Math Detours · Grade 10 · Connections · 16

Find the image with two rays

Where does a lens put the picture of a candle? Draw just two rays with a ruler. In the thin-lens model, the point where they cross is where the image forms.

Name ____________________Date ____________

Tonight's idea

Light leaves the tip of a candle flame in every direction. A convex lens (fatter in the middle) bends the rays that hit it. If the candle is far enough away, they all meet again at one point beyond the lens: the image of the tip. Two rays are enough to find it, and two are easy to draw.

A thin convex lens has focal length \(f\). Its axis is the line through its centre at right angles to it. Ray 1, through the centre of the lens, goes straight on. Ray 2, arriving parallel to the axis, bends at the lens and passes through the focus \(F\), on the axis \(f\) beyond the lens. The image is where they cross.

The lens formula agrees: \(\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}\), with \(u\) the distance from object to lens and \(v\) from lens to image. Positive \(v\) is beyond the lens. Negative \(v\) is on the object's side.

This is a model: a thin lens, and rays close to the axis. Solving a light problem by drawing rays is called ray tracing.

Watch one: a candle 30 cm from the lens

10 cm candle ray 2 ray 1 lens F image 5 cm 30 cm 15 cm
\(F\) is 10 cm beyond the lens. Heights are drawn large to be clear: in the model, a tip at any height gives the same \(v\).

A lens has \(f = 10\) cm. A candle stands 30 cm in front of it, and the tip of the flame is 10 cm above the axis. Measure \(x\) cm beyond the lens, and heights in cm above the axis.

Ray 1 drops 10 cm in the 30 cm from the tip to the centre of the lens, and carries on in the same line. Its height beyond the lens is \(-\frac{x}{3}\). Ray 2 runs level to the lens, 10 cm up, then heads for \(F\), dropping 10 cm in 10 cm. Its height is \(10 - x\).

They cross when \(10 - x = -\frac{x}{3}\), so \(10 = \frac{2x}{3}\) and \(x = 15\), at height \(-5\). The image is 15 cm beyond the lens, upside down and half as tall. Paper held there shows the picture: a real image.

Check: \(\frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}\), so \(v = 15\). Ray 1 is one straight line through the centre, so by similar triangles the image is \(\frac{v}{u} = \frac{15}{30}\) as tall as the object.

Your turn

  1. Keep the lens (\(f = 10\) cm) and the candle (tip 10 cm above the axis). Move the candle to 20 cm from the lens. (a) On spare paper draw it to scale, 1 cm for every 5 cm: axis, lens, \(F\), candle. (b) Draw rays 1 and 2. Where do they cross, in real cm? (c) Check with the formula. How tall is the image?
  2. Now the candle is 15 cm from the lens. (a) Use the formula to predict \(v\). (b) Check with a scale drawing. (c) How tall is the image, and which way up? (d) An object at 30 cm gave an image at 15 cm. An object at 15 cm gives an image at 30 cm. Is that luck?
  3. Move the candle to 5 cm from the lens. (a) Draw the two rays. Beyond the lens, do they get closer together or further apart? (b) Extend both rays backwards with dashed lines. Where do those cross? (c) What does the formula give for \(v\)? (d) Your eye looks back through the lens. What does it see?
  4. (a) Put the candle exactly 10 cm from the lens. How steeply does each ray drop beyond the lens? Where is the image? What does the formula say? (b) Find \(v\) and the height of the image for a candle at 11 cm, and then at 9 cm. Describe what the image does as the candle slides from 11 cm to 9 cm.

Stretch: why a magnifying glass is held close

A thin convex lens has focal length \(f\), and an object is at distance \(u\) with \(0 \lt u \lt f\). (a) Show that \(v = \frac{uf}{u - f}\), and say why it is negative. (b) Ray 1 makes the image \(\frac{\text{size of } v}{u}\) times as tall as the object. Show that this is \(\frac{f}{f - u}\), and that it is always more than 1. (c) With \(f = 10\) cm, where must the object go for an image 5 times as tall? (d) Why is there no upright image when \(u \gt f\)?

Answers

  1. (a) On paper \(F\) is 2 cm beyond the lens, and the candle is 4 cm in front of it with its tip 2 cm up. (b) The rays cross 4 cm beyond the lens and 2 cm below the axis. In real cm: 20 cm beyond the lens, 10 cm below the axis. (c) \(\frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}\), so \(v = 20\). The image is \(\frac{20}{20} = 1\) times as tall: 10 cm, the same size as the candle, upside down.
  2. (a) \(\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30}\), so \(v = 30\) cm. (b) On paper the candle is 3 cm in front of the lens, and the rays cross 6 cm beyond it, 4 cm below the axis. (c) \(\frac{v}{u} = \frac{30}{15} = 2\), so it is 20 cm tall, upside down. (d) Not luck. The formula treats \(u\) and \(v\) in the same way, so if one pair fits, the swapped pair fits too. Light can also run backwards along the same rays.
  3. (a) Further apart. Ray 1 drops 10 cm in 5 cm, which is 2 cm for each cm. Ray 2 drops only 1 cm for each cm and starts 10 cm higher. At \(x\) cm beyond the lens the gap is \(10 + x\) cm, so they never cross there. (b) 10 cm from the lens on the candle's side, 20 cm above the axis (on paper: 2 cm and 4 cm). (c) \(\frac{1}{v} = \frac{1}{10} - \frac{1}{5} = -\frac{1}{10}\), so \(v = -10\). The minus sign says the image is on the candle's side. (d) The rays reach the eye as if they came from that crossing point. So the eye sees a candle the right way up and twice as tall, 10 cm from the lens on the candle's side. No light really passes through that point, so paper held there shows nothing. It is called a virtual image, and this is a magnifying glass at work.
  4. (a) Ray 1 drops 10 cm in 10 cm, which is 1 cm for each cm. So does ray 2. They are parallel and never cross, forwards or backwards. There is no image. The formula gives \(\frac{1}{v} = \frac{1}{10} - \frac{1}{10} = 0\), and no number \(v\) has \(\frac{1}{v} = 0\). (b) At 11 cm: \(\frac{1}{v} = \frac{1}{10} - \frac{1}{11} = \frac{1}{110}\), so \(v = 110\) cm, and the image is \(\frac{110}{11} = 10\) times as tall: 100 cm, upside down. At 9 cm: \(\frac{1}{v} = \frac{1}{10} - \frac{1}{9} = -\frac{1}{90}\), so \(v = -90\) cm, and the image is \(\frac{90}{9} = 10\) times as tall: 100 cm, upright, on the candle's side. As the candle nears 10 cm the image races away beyond the lens and grows. At 10 cm there is none. Just inside, it is back, far away on the candle's side and the right way up.

Stretch. (a) \(\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{u - f}{uf}\), so \(v = \dfrac{uf}{u - f}\). The top is positive and the bottom is negative because \(u \lt f\). So the image is on the object's side: a virtual image. (b) The size of \(v\) is \(\dfrac{uf}{f - u}\). Divide by \(u\) to get \(\dfrac{f}{f - u}\). Since \(f - u\) is positive and less than \(f\), this is more than 1. (c) \(\dfrac{10}{10 - u} = 5\) gives \(10 - u = 2\), so \(u = 8\) cm. Check: \(v = \dfrac{80}{-2} = -40\) and \(\dfrac{40}{8} = 5\). (d) If \(u \gt f\) then \(v\) is positive, so the rays really cross beyond the lens. Ray 1 started above the axis and crossed it at the centre of the lens, so beyond the lens it is below the axis. The image is upside down. An upright image needs \(u \lt f\), and then it is always enlarged. That is why you hold a magnifying glass close. These ratios compare heights only. How big a thing looks to your eye also depends on how far away it is.

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