Reflection and spherical mirrors

The mirror that makes a face look bigger

Stand close to a shaving mirror and your face looks large. Step back and it turns upside down. A rear-view mirror in a car shows a wide road, but the cars look small and far. All three mirrors are curved, and each curve changes the image in its own way. In this lesson we find out where the image sits and how big it is.

The laws of reflection

A ray of light that hits a surface and bounces back is reflected. Two laws describe it. We measure every angle from the normal, which is the line at right angles to the surface at the point of incidence.

1. The incident ray, the reflected ray and the normal all lie in one plane.
2. The angle of incidence is equal to the angle of reflection: $i = r$.

These laws hold for every mirror, flat or curved. On a curved mirror we draw the normal at each point of the curve.

The image in a plane mirror

A flat mirror gives the simplest image. It has five properties.

• The image is as far behind the mirror as the object is in front of it.
• The image is virtual. The rays only seem to come from it. No light really passes through it, and a screen placed there stays dark.
• The image is upright and as large as the object.
• The image is laterally inverted. Your right hand becomes the left hand of the image.
• The line from the object to the image meets the mirror at a right angle.

Example. Ritu stands $2.0\ \text{m}$ in front of a wall mirror. Her image is $2.0\ \text{m}$ behind the glass, so the distance from Ritu to her image is $4.0\ \text{m}$. She walks toward the mirror at $1.2\ \text{m/s}$. The image walks toward the mirror at the same speed from the other side. Ritu and her image close in at $1.2 + 1.2 = 2.4\ \text{m/s}$.

How big must a mirror be? A person $1.70\ \text{m}$ tall wants to see the whole body. Rays from the top of the head and from the feet must reach the eyes. A little geometry gives the answer: the mirror needs only half the height of the person, $0.85\ \text{m}$. The distance from the mirror does not matter.

A turning mirror and two mirrors

Turn a plane mirror through an angle $\theta$ and keep the incident ray fixed. The normal turns through $\theta$, so the angle of incidence changes by $\theta$. The angle of reflection also changes by $\theta$. The reflected ray therefore turns through $2\theta$.

Example. A lamp and scale instrument, such as an old galvanometer, shines a narrow beam on a small mirror. The reflected beam falls on a scale $2.00\ \text{m}$ away, at right angles to the beam. The mirror turns through $1.5°$. The beam turns through $3.0°$, and the spot moves by $s = 2.00\ \text{m} \times \tan 3.0° = 0.105\ \text{m}$, which is $10.5\ \text{cm}$. A tiny turn of the mirror gives a large move on the scale.

Two plane mirrors that meet at an angle $\theta$ make several images. When $\dfrac{360°}{\theta}$ is a whole number, the number of images is $\dfrac{360°}{\theta} - 1$.

• At $90°$ there are $3$ images.
• At $72°$ there are $4$ images.
• At $60°$ there are $5$ images.
• Mirrors that face each other and are parallel give a long row of images, in principle without end.

Spherical mirrors

A spherical mirror is a small piece of a hollow sphere that is silvered on one side. A concave mirror reflects from the inside of the curve, like the inside of a spoon. A convex mirror reflects from the outside, like the back of a spoon.

We name the points and lines in the picture.

• The pole $P$ is the centre of the mirror surface.
• The centre of curvature $C$ is the centre of the sphere. The radius of curvature $R$ is the radius of that sphere.
• The principal axis is the line through $C$ and $P$.
• The principal focus $F$ is the point where a narrow beam parallel to the axis meets after reflection (concave), or seems to come from (convex). The focal length $f$ is the distance from $P$ to $F$.

Why $f = \dfrac{R}{2}$. A ray parallel to the axis hits the mirror at $M$. The line $CM$ is the normal, so the angle of incidence $i$ is the angle $\angle PCM$. The reflected ray makes the same angle with the normal, so the triangle $CMF$ has two equal angles, and $CF = FM$. For a narrow beam the point $M$ is very close to $P$, so $FM \approx FP$. Then $CF = FP$, and so $f = \dfrac{R}{2}$.

This holds only for rays close to the axis, called paraxial rays. Rays far from the axis meet closer to the mirror. This blur is spherical aberration. A parabolic mirror has no such blur for a distant source, so headlights, dishes and large telescopes use it.

Signs: the New Cartesian convention

We need a rule so that one formula works for every case. The rule used in NCERT and in most Indian textbooks is the New Cartesian sign convention.

• Light travels from left to right. The pole is the origin.
• Distances are measured from the pole. A distance in the direction of the incident light (to the right) is positive. A distance against the incident light (to the left) is negative.
• Heights above the principal axis are positive. Heights below it are negative.

A real object stands in front of the mirror, so the object distance $u$ is always negative. For a concave mirror, the focus and the centre of curvature are in front of the mirror, so $f$ and $R$ are negative. For a convex mirror, they are behind it, so $f$ and $R$ are positive.

Another convention. A-Level, IB and AP courses, and many books outside India, use the real-is-positive convention. The next table puts the two side by side.

The two formulas look the same, but the signs of the numbers are different. Both give the same physical image. When you read an answer, check which convention it uses. In this chapter we use New Cartesian unless we say so.

The mirror formula and the magnification

For paraxial rays, the object distance $u$, the image distance $v$ and the focal length $f$ obey

Here is the short route to it. Take a point object $O$ on the axis. A ray from $O$ hits the mirror at $M$ and is reflected to the image $I$. The normal at $M$ is the line $CM$. By the exterior angle theorem, the angle $\beta$ at $C$ is the sum of the angle $\alpha$ at $O$ and the angle of incidence $i$. In the same way, the angle $\gamma$ at $I$ is the sum of $\beta$ and the angle of reflection $r$. Because $i = r$, the two equations give $\alpha + \gamma = 2\beta$. For small angles each angle is its height divided by a distance along the axis. Put in the signs, and the heights cancel. What is left is $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{2}{R} = \dfrac{1}{f}$.

The linear magnification is the height of the image divided by the height of the object:

A negative $m$ means an inverted image. A positive $m$ means an upright image. A value of $|m|$ above $1$ means the image is larger than the object.

Example 1: a real image. A concave mirror has $R = -30\ \text{cm}$, so $f = -15\ \text{cm}$. An object $2.0\ \text{cm}$ tall stands $45\ \text{cm}$ in front of it, so $u = -45\ \text{cm}$.

1. We want $v$, so $v = \dfrac{uf}{u - f} = \dfrac{(-45\ \text{cm})(-15\ \text{cm})}{-45\ \text{cm} + 15\ \text{cm}} = -22.5\ \text{cm}$.
2. The magnification is $m = -\dfrac{v}{u} = -\dfrac{-22.5\ \text{cm}}{-45\ \text{cm}} = -0.50$.
3. The height of the image is $h' = mh = -0.50 \times 2.0\ \text{cm} = -1.0\ \text{cm}$.

The image is $22.5\ \text{cm}$ in front of the mirror, real, inverted and half the size. Check with the other convention. Now $u = +45\ \text{cm}$ and $f = +15\ \text{cm}$. Then $\dfrac{1}{v} = \dfrac{1}{15\ \text{cm}} - \dfrac{1}{45\ \text{cm}}$, so $v = +22.5\ \text{cm}$, and positive means real. It is the same image.

Example 2: a shaving mirror. The same mirror, $f = -15\ \text{cm}$, now has a face only $10\ \text{cm}$ from it, so $u = -10\ \text{cm}$.

1. $v = \dfrac{(-10\ \text{cm})(-15\ \text{cm})}{-10\ \text{cm} + 15\ \text{cm}} = +30\ \text{cm}$.
2. $m = -\dfrac{30\ \text{cm}}{-10\ \text{cm}} = +3.0$.

The image is $30\ \text{cm}$ behind the mirror, virtual, upright and three times as large. This is why a shaving mirror is concave and why you hold your face close to it.

Example 3: a rear-view mirror. A convex mirror has $f = +0.90\ \text{m}$. A car is $9.0\ \text{m}$ behind, so $u = -9.0\ \text{m}$.

1. $v = \dfrac{(-9.0\ \text{m})(0.90\ \text{m})}{-9.0\ \text{m} - 0.90\ \text{m}} = +0.82\ \text{m}$.
2. $m = -\dfrac{0.82\ \text{m}}{-9.0\ \text{m}} = +0.091$.

The image is virtual, upright and only about one eleventh as tall as the car. That is why the mirror can show a wide road, and why the label says that objects in the mirror are closer than they seem.

Ray diagrams

We find an image by drawing two or three rays whose paths we know.

• A ray parallel to the axis is reflected through $F$ (concave), or away as if from $F$ (convex).
• A ray through $F$ (or aimed at $F$) is reflected parallel to the axis.
• A ray to the pole $P$ is reflected at the same angle on the other side of the axis, because the axis is the normal at $P$.
• A ray through $C$ hits the mirror along the normal. It returns along the same path.

When the object is beyond $C$, the three rays meet in front of the mirror between $C$ and $F$. The image is real, inverted and smaller. The picture is drawn to scale, with the ray paths worked out from the mirror formula.

When the object is between $P$ and $F$, the reflected rays spread out. We trace them backward, the dashed lines, and they meet behind the mirror. The image is virtual, upright and larger.

A convex mirror always gives a virtual, upright and smaller image between $P$ and $F$, wherever the object is.

Start with the defaults. Is the image real or virtual? Now move the object slowly in, from $59\ \text{cm}$ to $5\ \text{cm}$. Where does the image jump from one side of the mirror to the other?

Practise with new numbers

Try it in Python

The program takes a concave mirror of focal length $15\ \text{cm}$ and moves an object in toward it, step by step. For each position it finds the image distance, the magnification and the kind of image.

Common mistakes

• Using $u$ as a positive number in the New Cartesian formula. A real object has a negative $u$.
• Giving a concave mirror a positive $f$. In New Cartesian a concave mirror has $f < 0$. In the real-is-positive form it has $f > 0$. Never mix the two in one calculation.
• Forgetting that $f = \dfrac{R}{2}$ only works for paraxial rays. It is not true for a wide beam.
• Saying that a virtual image cannot be seen. You can see it, in a plane mirror every day. It only cannot be caught on a screen.
• Turning the reflected ray by the angle of the mirror. A mirror that turns by $\theta$ turns the reflected ray by $2\theta$.

Remember these

• Reflection: $i = r$, and the incident ray, the reflected ray and the normal lie in one plane.
• A plane mirror gives a virtual, upright, same-size image as far behind as the object is in front.
• A spherical mirror has $f = \dfrac{R}{2}$. In New Cartesian, a concave mirror has $f < 0$ and a convex mirror has $f > 0$.
• Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$. Magnification: $m = -\dfrac{v}{u}$.
• Concave mirror: an object beyond $F$ gives a real, inverted image. An object between $F$ and $P$ gives a virtual, upright, larger image. A convex mirror always gives a virtual, upright, smaller image.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A girl stands $3.0\ \text{m}$ from a plane mirror. How far is she from her image? She moves $1.0\ \text{m}$ closer. What is the distance now?

2. A narrow beam hits a plane mirror and is reflected. The mirror turns through $8°$. Through what angle does the reflected beam turn?

3. A concave mirror has a radius of curvature of $40\ \text{cm}$. Where must an object be placed so that the image is real, inverted and of the same size?

4. A student says: a virtual image is not real, so no light is involved. Is this right? Use a plane mirror to explain.

5. A convex mirror has $f = +20\ \text{cm}$. An object is far from it. Why does the image always lie between the pole and the focus?

References

• NCERT Physics Part II (Class 12), Chapter 9, the sections on reflection of light by spherical mirrors.
• OpenStax University Physics Volume 3, Chapter 2, the sections on plane mirrors and spherical mirrors.
• OpenStax College Physics 2e, Chapter 25, the sections on the law of reflection and image formation by mirrors.