A I O real rays virtual rays B
Figure 3.14 Construction to find the image in a plane mirror
Examiner’s tips
䉴 You must be able to describe an experiment to demonstrate the refraction of light.
䉴 You should do this experiment in a darkened room. Direct a narrow beam of light at an angle to the side of a glass block placed on a large piece of paper. Mark on the paper the paths of the beams entering and leaving the block. By joining up the lines after removing the block, you can draw the path of the light as it travelled through the block. Figure 3.15 shows the paths of the rays in this experiment.
For refraction at A, where the ray enters the glass:
i = angle of incidence = angle between normal and incident ray r = angle of refraction = angle between normal and refracted ray
The ray is refracted towards the normal so the angle of refraction is less than the angle of incidence.
For refraction at B, where the ray leaves the glass:
a = angle of incidence = angle between normal and incident ray b = angle of refraction = angle between normal and refracted ray
The ray is refracted away from the normal so the angle of
refraction is greater than the angle of incidence. When the block is parallel-sided, the ray leaving is parallel to the ray entering.
Common error
✘ Finding the angles of incidence and refraction by measuring between the ray and the surface of the new material. ■ Sample question Copy and complete Figure 3.16a to show the path of the ray
through the glass prism as it is refracted twice. Show both normals. [4 marks] r i A a B b normal normal air glass
Figure 3.15 Refraction of a ray through a glass block
Student’s answer The student’s answer is shown by dashed lines in Figure 3.16b. [3 marks]
Examiner’s comments The student has correctly drawn the refracted ray within the prism and both normals. However, the student’s second refraction, as the ray left the prism, was towards the normal. When a ray moves into a region where it travels faster, it is refracted away from the normal.
Correct answer The correct rays are shown by solid lines in Figure 3.16b.
[4 marks]. The amount of refraction is determined by the refractive index – the ratio of speed of light in air to speed of light in glass.
Refractive index, n =
Sample question Light travels at 3 × 108m/s in air and 2.25 × 108m/s in water. Calculate
a)the refractive index, n, of water,
b)the angle of refraction for a ray approaching water with an angle of incidence of 55°. [4 marks]
Student’s answer a) n = = 1.33 [2 marks]
b) r = 34° [0 marks]
Examiner’s comments a) Correct answer with working.
b) The answer is only slightly inaccurate but there is no working, which means the examiner has no way of knowing whether the student made a small mistake or was completely wrong and simply close to the correct answer by luck.
Correct answer a) n = = 1.33 [2 marks]
b) sin r = = = 0.616 r = 38.0° [2 marks] 0.819 ––––– 1.33 sin i –––– 1.33 3× 108 ––––––––– 2.25 × 108 3× 108 ––––––––– 2.25× 108
speed of light in air –––––––––––––––––– speed of light in glass
student’s ray after second refraction
correct ray after second refraction student’s correct ray
after first refraction
Figure 3.16b
Examiner’s tip
䉴 You must be able to recall and use the equation: ––––sin i = n sin r
●
Try this
The answers are given on p. 103.3 Figure 3.17 shows a light ray entering a glass block of refractive index 1.5.
The angle of incidence is changed so that angle A is now 60°.
a) Which of angles A–F is the angle of incidence? b) Which of angles A–F is the angle of refraction? c) Work out the values of angles B–F.
Internal reflection and critical angle
When a ray inside a block of glass or tank of water passes out into the air, some light is reflected internally as well as being refracted away from the normal.
i = angle of incidence = angle between normal and incident ray r = angle of internal reflection = angle between normal and
internally reflected ray R = angle of refraction = angle between normal and refracted ray The law of reflection still applies, so i = r.
The greater the angle of incidence, the more energy goes into the internally reflected ray, which becomes brighter. The greatest angle of incidence when refraction can still occur is called the critical angle; in this case (shown in Figure 3.19) the angle of
i r
R air
water
incident ray internallyreflected ray
refracted ray
Figure 3.18 Rays at a water–air boundary
C refracted ray F A B E normal incident ray D Figure 3.17
angle of incidence, i = c, critical angle angle of refraction, R = 90°
If the angle of incidence is greater than the critical angle, there is no refracted ray and all the energy is in the bright internally reflected ray. This is called total internal reflection (Figure 3.20).
Water has a refractive index of 1.33 and a critical angle of 48.5°. The following calculation confirms that there can be no refracted ray with an angle of incidence of 50°.
For light leaving water: n = = 0.75 i = 50°, sin i = 0.766
sin r = = = 1.02
1.02 is a non-existent value for the sine of an angle.
● Optical fibres Each time the light strikes the wall of the optical fibre, the angle of incidence is greater than the critical angle and so total internal reflection occurs. There is very little loss of energy. The light can be considered ‘trapped’ in the optical fibre and can travel long distances, even if the fibre is bent in order to carry information or illuminate and view inaccessible places.
0.766––––– 0.75 sin i –––n 1 –––– 1.33
Figure 3.21 Length of optical fibre
incident ray bright totally internally reflected ray no refracted ray i r i > c air water
Figure 3.20 Total internal reflection
R air water incident ray internally reflected ray refracted ray i = c r = c = 90°
Formation of a real image by a converging lens
All rays of light parallel to the principal axis are refracted by the lens to pass through the principal focus, F. The distance between F and the optical centre, C, is called the focal length, f.
If the object is placed more than one focal length behind the lens, the image will always be real and inverted. Depending on where the object is placed, the image may be magnified, the same size as the object or diminished.
Common error
✘ Inaccurate drawing work leading to very poor answers. ✓ Because rays intersect at small angles, errors can easily be
magnified. It is essential to work carefully with a sharpened
O C
F I
image B A
Figure 3.23 Ray diagram for a converging lens
Examiner’s tip
䉴 You must be able to draw ray diagrams to illustrate the formation of an image by a converging lens. Follow these steps:
1) Preliminary – draw the principal axis, the object, a vertical line for