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Cambridge IGCSE0625

Light

Physics 0625 Chapter Notes

What this chapter covers

Light - Reflection of lightLight - Refraction of lightLight - Thin lensesLight - Dispersion of light
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In this chapter

  1. 1Normal, angle of incidence and angle of reflection
  2. 2The image in a plane mirror
  3. 3The law of reflection
  4. 4Constructing the image and the ray path
  5. 5Normal, angle of incidence and angle of refraction
  6. 6Light crossing a boundary
  7. 7Critical angle and total internal reflection
  8. 8Refractive index as a ratio
  9. 9Using the sines of the angles
  10. 10Refractive index and the critical angle
  11. 11Optical fibres
  12. 12Converging and diverging lenses
  13. 13Real images and ray diagrams
  14. 14Virtual images
  15. 15Virtual images and the magnifying glass
  16. 16Correcting short and long sight
  17. 17Dispersion by a glass prism
  18. 18Which colour bends most
  19. 19The seven colours in order
  20. 20Monochromatic light

1. Normal, angle of incidence and angle of reflection

The normal is the line drawn at right angles to the mirror, at the point where the ray hits it. The incident ray is the ray that arrives at the mirror. The reflected ray is the ray that leaves it.

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. Neither angle is measured from the mirror surface.

Examiner insight

“A large number of candidates correctly stated that the dashed line was the normal. However, a variety of incorrect answers were also seen, such as ‘mirror line’, ‘refraction’ or ‘reflection’.” “A common error was selecting the angles between the mirror and the rays rather than the angles between the rays and the normal.”

Worked example 11 mark

A light ray strikes a plane mirror and is reflected. Which angle is always equal in size to the angle of reflection?

A the angle between the incident ray and the mirror · B the angle between the incident ray and the normal to the mirror · C the angle between the reflected ray and the mirror · D the angle between the reflected ray and the incident ray

  1. 1

    B: the angle between the incident ray and the normal to the mirror. [1]

2. The image in a plane mirror

A plane mirror forms an optical image of an object. The image is the same size as the object. It is the same distance behind the mirror as the object is in front.

The image is virtual. It is also upright.

Examiner insight

“Most candidates knew that the image was the same size as the object but had the misconception that the image was real, so chose option B.” “A common error was to say that the image was the same shape rather than the same size, and to simply state that the image was the same distance, i.e. not stating that the image was the same distance from the mirror as the object.”

Worked example 11 mark

Which description of the image of an object in a plane mirror is correct?

A upright and real · B upright and virtual · C upside down and real · D upside down and virtual

  1. 1

    B: upright and virtual. [1]

3. The law of reflection

For reflection, the angle of incidence is equal to the angle of reflection. Both angles are measured from the normal.

If a question gives the angle between the ray and the mirror, subtract it from 90° first.

i = r

Examiner insight

“Very few candidates were able to state the law of reflection. The most common error was to attempt some form of Snell’s law.” “A small number of candidates failed to obtain credit for the last part stating that the incident ray was equal to the reflected ray rather than referring to the angles of incidence and reflection.”

Worked example 11 mark

Figure 6.1 shows a ray of red light reflected by a plane mirror. The value of angle P on Figure 6.1 is 50°. Determine the value of the angle of reflection for the ray of red light.

  1. 1

    P is measured from the mirror, so angle of incidence = 90° − 50° = 40°. Angle of reflection = 40° [1]

4. Constructing the image and the ray path

To find the image, draw a line from the object at right angles to the mirror and extend it behind. Mark the image the same distance behind the mirror as the object is in front.

To draw the ray from the object to the eye, join the image to the eye with a ruler. The ray is reflected at the mirror where this line crosses it. Draw the ray from the object to that point, then on to the eye, with arrows. Check that angle of reflection = angle of incidence.

Examiner insight

“Some common errors included the omission of, or wrong direction of, arrows on the rays, or the rays not reaching the eye.” “Stronger candidates carefully marked with an X the position of the image as far behind the mirror as the clock was in front of the mirror along a line from the clock perpendicular to the mirror.”

Worked example 13 marks

An image of object O is formed by a plane mirror. A person views the image. Figure 6.1 shows a scale diagram of the arrangement.

(i) Mark the position of the image. Label it I.

(ii) Draw on Figure 6.1 to show the path of the light from object O to the person's eye.

  1. 1

    I is on the line through O at right angles to the mirror, 80 + 240 = 320 cm behind the mirror [1]

  2. 2

    ray from O reflected at the mirror and goes to the eye [1]

  3. 3

    angle of reflection = angle of incidence [1]

Recap

  • The normal is at right angles to the mirror. Both angles are measured from it.
  • Angle of incidence = angle of reflection.
  • A plane mirror image is the same size, the same distance behind the mirror, and virtual.
  • Extended: the ray is reflected where the line from the image to the eye crosses the mirror.

5. Normal, angle of incidence and angle of refraction

The normal is the line at right angles to the surface at the point where the ray meets it. The angle of incidence is the angle between the incident ray and the normal. The angle of refraction is the angle between the refracted ray and the normal.

To show refraction, shine a narrow ray of light at transparent blocks of different shapes, such as a rectangular block, a semicircular block or a prism. Draw the normal where the ray meets each surface.

Examiner insight

“Incorrect answers included drawing the normal at right angles to the incident ray, only drawing a refracted ray or drawing a continuation of the incident ray.”

Worked example 11 mark

A narrow beam of light travels through glass. It reaches the edge of the glass and refracts into the air. What is the angle of refraction?

  1. 1

    the angle of refraction is between the refracted ray and the normal

  2. 2

    D [1]

6. Light crossing a boundary

When light passes from one medium into another, its speed changes. The ray usually changes direction. This is refraction.

If the speed of light is less in the second medium, the ray refracts towards the normal. When it leaves, it refracts away from the normal.

A ray that enters along the normal does not change direction. Its angle of incidence is 0°.

Examiner insight

“Few candidates were able to specify in which direction the ray of light was refracted.” “Common errors included refracting the ray in glass beyond the normal and having the emergent ray refract towards the normal instead of away from the normal.”

Worked example 11 mark

A light ray travelling in air hits the surface of a transparent medium. The speed of light in air is greater than the speed of light in the medium. What happens to the ray?

A The ray is refracted with an angle of refraction that is greater than the angle of incidence. · B The ray is totally internally reflected. · C The ray refracts away from the normal. · D The ray refracts towards the normal.

  1. 1

    the speed is less in the medium

  2. 2

    D: the ray refracts towards the normal. [1]

7. Critical angle and total internal reflection

The critical angle is the angle of incidence in the denser medium when the angle of refraction is 90°.

For total internal reflection the ray must travel in a more dense medium towards a less dense medium. The angle of incidence must be greater than the critical angle. The ray is then totally reflected: there is no refracted ray, and the angle of reflection equals the angle of incidence. If some of the light still refracts out, the reflection is not total.

You can see it in a semicircular glass block, by raising the angle of incidence past the critical angle. Everyday examples are a glass prism and an optical fibre.

Examiner insight

“Candidates were required to state that for total internal reflection, a ray had to be travelling in a more dense medium towards a less dense medium.” “An important requirement in this case was to make the point that the critical angle is the angle of incidence, not simply “the angle”.”

Worked example 11 mark

State the meaning of critical angle.

  1. 1

    (value of the) angle of incidence when the angle of refraction = 90° [1]

8. Refractive index as a ratio

Refractive index, n, is the ratio of the speeds of a wave in two different regions. For light going from air into a medium, divide the speed in air by the speed in the medium. The same ratio works for the wavelengths. The light has a shorter wavelength in the medium.

n = (speed in air)/(speed in glass)

n = λ_air/λ_glass

Worked example 11 mark

When red light travels from air into glass, the speed of the light decreases from 3.0 × 10⁸ m/s to 2.1 × 10⁸ m/s. The wavelength of the light in air is 620 nm. What is the wavelength of the light in the glass? A 430 nm · B 560 nm · C 620 nm · D 890 nm

  1. 1

    n = (3.0 × 10⁸)/(2.1 × 10⁸) = 1.43

  2. 2

    λ_glass = 620/1.43 = 430 nm (A) [1]

9. Using the sines of the angles

To find n from angles, use the sines of the angle of incidence and the angle of refraction. Both angles are measured from the normal. If a question gives the angle with the surface, take it from 90° first.

n = (sin i)/(sin r)

Examiner insight

“Over a third of the candidates took the angle of incidence as 40° instead of using that value to calculate the angle of incidence relative to the normal.”

Worked example 11 mark

The diagram shows a ray of light entering a glass block. What is the refractive index of the glass?

A 1.25 · B 1.29 · C 1.58 · D 1.64

  1. 1

    angle of incidence = 90° - 44° = 46°

  2. 2

    n = (sin i)/(sinr) = (sin 46°)/(sin 26°) = 1.64 (D) [1]

10. Refractive index and the critical angle

For light going from a more dense medium to a less dense medium, the sine of the critical angle is the denominator.

Find the critical angle with the inverse sine. Refractive index has no unit.

n = 1/(sin c)

sin c = 1/n

Examiner insight

“Weaker candidates often scored partial credit for a correct equation but if n = sin i / sin r was used they did not realise that the sine of the critical angle should have been the denominator for light travelling from a denser to a rarer medium.” “Weaker answers were those where the equation was incorrectly rearranged to c = 1 / sin n.”

11. Optical fibres

An optical fibre is made of glass. Light travels along it by total internal reflection. At each reflection the angle of reflection equals the angle of incidence.

Optical fibres are used in telecommunications, for example high-speed broadband. The glass is transparent to the radiation, and the fibre carries a high rate of data.

Examiner insight

“Others showed total internal reflection but with unequal angles of incidence and reflection.” “Correct answers needed to refer to the transparency to the radiation and also that high-speed broadband relies on the optical fibre carrying a high rate of data.”

Recap

  • Angles of incidence and refraction are measured from the normal, never from the surface.
  • Into a slower medium the ray refracts towards the normal. Out of it, away from the normal.
  • Critical angle: the angle of incidence when the angle of refraction is 90°.
  • Total internal reflection: more dense to less dense, with angle of incidence greater than the critical angle.
  • Extended: n = sin i / sin r, n = 1 / sin c, and n is the ratio of the speeds of a wave in two regions.
  • Extended: optical fibres carry telecommunications data by total internal reflection.

12. Converging and diverging lenses

A thin converging lens makes a parallel beam of light converge to a point. The light refracts as it passes through the lens. A thin diverging lens makes a parallel beam diverge (spread out).

In a ray diagram the principal axis is the straight line drawn through the centre of the lens, at right angles to it. The principal focus (focal point) is a point on the principal axis where rays parallel to the principal axis converge after passing through the lens.

The focal length is the distance from the centre of the lens to the principal focus. A converging lens has two principal foci, one on each side.

Worked example 12 marks

Complete the definition of the ‘principal focus’ of a lens.

The principal focus of a lens is a point on the principal axis where …

  1. 1

    rays parallel to the principal axis [1]

  2. 2

    converge after passing through the lens [1]

13. Real images and ray diagrams

To find the image of a point on an object, draw two rays from it. One ray runs parallel to the principal axis and is refracted through the principal focus. The other goes through the centre of the lens undeviated. The image of that point is where the rays cross.

A real image is formed where the rays actually meet. It can be projected on a screen.

Describe an image with one term from each group: enlarged, same size or diminished; upright or inverted; real or virtual. If the object moves closer to the lens but stays outside the principal focus, the image gets larger. An object at twice the focal length from the lens (2F) gives an image the same size, also at 2F.

Examiner insight

“A common error was not drawing the ray through the centre as a continuous straight line.” “Some candidates repeated the characteristic given in the question stem.”

Worked example 12 marks

Figure 6.1 is a full-scale drawing of an object O, a thin converging lens L and the image I of object O. The image in Figure 6.1 is inverted.

State two other characteristics of the image.

  1. 1

    real [1]

  2. 2

    same size [1]

14. Virtual images

A virtual image is formed when diverging rays are extrapolated backwards. The rays only seem to come from the image.

A virtual image does not form a visible projection on a screen. For the mark, say that it cannot be formed on a screen or that the light rays do not actually meet at the image.

Examiner insight

“Stronger candidates stated clearly that either the virtual image could not be formed on a screen or that the light rays do not intersect at the image.” “Answers such as “upright” or “on the same side as the lens” were not awarded credit as the question was referring to virtual images in general.”

15. Virtual images and the magnifying glass

A converging lens forms a virtual image when the object is between the lens and the principal focus. The rays diverge after the lens. Extend them backwards to find where they seem to come from: that is the image.

The image is virtual and upright. It is on the same side of the lens as the object and further away from the lens than the object.

A single lens used like this is a magnifying glass. The object is closer to the lens than the principal focus, and the image is enlarged.

Examiner insight

“Some candidates placed X at or directly above F.” “A common mistake was to suggest that the image is on the opposite side of the lens.”

Worked example 11 mark

Figure 6.2 shows a different converging lens with a focal length F. The lens in Figure 6.2 is used as a magnifying glass.

Put a tick alongside the two statements which describe the position of the magnified image.

on the opposite side of the lens to the object · on the same side of the lens as the object · the same distance from the lens as the object · closer to the lens than the object · further away from the lens than the object

  1. 1

    on the same side of the lens as the object and further away from the lens than the object [1]

16. Correcting short and long sight

A short-sighted eye sees distant objects blurred and close objects clearly. The image of a distant object forms in front of the retina, because the eye bends the parallel light too much.

A diverging lens corrects short-sightedness. It spreads the light out a little before the eye bends it, so the rays meet on the retina.

A long-sighted eye sees distant objects clearly and close objects blurred. The rays would not meet until behind the retina. A converging lens corrects long-sightedness: it makes the rays meet on the retina.

Examiner insight

“This is because parallel light is bent too much, so it needs to be corrected by a diverging lens to ‘unbend’ it a little before the eye does its bending.” “A common misconception was to think that long-sightedness is when people see faraway objects less clearly.”

Worked example 11 mark

A student is short-sighted.

The student is standing in a park looking at a tree in the distance.

Where, in relation to the retina, does the eye form an image of the tree, and which type of lens is used to correct the image?

location of imagetype of lens
Abehind retinaconverging
Bbehind retinadiverging
Cin front of retinaconverging
Din front of retinadiverging
  1. 1

    D: in front of retina, diverging lens. [1]

Recap

  • A converging lens brings a parallel beam to the principal focus. A diverging lens makes it diverge.
  • Focal length = distance from the centre of the lens to the principal focus.
  • Rays: one parallel to the principal axis then through the principal focus, one through the lens centre.
  • Describe an image: enlarged / same size / diminished, upright / inverted, real / virtual.
  • A real image can be projected on a screen. A virtual image cannot, and its rays do not actually meet.
  • Extended: an object inside the principal focus gives a virtual, upright image: a magnifying glass.
  • Extended: short sight (image in front of the retina) is corrected by a diverging lens. Long sight (rays would meet behind the retina) is corrected by a converging lens.

17. Dispersion by a glass prism

When white light passes through a glass prism it is refracted and separated into the colours of the spectrum. This separation is dispersion.

The colours separate because they are refracted by different amounts. The dispersion starts as the light enters the prism, not only as it leaves.

A ray of red light alone will not disperse.

Examiner insight

“Weak answers were too vague just referring to different colours appearing or light being split into a rainbow.” “Candidates need to be aware that a ray of red light will not disperse.”

Worked example 11 mark

White light enters a glass prism and emerges separated into the seven colours of the visible spectrum. Which term describes this effect? A dispersion B reflection C total internal reflection D transmission

  1. 1

    A: the separation of white light into colours is dispersion [1]

18. Which colour bends most

The colours are refracted by different amounts. Violet is refracted more than the other colours. Red is refracted least.

On a diagram of a prism, the red ray is the one nearest the original direction of the beam. The violet ray is bent furthest from it.

Examiner insight

“Only stronger candidates recognised that violet light is refracted more than other colours.” “Very few candidates knew that the ray of blue light would refract more than the ray of red light.”

Worked example 14 marks

A beam of white light is dispersed in a glass prism. Figure 9.1 shows the paths taken by green, indigo, red and yellow light. State which of the rays, 1, 2, 3 or 4, is the green ray and the red ray.

The light leaves the prism. On Figure 9.1, continue ray 1 and ray 4 to show the path of light in the air.

  1. 1

    the green ray is ray 3 [1]

  2. 2

    the red ray is ray 1, bent least [1]

  3. 3

    ray 1 is refracted away from the normal as it leaves the prism [1]

  4. 4

    ray 1 and ray 4 diverge in the air [1]

19. The seven colours in order

The traditional seven colours are red, orange, yellow, green, blue, indigo and violet. Do not add white, black or pink. Do not give both purple and violet.

In order of increasing frequency the list runs red to violet. Red has the lowest frequency and violet the highest.

In order of increasing wavelength it runs the other way. Red has the longest wavelength. In a vacuum all colours travel at the same speed.

Examiner insight

“More candidates chose option A than the correct option (D), indicating that they had probably learned the ROYGBIV mnemonic by rote, without recognising that red light has a lower frequency and a higher wavelength than violet light (as specified in Syllabus Section 3.2.4.2).” “A considerable number of candidates correctly named all seven colours but the knowledge that violet has the highest frequency was less secure.”

Worked example 11 mark

Which list states three colours of the visible spectrum in order of decreasing frequency? A red, yellow, blue B orange, green, blue C indigo, yellow, green D violet, orange, red

  1. 1

    D: violet has the highest frequency and red the lowest, so violet, orange, red decreases in frequency [1]

20. Monochromatic light

Visible light of a single frequency is described as monochromatic. Light of a single wavelength also counts.

One colour is not enough. Each colour of light covers a range of frequencies, so a beam of one colour can still contain many frequencies.

Examiner insight

“Candidates should know that light of one colour can still have a range of frequencies and so is an inadequate definition of monochromatic light.” “Weaker candidates often gave the insufficient answer that it is light of a single colour.”

Worked example 11 mark

Complete the sentence. Light of a single frequency is described as ........ light.

  1. 1

    monochromatic [1]

Recap

  • Dispersion is the separation of white light into its colours, for example by a glass prism.
  • The colours are refracted by different amounts. Violet is refracted most, red least.
  • A ray of red light alone will not disperse.
  • Seven colours: red, orange, yellow, green, blue, indigo, violet.
  • Red to violet is increasing frequency. Red has the longest wavelength.
  • Extended: visible light of a single frequency is monochromatic. One colour is not enough.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A ray of light hits a plane mirror. Name the line drawn at right angles to the mirror at the point where the ray hits it.1 mark
  2. State the law of reflection.1 mark
  3. State two characteristics of the image formed by a plane mirror.2 marks
  4. A ray of light hits a plane mirror. The angle between the incident ray and the surface of the mirror is 35°. Calculate the angle of reflection.2 marks
  5. The angle between an incident ray and the reflected ray at a plane mirror is 70°. Calculate the angle of incidence.2 marks
  6. A boy stands 1.5 m in front of a plane mirror. Calculate the distance between the boy and his image.2 marks
  7. A student stands 3.0 m in front of a plane mirror and faces it. A clock hangs on the wall 1.0 m directly behind the student. Calculate the distance from the student to the image of the clock.3 marks
  8. Two plane mirrors meet at right angles. A ray of light hits the first mirror with an angle of incidence of 30°. It is reflected on to the second mirror. Calculate the angle of incidence at the second mirror.3 marks
  9. A stage technician aims a horizontal laser beam at a small plane mirror. The mirror turns the beam through 50°, so the reflected beam travels at 50° to the original direction of the beam. (a) The incident ray and the reflected ray meet at the mirror. Calculate the angle between these two rays. (b) Calculate the angle of incidence. (c) Calculate the angle between the incident ray and the surface of the mirror.4 marks
  10. A lamp L and a camera C are in front of a long plane mirror on a shop wall. L is 60 cm from the mirror and C is 180 cm from the mirror. Measured parallel to the mirror, L and C are 180 cm apart. A ray of light from L is reflected by the mirror at point M and enters the camera. (a) State the distance of the image of L from the mirror. (b) N is the point on the mirror closest to L. Calculate the distance NM. You may use a scale drawing. (c) Calculate the distance from the camera to the image of L.6 marks
  11. State what is meant by the normal at the point where a ray of light meets a glass surface.1 mark
  12. A ray of light hits the surface of a glass block. The angle between the ray and the surface is 35°. Calculate the angle of incidence.2 marks
  13. Light travels from air into glass, where its speed is less. (a) State how the ray changes direction as it enters the glass. (b) State how it changes direction as it leaves the glass into air.2 marks
  14. A ray of light in air hits a glass block at an angle of incidence of 70°. A student says it is totally internally reflected. Explain why the student is wrong.2 marks
  15. The critical angle for a piece of glass is 42°. A ray in the glass meets a glass–air boundary. State what happens to the ray when the angle of incidence is (a) 30°, (b) 42°, (c) 55°.3 marks
  16. Light travels from air into glass. Its speed falls from 3.0 × 10⁸ m/s to 2.0 × 10⁸ m/s. (a) Calculate the refractive index of the glass. (b) The wavelength of the light in air is 600 nm. Calculate its wavelength in the glass.3 marks
  17. A ray of light in air makes an angle of 50° with the surface of a glass block. The angle between the refracted ray and the normal is 28°. Calculate the refractive index of the glass.3 marks
  18. A ray of light in air strikes a plastic block at an angle of incidence of 60°. The refractive index of the plastic is 1.50. Calculate the angle of refraction.3 marks
  19. A diver at the bottom of a pool shines a torch up at the water–air surface. She tilts the torch until the refracted ray in the air just travels along the surface. The ray in the water then makes an angle of 41° with the surface. The speed of light in air is 3.0 × 10⁸ m/s. (a) Determine the critical angle for the water. (b) Calculate the refractive index of the water. (c) Calculate the speed of light in the water. (d) She tilts the torch until the ray in the water makes an angle of 30° with the surface. State and explain what happens to the ray at the surface.7 marks
  20. A straight glass optical fibre is in air. Its flat end face is at right angles to its sides. A ray of light in air meets the end face at an angle of incidence of 40° and refracts into the glass. The refracted ray then meets the side of the fibre at a point P. The refractive index of the glass is 1.50. (a) Calculate the angle of refraction at the end face. (b) Determine the angle of incidence at P. (c) Calculate the critical angle for the glass. (d) State and explain what happens to the ray at P. (e) The fibre is used for high-speed broadband. State two reasons why a glass fibre is suitable.9 marks
  21. State what is meant by the focal length of a thin lens.1 mark
  22. Describe what a thin converging lens does to a parallel beam of light.2 marks
  23. The two principal foci of a thin converging lens are 24 cm apart. State the focal length of the lens.2 marks
  24. Light from a distant tree passes through a thin converging lens. A screen shows a clear image of the tree. The image is smaller than the tree and upside down. Describe the image using the terms enlarged / same size / diminished, upright / inverted and real / virtual.3 marks
  25. A student draws a ray diagram for a thin converging lens and an object. Describe the two rays she draws from the tip of the object and how she finds the tip of the image.3 marks
  26. State two differences between a real image and a virtual image.2 marks
  27. A thin converging lens has a focal length of 10 cm. A lamp is 40 cm from the lens and a sharp image of the lamp forms on a screen. The lamp is moved to 25 cm from the lens and the screen is moved until the image is sharp again. State and explain what happens to the size of the image, and state whether the image is real or virtual.3 marks
  28. A thin converging lens forms a virtual image of an object. Describe where the object is and explain how the virtual image is formed.3 marks
  29. A stamp collector holds a thin converging lens above a stamp and holds a card above the lens. With the lens 30 cm from the stamp, a sharp image of the stamp forms on the card. The image is the same size as the stamp. (a) Determine the focal length of the lens. (b) She moves the lens to 20 cm from the stamp and moves the card until the image is sharp again. State three characteristics of this image. (c) She moves the lens to 10 cm from the stamp to use it as a magnifying glass. Explain why no image can now be formed on the card, and state which side of the lens the image is on.7 marks
  30. Asha can read a distant road sign clearly, but the words in a book held close to her eyes are blurred. Her brother Ravi sees the book clearly, but the road sign is blurred. (a) Name the eye condition each of them has. (b) Asha looks at the book without glasses. State where the rays from a point on the page would meet, in relation to her retina. (c) State the type of lens in each person's glasses. (d) Asha says her lens works by magnifying the page. Describe what her lens really does to the rays. (e) By mistake Asha puts on Ravi's glasses to read the book. Explain whether the page looks sharper or more blurred.7 marks
  31. A glass prism separates a beam of white light into colours. Name this effect.1 mark
  32. List the seven colours of the visible spectrum in order of increasing frequency.2 marks
  33. White light passes through a glass prism. State which colour is refracted most and which colour is refracted least.2 marks
  34. A student writes the colours of the spectrum in order of decreasing wavelength as: violet, indigo, blue, green, yellow, orange, red. Explain what is wrong and give the correct order.2 marks
  35. A student shines a ray of red light into a glass prism and says that it will split into the colours of the spectrum. Is she correct? Give a reason.2 marks
  36. A student draws white light entering a glass prism. She draws the ray inside the prism staying white and only splitting into colours as it leaves. State where the colours start to separate, and give the reason they separate.2 marks
  37. White light is dispersed by a glass prism. Ray X is blue light and ray Y is red light. State which ray is refracted more, which has the higher frequency and which has the longer wavelength.3 marks
  38. A student says that monochromatic light is light of one colour. Explain why this answer is not enough.2 marks
  39. Four narrow beams of light are shone in turn at a triangular glass prism. Beam P leaves as a spread of seven colours. Beam Q leaves as two rays, one red and one violet. Beam R leaves as two rays, one blue and one violet. Beam S leaves as a single red ray. (a) State which beam is white light. (b) State which beam could be monochromatic. Give a reason. (c) A student says beam R is monochromatic because its two rays are almost the same colour. Explain why she is wrong.4 marks
  40. Lamp L gives white light. Lamp M gives light of only two frequencies, one orange and one indigo. A narrow beam from lamp M enters a glass prism. Two rays, X and Y, travel through the glass and on to a screen. Ray Y stays closer to the beam's original direction than ray X. (a) State the colour of ray X. Explain your answer. (b) State whether the beam from lamp M is monochromatic before it enters the prism, and whether ray X is monochromatic. Give a reason for each. (c) Lamp L replaces lamp M. Nothing else is moved. Name the colours that reach the screen between the places where X and Y landed, and the colour that lands beyond X.6 marks

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