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Rays, Reflection and the Plane Mirror

14 min read4 ideas6 checks2 figures to playBy Asheshanand Acharya
What you will learn 0 / 4
Light travels in rays
Maths you need
Look around
Round spots under a neem tree

On a sunny afternoon, look at the ground under a neem tree. The shade is sprinkled with bright spots of sunlight. The gaps between the leaves have every kind of shape, yet almost every spot is a round or oval patch. During a partial solar eclipse the same spots become small crescents.

you will see why every spot is round, whatever the shape of the gaps between the leaves. You will also estimate the size of the spots.

Light goes straight

The waves of the unit on Waves spread round corners, because their wavelengths are comparable to doors and walls. Light is also a wave, but its wavelength is tiny. Visible light has wavelengths from about \(400\,\text{nm}\) (violet) to \(700\,\text{nm}\) (red). A gap of \(1\,\text{mm}\) is \(2000\) times as wide as a wavelength of \(500\,\text{nm}\). Through such openings light bends round the edges very little, and it travels in straight lines.

This straight-line travel is the whole basis of geometrical optics. We draw the path of light as a ray, a straight line with an arrow along the direction in which the light energy travels. A bundle of rays is a beam. The spreading of light round an edge, called diffraction, is left to the chapter on diffraction later in this unit.

definition 1.1
A ray of light

A ray is a straight line drawn along the direction in which light energy travels. A ray changes direction only where the light meets a new surface or a new medium.

A beam can have three shapes. In a parallel beam the rays never meet; sunlight reaching the Earth and the light of a laser pointer are parallel beams. In a diverging beam the rays spread out from a point, like the light of a bulb or a diya. In a converging beam the rays close in towards a point, like sunlight after a magnifying glass.

A pinhole makes an inverted imageAn upright arrow, the object, stands 30 cm to the left of an opaque plate with a tiny hole in it. A screen stands 20 cm to the right of the plate. A ray from the tip of the arrow passes through the hole and reaches the screen 2.67 cm below the axis; a ray from the foot of the arrow reaches the screen on the axis. The screen is lit only between these two points, where an inverted image of the arrow forms. pinhole screen object image
Fig 1.1An object 4 cm tall stands 30 cm in front of a pinhole, and a screen stands 20 cm behind it. The ray from the tip goes straight through the hole and lands 2.67 cm below the axis. Only rays that pass the hole reach the screen, so an inverted image forms there.

A tiny hole in a card makes an image with no lens at all. Each point of the object sends light in all directions. Only one narrow ray from each point gets through the hole, and it goes straight on. The ray from the top of the object therefore lands low on the screen, and the ray from the bottom lands high. In Fig 1.1 the image is \(4 \times 20/30 = 2.67\,\text{cm}\) tall, and it is inverted. This is a pinhole camera.

CheckpointUnanswered

Light from a distant star reaches a telescope on the Earth. What kind of beam enters the telescope?

Real and virtual objects and images

How the eye locates a point

The eye does not reach out to the object. Each point of an object sends out a diverging cone of rays, and the eye catches a thin part of that cone. The eye and brain trace the caught rays backwards, in straight lines, to the point from which they seem to come. We see the point there. The eye has no way to know what happened to the rays before they arrived.

This leads to two kinds of image. A real image is formed where rays actually meet after reflection or refraction. Light energy reaches that point, so a screen placed there shows the image. A virtual image is formed where rays only seem to meet. The rays diverge from the mirror, and their backward extensions, drawn dashed, meet behind it. No light reaches a virtual image, so a screen there stays dark.

Property
Real image
Virtual image
The rays
actually meet at the image
only seem to come from the image
On a screen
can be caught
cannot be caught
Drawn in a ray diagram
with solid rays
with dashed backward extensions
Seen by an eye
yes, from beyond the image
yes, looking into the mirror

The same idea applies to objects. A real object is a point from which rays actually diverge, like a lamp or a lit face. Sometimes a mirror or a lens receives a converging beam before its rays have met. The point where they would have met is then a virtual object. Virtual objects matter when two lenses or mirrors act one after the other, in the chapter on lenses.

principle 1.1
The principle of reversibility

If the direction of a ray is reversed, the ray retraces its path exactly, through every reflection and refraction on the way.

CheckpointUnanswered

Can a virtual image be photographed?

Laws of reflection

The two laws of reflection

When a ray meets a polished surface, draw the normal, the line perpendicular to the surface at the point of incidence. The angle of incidence \(i\) is the angle between the incident ray and the normal. The angle of reflection \(r\) is the angle between the reflected ray and the normal. Both angles are measured from the normal, never from the surface.

The law of reflectionA horizontal plane mirror with its hatched back below. A dashed normal rises from the point O where a ray meets the mirror. The incident ray comes down from the upper left and the reflected ray leaves to the upper right. The angle i between the incident ray and the normal equals the angle r between the reflected ray and the normal, 40 degrees each. The two rays lie on opposite sides of the normal. i r O normal incident ray reflected ray mirror
Fig 1.2A ray meets a plane mirror at O. The dashed normal is perpendicular to the mirror at O. The angle of incidence \(i\) and the angle of reflection \(r\) are equal, here \(40°\) each. The two rays lie on opposite sides of the normal.
law 1.1
The laws of reflection

First, the incident ray, the reflected ray and the normal at the point of incidence lie in one plane, the plane of incidence. Second, the angle of reflection equals the angle of incidence, \(r = i\), with the reflected ray on the other side of the normal.

The first law says that reflection never throws a ray sideways out of the plane of incidence. The second law fixes the direction inside that plane. Together they give the reflected ray completely. The laws hold at every point of every surface, flat or curved, because each tiny patch of a surface is flat.

Common mistake

A question may give the angle between the ray and the mirror itself. Subtract it from \(90°\) before using \(r = i\). A ray at \(40°\) to the mirror has \(i = 90° - 40° = 50°\), and \(r = 50°\) too.

A mirror gives regular reflection: its surface is smooth, all the normals are parallel, and a parallel beam stays parallel. A sheet of paper gives diffuse reflection. Each tiny patch still obeys \(r = i\), but the patches tilt every way, so the reflected rays scatter in all directions. That is why you can read a page from any seat in a room.

Checkpoint · numericUnanswered

A ray meets a plane mirror, making an angle of \(35°\) with the surface of the mirror. Find the angle between the incident ray and the reflected ray, in degrees.

°± 2 %
Image in a plane mirror

The image in a plane mirror

Place a point object O at a distance \(d\) in front of a plane mirror. Take two rays from O that reflect and reach the eye. The reflected rays diverge, so they never meet in front of the mirror. Extend them backwards, dashed, behind the mirror. They meet at a point I, and the eye sees O there.

Where exactly is I? Each reflected ray makes the same angle with the mirror as its incident ray. So each ray, with its extension, forms two triangles on either side of the mirror that are congruent. Every such pair gives the same point I: on the normal through O, at the same distance \(d\) behind the mirror. Every ray from O, extended backwards, passes through I, whatever its angle.

The image in a plane mirror stays putA vertical plane mirror with an object point O 3 cm in front of it. An eye on the left looks at the mirror; a slider moves the eye up and down. Two rays leave O, reflect from the mirror and enter the eye. Their dashed backward extensions meet behind the mirror at the image I, 3 cm behind it and level with O. However the eye is moved, the rays reflect from different parts of the mirror, but the extensions always meet at the same point I. A readout gives the angle of incidence of the rays, always equal to the angle of reflection. O I 3 cm 3 cm i = r = 7.2°
SimA point object O is 3 cm in front of a plane mirror. Move the eye with the slider. The two rays that enter the eye reflect from different parts of the mirror, and the readout shows that \(i = r\) for each. Their dashed extensions always meet at the same point I, 3 cm behind the mirror.

That fixed meeting point is what makes I an image. An eye anywhere in front of the mirror traces the rays back to the same place. The image of a body is the set of images of all its points. So the image is virtual, erect and the same size as the object. It lies as far behind the mirror as the object is in front, on the same normal.

RememberImage in a plane mirror
\[ \text{image distance behind the mirror} = \text{object distance in front}, \qquad m = +1 \]
Holds whena plane mirror and any object in front of it, real or extended; the image lies on the normal through each point of the object.
Symbols\(m\) the ratio of the image height to the object height
Unitsdistances in any one unit

One more property surprises people. Raise your right hand in front of a mirror. The image raises the hand on its right side as you look at it, which is the image's left hand. A plane mirror reverses front and back along its normal, and this makes right and left swap. The effect is called lateral inversion. It is why the word AMBULANCE is painted reversed on the front of an ambulance: in a car's mirror it reads correctly.

A clock seen in a mirror is laterally inverted about its 12 to 6 line. The hour hand of the image points to the mirror position of the real hand. To read the real time, subtract the image reading from 11:60. An image showing 4:30 means the real time is \(11{:}60 - 4{:}30 = 7{:}30\).

ExampleUsing the image to find where a ray reflects

A plane mirror lies along the \(x\) axis. A point source S is at \((0, 3)\) and a point P is at \((8, 3)\), in metres. A ray from S reaches P after one reflection at Q on the mirror. Find Q and the length of the path SQP.

Reveal one step at a time and finish the rest yourself.
Deeper

The example hides a rule found by Hero of Alexandria in the first century. Of all paths from S to the mirror and on to P, the path that obeys \(r = i\) is the shortest. The straight line from the image S′ to P proves it, since a straight line is the shortest distance. In the chapter on refraction the same idea becomes Fermat's principle of least time.

Checkpoint · numericUnanswered

A girl stands \(2\,\text{m}\) in front of a plane mirror and then walks \(0.5\,\text{m}\) towards it. What is the distance between the girl and her image now, in metres?

m± 2 %
CheckpointUnanswered

The image of a wall clock in a plane mirror seems to show 2:15. What is the real time?

Story · Ibn al-Haytham, 1021
Light comes into the eye

For centuries, many scholars believed that the eye sends out rays that feel the objects it sees. Ibn al-Haytham, working in Cairo, finished his Book of Optics in about 1021. He argued that light travels in straight lines from every point of an object into the eye. He tested the idea with a darkened room and a small hole, the pinhole camera of this lesson. He also studied reflection from plane and curved mirrors with the same care.

Now back to the neem tree. Each gap between the leaves acts as a pinhole, and each spot on the ground is an image of the Sun, not of the gap. The Sun is round, so the spots are round. The Sun's diameter is about \(1.39 \times 10^9\,\text{m}\), at a distance of about \(1.5 \times 10^{11}\,\text{m}\). A gap \(5\,\text{m}\) above the ground therefore gives a spot about \(5 \times 1.39/150 \approx 0.046\,\text{m}\) across, about \(4.6\,\text{cm}\). During an eclipse the Sun itself is a crescent, and so are its images.

EstimateUnanswered

A leaf gap is \(10\,\text{m}\) above the ground. Estimate the diameter of the spot of sunlight it makes on the ground, in cm. The Sun's diameter is about \(1/108\) of its distance from the Earth.

cmrough value is enough
Exam traps
  1. Measuring the angle of incidence from the surface of the mirror.
    Draw the normal first. Both \(i\) and \(r\) are measured from the normal, so a ray at \(40°\) to the mirror has \(i = 50°\).
  2. Saying that a virtual image cannot be seen.
    A virtual image cannot be caught on a screen, but an eye or a camera looking into the mirror sees it clearly.
  3. Putting the image of a plane mirror on the mirror's surface.
    The image is as far behind the mirror as the object is in front of it, on the same normal.