A barber's shop in any Indian town has a large mirror in front of the chair and another on the wall behind it. Sitting in the chair, you see yourself, then the back of your head, then yourself again, smaller and dimmer, in a long curving corridor. The corridor seems to go on for ever.
Half your height is enough
A dressing mirror hangs on a wall. How tall must it be to show your whole body? Let the top of your head be H, your eyes E and your feet F. The eye sees the top of the head by a ray from H that reflects into E. By \(r = i\), that ray meets the mirror at the height halfway between H and E. The eye sees the feet by a ray that meets the mirror halfway between E and F.
So the mirror must run from half of EF above the floor to half of EH above the eye. Its length is \(\tfrac12 EF + \tfrac12 EH = \tfrac12 HF\), half your height. The distance from the mirror never entered the argument. A person \(180\,\text{cm}\) tall with the eyes \(10\,\text{cm}\) below the top needs \(90\,\text{cm}\) of mirror. Its lower edge must be \(170/2 = 85\,\text{cm}\) above the floor, and its upper edge \(170 + 5 = 175\,\text{cm}\).
A child \(120\,\text{cm}\) tall has eyes \(110\,\text{cm}\) above the floor. How high above the floor must the lower edge of a mirror be for the child to see the feet, in cm?
Two mirrors at an angle
Put an object between two plane mirrors that meet at an angle \(\theta\). Each mirror forms an image of the object. Each image then acts as an object for the other mirror, which forms an image of the image. The chain goes on while the latest image is still in front of the next mirror. It stops when an image lies behind both mirrors, since no mirror faces it. Name each image by its mirrors in order: \(I_{12}\) is the image in \(M_1\), then in \(M_2\).
Every reflection keeps the distance from the line where the mirrors meet. So the object and all its images lie on one circle about that line. Take mirrors at \(90°\) along the \(x\) and \(y\) axes, and the object at \((2, 1)\) cm. The image in \(M_1\) is at \((2, -1)\) and the image in \(M_2\) at \((-2, 1)\). Each of these is reflected in the other mirror, and both give the same third image at \((-2, -1)\). There are three images.
The count follows a rule in \(n = 360°/\theta\). If \(n\) is even, there are \(n - 1\) images wherever the object is: \(90°\) gives \(3\), and \(60°\) gives \(5\). If \(n\) is odd, the two last images coincide only for an object on the bisector. So there are \(n - 1\) images on the bisector and \(n\) off it: \(72°\) gives \(4\) or \(5\).
When \(360°/\theta\) is not a whole number, the count is its whole part or one more, depending on the position of the object. Many books quote the whole part alone. For \(\theta = 50°\), \(360°/50° = 7.2\). Following the chains gives \(7\) images for an object within \(20°\) of either mirror. For an object within \(5°\) of the bisector it gives \(8\).
Two plane mirrors are inclined at \(45°\). How many images of an object placed between them are formed?
As the angle closes to zero the mirrors become parallel, and \(360°/\theta\) grows without limit. Two parallel mirrors give endless images, which is the corridor of the barber's shop. Take mirrors \(3\,\text{m}\) apart with the object \(1\,\text{m}\) from \(M_1\). The nearest image in \(M_1\) is \(1\,\text{m}\) behind \(M_1\), \(2\,\text{m}\) from the object. The nearest image in \(M_2\) is \(2\,\text{m}\) behind \(M_2\), \(4\,\text{m}\) from the object, and the two are \(6\,\text{m}\) apart.
Two plane mirrors are inclined at \(72°\), and a small lamp is placed on the bisector of the angle between them. How many images of the lamp are seen?
In 1816 the Scottish physicist David Brewster was studying how light reflects from glass. He placed two strips of mirror at an angle inside a tube, with coloured fragments at one end. At \(60°\) the fragments and their five images filled a circle with a pattern of six matching sectors. He patented the kaleidoscope in 1817, and within months it was sold in huge numbers across Britain.
Moving images
The image is always as far behind the mirror as the object is in front. Suppose the object moves \(1\,\text{m}\) towards a fixed mirror. The image must then move \(1\,\text{m}\) towards the mirror too, from the other side. So along the normal the image moves at the same speed as the object, in the opposite direction. Parallel to the mirror, the image copies the object exactly.
Now move the mirror instead. Keep the object still and move the mirror a distance \(s\) towards it. The object is now \(s\) nearer the mirror, so the image is \(s\) nearer the mirror too. But the mirror itself moved \(s\), so the image moved \(2s\). A mirror moving at \(2\,\text{m/s}\) moves the image at \(4\,\text{m/s}\), in the same direction as the mirror.
The formula handles every case. Take the direction from the object towards the mirror as positive. An object at \(+5\,\text{m/s}\) and a mirror at \(-2\,\text{m/s}\) give \(v_I = 2(-2) - 5 = -9\,\text{m/s}\). The image moves at \(9\,\text{m/s}\) towards the object, and the two close in at \(5 + 9 = 14\,\text{m/s}\).
A plane mirror lies in the \(y\)-\(z\) plane. An object moves with \(\vec v_O = (3\hat i + 4\hat j)\,\text{m/s}\). Find the velocity of the image, its velocity relative to the object, and the image velocity if the mirror also moves at \(2\hat i\,\text{m/s}\).
(a) The normal is the \(x\) axis. The \(x\) component is reversed and the \(y\) component is copied: \(\vec v_I = (-3\hat i + 4\hat j)\,\text{m/s}\), with speed \(\sqrt{9 + 16} = 5\,\text{m/s}\), as fast as the object.
(b) \(\vec v_I - \vec v_O = (-3 - 3)\hat i + (4 - 4)\hat j = -6\hat i\,\text{m/s}\). The object and the image approach each other along the normal at \(6\,\text{m/s}\).
(c) Along the normal, \(v_{Ix} = 2(2) - 3 = 1\,\text{m/s}\); along the mirror, \(v_{Iy} = 4\,\text{m/s}\). So \(\vec v_I = (\hat i + 4\hat j)\,\text{m/s}\).
An object is at rest. A plane mirror moves towards it along its normal at \(3\,\text{m/s}\). At what speed does the image move relative to the ground, in m/s?
An object moves at \(4\,\text{m/s}\) parallel to a plane mirror that is at rest. Which statements are true?
Now the barber's shop is clear. The two mirrors are nearly parallel, so each image is an object for the other mirror, and the chain has no end. Each image of an image sits farther away, so it looks smaller. Each reflection loses some of the light, so each is also dimmer than the last. The mirrors are rarely exactly parallel, so the corridor curves slowly away.
- Thinking a farther mirror must be taller to show your whole body.The points of reflection sit halfway between the levels of your head, eyes and feet, whatever the distance. Half your height is always enough.
- Using \(360°/\theta - 1\) for every angle.That holds for an even \(360°/\theta\). For an odd value, off the bisector there are \(360°/\theta\) images, and for a value that is not whole the count depends on the position.
- Saying the image of a moving mirror moves at the mirror's speed.A mirror moving along its normal moves the image at twice its speed: \(v_I = 2v_M - v_O\).