A boy walks home at night along a lane of his colony. He passes a street lamp and walks on, and his shadow stretches out in front of him. The farther he walks, the longer the shadow grows. The lamp is \(6\,\text{m}\) high and the boy is \(1.5\,\text{m}\) tall. How long is his shadow when he is \(3\,\text{m}\) from the foot of the lamp? Nobody needs to measure it.
Triangles of the same shape
Draw the lamp, the boy and the ray of light that just grazes the top of his head. That ray reaches the ground at the tip of his shadow. Two right triangles appear. The large one has the lamp as its upright side, and the small one has the boy. Both have a right angle on the ground, and both have the same angle at the tip of the shadow.
Two triangles whose three angles are equal are called similar triangles. They have the same shape, though not the same size. One is an enlarged copy of the other. So every side of the large triangle is the same multiple of the matching side of the small one. The matching sides are the sides that face equal angles.
Two triangles are similar when the three angles of one are equal to the three angles of the other. Then the sides that face equal angles are in one ratio. If one side of the second triangle is \(k\) times its match in the first, every side is \(k\) times its match. The number \(k\) is the scale factor.
Two equal angles are enough to decide. The three angles of every triangle add up to \(180°\). So if two angles of one triangle equal two angles of another, the third angles are equal too. The lamp and the boy show it at once: the right angle on the ground and the shared angle at the tip.
The areas follow the sides in a squared way. Enlarge a triangle by the factor \(k\), and its base and its height both grow \(k\) times. So its area, half the base times the height, grows \(k^2\) times. A triangle with sides three times as long has nine times the area.
The length of the shadow
Now the boy's shadow follows in four lines. Let the lamp have the height \(H\) and the boy the height \(h\). Let him stand at a distance \(d\) from the foot of the lamp, with a shadow of length \(s\). The small triangle has the height \(h\) and the base \(s\). The large triangle has the height \(H\) and the base \(d + s\), not \(d\).
\[ \dfrac{s}{h} = \dfrac{d + s}{H} \]
Why. the two triangles are similar, so base over height is the same number in both.
\[ Hs = h\,(d + s) \]
Why. multiply both sides by \(hH\).
\[ Hs - hs = h\,d \]
Why. collect the terms that contain \(s\) on the left.
\[ s = \dfrac{h\,d}{H - h} \]
Read it. take \(s\) out as a factor and divide by \(H - h\). This is positive, because the lamp is taller than the boy.
For the lamp of \(6\,\text{m}\) and the boy of \(1.5\,\text{m}\), \(H - h = 4.5\,\text{m}\). At \(d = 3\,\text{m}\), the shadow is \(s = 1.5 \times 3/4.5 = 1\,\text{m}\). In fact \(s = d/3\) at every distance. The shadow is always one third of the boy's distance from the lamp.
The tip of the shadow is at \(d + s = 4d/3\) from the lamp. So when the boy walks at \(1\,\text{m/s}\), the tip of his shadow runs ahead at \(1.33\,\text{m/s}\). The tip always moves faster than the boy, and the shadow grows as he walks.
A lamp post is \(5\,\text{m}\) high. A girl \(1.25\,\text{m}\) tall stands \(6\,\text{m}\) from its foot. How long is her shadow, in metres?
The pinhole camera
A pinhole camera is a dark box with a tiny hole in one face and a screen on the opposite face. Light from the top of an object passes through the hole in a straight line and reaches the screen below the axis. Light from the foot reaches it above. So the image is upside down.
The object and the image make two triangles that meet at the hole. Their angles at the hole are vertically opposite, so they are equal. Both triangles also have a right angle on the axis. So they are similar, and the heights are in the ratio of the distances. \[ \dfrac{h_i}{h_o} = \dfrac{v}{u} \] Here \(h_o\) and \(h_i\) are the heights of the object and the image. The object is a distance \(u\) from the hole, and the screen a distance \(v\).
Stand under a neem tree at noon and look at the ground. The round spots of light are pinhole images of the Sun, made by gaps in the leaves. The Sun is \(1.39 \times 10^9\,\text{m}\) across and \(1.50 \times 10^{11}\,\text{m}\) away, a ratio of about \(1/108\). A gap \(5\,\text{m}\) above the ground makes a spot \(5/108\,\text{m} = 4.6\,\text{cm}\) across. The spots are round even when the gaps are not, because each spot is an image of the round Sun.
A tree \(6\,\text{m}\) tall stands \(20\,\text{m}\) in front of a pinhole camera. The screen is \(10\,\text{cm}\) behind the hole. How tall is the image of the tree, in centimetres?
The screen of a pinhole camera is moved from \(10\,\text{cm}\) to \(20\,\text{cm}\) behind the hole. What happens to the area of the image?
Back to physics: the image in a concave mirror
The chapter Reflection: Plane and Spherical Mirrors finds the mirror formula from two pairs of similar triangles. Here is the same argument with plain lengths, for one mirror. An object AB stands on the axis of a concave mirror, with its foot B on the axis. One ray from the tip A runs parallel to the axis, meets the mirror at M and reflects through the focus F. A second ray goes to the pole P and reflects symmetrically. The two meet at the tip A′ of the image.
The pair at F is A′B′F and MPF. Both have a right angle on the axis, and their angles at F are vertically opposite. The pair at P is A′B′P and ABP. Both have a right angle on the axis, and their angles at P are equal by the law of reflection. For a small mirror, M is almost on the upright line through P, so PM equals the object height AB.
A concave mirror has a radius of curvature of \(40\,\text{cm}\). An object is placed \(30\,\text{cm}\) in front of it. Using only the two pairs of similar triangles, find where the image forms and how large it is.
The focal length is half the radius, \(20\,\text{cm}\). Let the image form a distance \(x\) in front of the mirror. Let \(h\) and \(h'\) be the heights of the object and the image, as plain lengths.
The pair at P gives \(h'/h = x/30\). The pair at F, with \(PM = h\), gives \(h'/h = (x - 20)/20\), since \(B'F = x - 20\).
\(x/30 = (x - 20)/20\) gives \(20x = 30x - 600\), so \(x = 60\,\text{cm}\). Then \(h'/h = 60/30 = 2\).
The image forms \(60\,\text{cm}\) in front of the mirror. It is inverted and twice as tall. With the signs of the physics chapter this is \(v = -60\,\text{cm}\) and \(m = -2\), exactly its answer.
The street lamp, the pinhole and the mirror are one idea. In each case two triangles share their angles, so their sides share one ratio. The shadow of the boy is \(1\,\text{m}\) at \(3\,\text{m}\) from the lamp. The Sun's image is \(4.6\,\text{cm}\) under a gap \(5\,\text{m}\) up. The mirror puts the image of an object at \(30\,\text{cm}\) at \(60\,\text{cm}\), twice as tall.
An object \(1.5\,\text{cm}\) tall stands \(30\,\text{cm}\) in front of a thin convex lens, and a sharp image forms on a screen \(60\,\text{cm}\) behind it. The ray through the optical centre goes straight on. How tall is the image, in centimetres?
A man \(1.8\,\text{m}\) tall stands \(2\,\text{m}\) in front of a plane mirror, and his image stands \(2\,\text{m}\) behind it. The rays from the image of his head and of his feet to his eye cross the mirror halfway along their length. What is the shortest mirror that shows his whole height?
Similar triangles need equal angles, not only a similar look. A lamp that is a long tube, not a point, or a shadow on sloping ground, spoils the equal angles and the ratio. The mirror's triangles are similar only for a small mirror, where PM equals AB. For a wide mirror the pair at F and the pair at P give different ratios, and the image blurs.
The Greek writers tell of Thales, who visited Egypt and wished to know the height of the Great Pyramid. He could not climb it with a measuring rope. So he waited until the shadow of his own body was exactly as long as he was tall. At that moment the shadow of the pyramid was also as long as the pyramid was tall. He measured the shadow on the sand, from the middle of the base, and had the height.
The chapter Reflection: Plane and Spherical Mirrors uses similar triangles three times. They derive the mirror formula, give the linear magnification \(m = -v/u\) and show that a mirror half your height shows your whole body. Spherical Surfaces and Lenses finds the magnification of a thin lens, \(m = v/u\), from the two triangles on either side of the optical centre. Rolling Motion places the instantaneous axis of a rolling wheel by proportion, when two points have parallel velocities of different sizes.
- Taking \(d\) as the base of the lamp's triangle.The large triangle runs from the foot of the lamp to the tip of the shadow, so its base is \(d + s\).
- Matching the wrong sides.Pair the sides that face equal angles: height with height, base with base.
- Doubling an area when the lengths double.Areas of similar figures go as the square of the scale factor, so double the lengths gives four times the area.