y_E) \]

Equivalently:

\[ \boxed{y_M=\frac{ay_E+dy_O}{a+d}} \]

This equation is useful when checking whether a particular point on an object can be seen in a finite mirror.

Remember that all heights and distances must use consistent units.

The important test is whether \(y_M\) falls within the physical height of the mirror.

This distinction becomes particularly interesting when the observer moves. The image can remain fixed while different parts of an object become visible or disappear.

05Worked examples

06Questions and solutions

Question 1

A small illuminated object is placed 0.45 m in front of an ideal plane mirror.

Determine:

a) How far behind the mirror its image appears.

b) Whether a white screen placed at the image position would display a sharp image.

Explain your reasoning.

Solution 1

The image appears 0.45 m behind the mirror, and a screen placed there will not catch a sharp image.

For an ideal plane mirror, the perpendicular image distance equals the perpendicular object distance.

Therefore:

\[ d_i=d_o \]

\[ d_i=0.45\ \mathrm{m} \]

The image is virtual because the reflected rays do not actually converge behind the mirror.

Their backward extensions meet at the apparent image position.

A screen requires actual light rays to reach it. Since no reflected light travels behind the opaque mirror, the screen remains dark.

Question 2

A plane mirror is tilted so that its surface makes an angle of \(20^\circ\) anticlockwise from the positive horizontal direction.

A light ray approaches the mirror from the upper-left, travelling downwards and to the right at \(40^\circ\) below the horizontal.

Determine:

a) The angle of incidence.

b) The direction of the reflected ray, measured anticlockwise from the positive horizontal direction.

c) The media through which the incident and reflected rays travel.

Assume the mirror is ideal.

Solution 2

The angle of incidence is \(30^\circ\), and the reflected ray travels at \(80^\circ\) anticlockwise from the positive horizontal direction. Both incident and reflected rays travel in air.

Step 1: Determine the angle between the incident ray and the mirror surface.

The mirror is at \(20^\circ\) above the horizontal.

The incoming ray travels at \(40^\circ\) below the horizontal.

The acute angle between the incident ray’s line and the mirror surface is:

\[ 20^\circ+40^\circ=60^\circ \]

Step 2: Calculate the angle of incidence.

The normal is perpendicular to the mirror.

Therefore:

\[ i=90^\circ-60^\circ \]

\[ \boxed{i=30^\circ} \]

Step 3: Apply the law of reflection.

\[ r=i \]

\[ \boxed{r=30^\circ} \]

The reflected ray must make the same angle with the normal, but on the opposite side.

Step 4: Find the reflected direction.

The reflection can be understood as a reflection of the incoming direction across the mirror’s surface.

Let:

• \(\phi=20^\circ\) be the mirror’s direction.
• \(\theta_{\text{in}}=-40^\circ\) be the incoming ray direction.

The reflected direction is:

\[ \theta_{\text{out}}=2\phi-\theta_{\text{in}} \]

Substitute:

\[ \theta_{\text{out}} =2(20^\circ)-(-40^\circ) \]

\[ \boxed{\theta_{\text{out}}=80^\circ} \]

Step 5: Identify the media.

Both rays travel through air on the incident side of the mirror.

There is no transmitted ray entering glass in this idealised reflection model.

A common mistake is to use the incoming ray’s \(40^\circ\) angle from the horizontal as its angle of incidence. The angle of incidence is measured from the normal, which must first be constructed.

Question 3

An ideal plane mirror is initially positioned at \(x=0\).

A point object is fixed at:

\[ x_O=-0.80\ \mathrm{m} \]

The mirror then moves 0.10 m towards the object, finishing at:

\[ x_M=-0.10\ \mathrm{m} \]

The object remains stationary.

Using positive \(x\) to the right, determine:

a) The initial position of the image.

b) The final position of the image.

c) The displacement of the image.

d) The change in separation between the object and its image.

Explain why the image does not move by the same distance as the mirror.

Solution 3

The image moves 0.20 m to the left, twice the mirror’s displacement. The object-image separation decreases from 1.60 m to 1.40 m.

Step 1: Establish the relationship between the object, mirror, and image.

For a plane mirror, the mirror lies halfway between the object and its image.

Using horizontal coordinates:

\[ x_M=\frac{x_O+x_I}{2} \]

Rearranging:

\[ \boxed{x_I=2x_M-x_O} \]

Step 2: Calculate the initial image position.

Initially:

\[ x_M=0,\qquad x_O=-0.80 \]

Therefore:

\[ x_I=2(0)-(-0.80) \]

\[ \boxed{x_I=+0.80\ \mathrm{m}} \]

Step 3: Calculate the final image position.

The mirror moves to \(x_M=-0.10\ \mathrm{m}\).

The object remains at \(x_O=-0.80\ \mathrm{m}\).

Substitute:

\[ x_I=2(-0.10)-(-0.80) \]

\[ x_I=-0.20+0.80 \]

\[ \boxed{x_I=+0.60\ \mathrm{m}} \]

Step 4: Calculate the image displacement.

\[ \Delta x_I=x_{\text{final}}-x_{\text{initial}} \]

\[ \Delta x_I=0.60-0.80 \]

\[ \boxed{\Delta x_I=-0.20\ \mathrm{m}} \]

The negative sign means the image moves to the left.

Step 5: Calculate the change in object-image separation.

Initially:

\[ D_1=0.80+0.80 \]

\[ D_1=1.60\ \mathrm{m} \]

Finally:

\[ D_2=0.80+0.60 \]

\[ D_2=1.40\ \mathrm{m} \]

Therefore:

\[ \Delta D=1.40-1.60 \]

\[ \boxed{\Delta D=-0.20\ \mathrm{m}} \]

The tempting prediction is that the image moves 0.10 m because the mirror moves 0.10 m.

But the mirror must remain halfway between the object and image. Since the object is fixed, moving the midpoint by 0.10 m requires the image to move by 0.20 m.

Question 4

A point object is positioned 1.20 m in front of a vertical plane mirror at a height of 1.60 m above the floor.

An observer’s eye is positioned 0.80 m in front of the mirror at a height of 0.40 m.

The mirror extends vertically from 0.20 m to 1.20 m above the floor.

Determine whether the observer can see the object’s virtual image.

If it is visible, calculate the height at which the reflected ray strikes the mirror.

Assume an ideal mirror and a point-like observer’s eye.

Solution 4

Yes, the observer can see the virtual image because the ray intersects the mirror at a height of 0.88 m, within its physical boundaries.

Step 1: Identify the relevant coordinates.

The object distance is:

\[ a=1.20\ \mathrm{m} \]

The observer’s distance is:

\[ d=0.80\ \mathrm{m} \]

The object height is:

\[ y_O=1.60\ \mathrm{m} \]

The eye height is:

\[ y_E=0.40\ \mathrm{m} \]

Step 2: Calculate the mirror intersection height.

Use the straight-line relationship:

\[ y_M=\frac{ay_E+dy_O}{a+d} \]

Substitute:

\[ y_M= \frac{(1.20)(0.40)+(0.80)(1.60)} {1.20+0.80} \]

\[ y_M=\frac{0.48+1.28}{2.00} \]

\[ \boxed{y_M=0.88\ \mathrm{m}} \]

Step 3: Check whether the ray intersects the physical mirror.

The mirror extends from:

\[ 0.20\ \mathrm{m}\leq y\leq1.20\ \mathrm{m} \]

The calculated intersection is:

\[ y_M=0.88\ \mathrm{m} \]

Therefore, the ray strikes the mirror within its reflecting surface.

The image is visible.

The subtle point is that the virtual image is at a height of 1.60 m, above the mirror’s top edge.

You might therefore predict that it cannot be seen.

But the image does not need to sit directly behind the physical reflecting section. The ray from the eye to the image only needs to intersect the mirror within its boundaries.

Question 5

A vertical plane mirror is positioned at \(x=0\). Its reflecting surface extends from a height of 0 m to 1.00 m.

A narrow illuminated vertical rod is placed parallel to the mirror, 2.00 m in front of it.

The rod extends from:

\[ y=0.50\ \mathrm{m} \]

to:

\[ y=1.50\ \mathrm{m} \]

An observer’s eye remains at a height of 0.50 m.

Initially, the observer stands 1.00 m in front of the mirror. The observer then walks directly backwards until they are 4.00 m in front of it.

Assume:

• The mirror is ideal and perfectly reflecting.
• The object and mirror remain stationary.
• The observer has a point-like eye.
• The mirror has no frame that further restricts visibility.
• All distances are measured in metres.

Determine:

a) The position and vertical extent of the virtual image.

b) Whether the entire rod is visible from the initial observer position.

c) The greatest observer distance at which the entire rod remains visible.

d) The portion of the rod visible when the observer is 4.00 m from the mirror.

e) Whether the virtual image moves as the observer walks backwards.

Explain why the upper part of the rod can disappear even though the image remains in exactly the same position.

Solution 5

The entire rod is visible initially. At the final observer position, only the section from 0.50 m to 1.25 m is visible, giving a visible length of 0.75 m. The virtual image does not move at all.

This problem combines image formation with the geometry of a finite mirror. We need to track the image and the actual ray paths separately.

Step 1: Locate the virtual image.

The rod is 2.00 m in front of the mirror.

For a plane mirror, the image is the same perpendicular distance behind the reflecting surface.

Therefore, the image is at:

\[ x_I=+2.00\ \mathrm{m} \]

Its vertical extent is unchanged:

\[ 0.50\leq y_I\leq1.50\ \mathrm{m} \]

The virtual image is therefore a 1.00 m tall rod, 2.00 m behind the mirror.

Step 2: Check whether the entire rod is visible initially.

The observer is initially 1.00 m from the mirror.

Therefore:

\[ a=2.00,\qquad d=1.00 \]

The eye is at:

\[ y_E=0.50\ \mathrm{m} \]

Consider the upper endpoint of the virtual image:

\[ y_O=1.50\ \mathrm{m} \]

The mirror intersection height is:

\[ y_M=\frac{ay_E+dy_O}{a+d} \]

Substitute:

\[ y_M= \frac{(2.00)(0.50)+(1.00)(1.50)} {2.00+1.00} \]

\[ y_M=\frac{1.00+1.50}{3.00} \]

\[ \boxed{y_M=0.833\ \mathrm{m}} \]

This is within the mirror’s height of 0 to 1.00 m.

For the lower endpoint, \(y_O=0.50\ \mathrm{m}\):

\[ y_M= \frac{(2.00)(0.50)+(1.00)(0.50)} {3.00} \]

\[ \boxed{y_M=0.50\ \mathrm{m}} \]

This also falls within the mirror.

Since the mirror is vertical and the ray intersection changes continuously between the endpoints, the entire rod is visible.

Step 3: Find the greatest observer distance for which the entire rod remains visible.

The upper endpoint is the limiting point.

At the greatest viewing distance, its reflected ray must just reach the upper edge of the mirror:

\[ y_M=1.00\ \mathrm{m} \]

Let \(d\) represent the observer’s distance from the mirror.

Using the upper endpoint:

\[ 1.00= \frac{(2.00)(0.50)+d(1.50)} {2.00+d} \]

Multiply both sides:

\[ 2.00+d=1.00+1.50d \]

Rearrange:

\[ 1.00=0.50d \]

Therefore:

\[ \boxed{d=2.00\ \mathrm{m}} \]

At exactly 2.00 m, the upper endpoint is just visible at the mirror’s top edge.

If the observer moves further away, the ray to the upper endpoint misses the mirror.

Step 4: Determine what is visible when the observer is 4.00 m away.

Now:

\[ a=2.00,\qquad d=4.00 \]

First, consider the upper endpoint at \(y_O=1.50\ \mathrm{m}\).

\[ y_M= \frac{(2.00)(0.50)+(4.00)(1.50)} {2.00+4.00} \]

\[ y_M=\frac{1.00+6.00}{6.00} \]

\[ \boxed{y_M=1.167\ \mathrm{m}} \]

This is above the mirror’s upper edge at 1.00 m.

The upper endpoint is therefore not visible.

The lower endpoint remains visible:

\[ y_M= \frac{(2.00)(0.50)+(4.00)(0.50)} {6.00} \]

\[ \boxed{y_M=0.50\ \mathrm{m}} \]

Step 5: Find the highest point on the rod that remains visible.

The limiting point is the one whose ray just intersects the mirror’s top edge.

Set:

\[ y_M=1.00\ \mathrm{m} \]

Let \(y_O\) be the height of the highest visible point.

Using the same relationship:

\[ 1.00= \frac{(2.00)(0.50)+(4.00)y_O} {2.00+4.00} \]

Multiply both sides:

\[ 6.00=1.00+4.00y_O \]

Rearrange:

\[ 5.00=4.00y_O \]

\[ \boxed{y_O=1.25\ \mathrm{m}} \]

Therefore, the visible portion of the original rod extends from 0.50 m to 1.25 m.

Its visible length is:

\[ L=1.25-0.50 \]

\[ \boxed{L=0.75\ \mathrm{m}} \]

The upper 0.25 m of the rod is no longer visible.

Step 6: Explain whether the image moves.

The image does not move.

The object and mirror remain stationary, so the geometrical image position remains:

\[ x_I=+2.00\ \mathrm{m} \]

Its vertical extent also remains unchanged.

What changes is the observer’s position. As the observer moves backwards, the straight lines connecting their eye to the virtual image intersect the mirror at different heights.

Eventually, the ray from the upper endpoint misses the mirror altogether.

This is the crucial distinction: an image can remain in the same position while becoming partially or completely invisible to an observer.

The next useful decision in geometrical optics is determining whether light reflects from a surface or enters a different medium and refracts. Reflection uses \(i=r\), while refraction requires Snell’s law and the refractive indices of the two media. For that next step, see Refraction and Image Formation in Lenses for HSC Physics.