Tasks developed by Dave Sedgman, Knox Grammar School, 2013
NSW da Vinci Decathlon
An academic gala day
for years 9 and 10
Mathematics Challenge
Session 2
Team Number
_____________
Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Mathematics of Light and Colour
Mathematics of Light by Chin Chih Yang
In this section we investigate some aspects of the mathematics of light and colour.
A note on units.
Speed of light = 300 000kilometres /second =
Use 1 year = 365 days.
3  108 metres/second
Please show all working as marks will be awarded for thinking!
You can use a calculator where it is necessary.
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Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Question 1. The Light Year
(i) What is meant by the term light year?
1 mark
(ii) How far does light travel in a year?
Give your answer to 3 significant figures in scientific notation ( standard form).
2 marks
(iii) Suppose that a child is born on Earth in the year 2000.
You are on an imaginary planet that is 94.6 trillion kilometres away from
Earth and looking through a very high-powered telescope and you witness
this child's third birthday party.
How old, in years, is that child on Earth at the time you are watching the child's
third birthday party? Explain your answer.
3
2 marks
Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Question 2. Pentagonal Colours
In how many different ways can I colour the five edges of a pentagon red, blue
3 marks
and green so that no two adjacent edges are the same colour?
Reflections and rotations are to be accepted as different solutions. ( The same problem could be more
practically understood by imagining painting a room with 5 walls so that no two adjacent walls will
be of the same colour.)
( A tree diagram may be helpful.)
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Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Question 3. Travelling at speeds approaching the speed of light.
When objects start to travel at very large speed approaching that of the speed of light some very unusual
things happen. The mass of the object increases according to the formula.
M
M0
v2
1 2
c
Where M is the mass (kg) of the object travelling at speed v metres per second.
M 0 is the original mass (kg) of the object when it is stationary.
c is the speed of light in metres per second.
(i)
Show that when the object is travelling at three quarters of the speed of light
2 marks
that the mass M ( kg) of the object can be given as
M
4M 0
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(ii) Show that this represents an increase in mass, compared to stationary, of
approximately 51% .
(iii) What would be the situation for the mass M kg when its speed actually approached
the speed of light ?
(This can be put another way. i.e. evaluate limit
v c
5
M0
v2
1 2
c
)
2 marks
1 mark
Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Question 4. Lenses, Focal Length, Aperture and Camera f stops
The focal length of a lens is the distance between the optical centre of the lens and the points where
a clear image is formed. This is shown in figure 1 below. The focal length of modern camera lenses
and telescopes is usually measured in millimetres.
Focal length is very important in photography. Camera lenses each have a specific focal length. These
range from 18 mm to over 500 mm
Adjustable Camera Apertures - f-stops
Unlike telescopes, you can adjust the aperture (or diameter) of many camera lenses as shown in
Figure 2 using a ring on the lens or an electronic button. This is one of the ways to get the proper
amount of light for the photograph..
Figure 2: Adjustable Aperture Lens.
( Changing the diameter of the opening of the lens to vary the amount of light that can pass through.)
Large diameter (aperture) means a small f-stop ( f1.4) that allows maximum amount of light to pass
through the lens into the camera.
Small diameter (aperture) means a large f-stop( f16) that allow very little light into the camera.
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Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Question 4 continued.
(i)
Photographers use a formula to find the f-stop.
f stop 
focal length of lens
diameter of lens
Find the diameter of the aperture of a lens which has a 28 mm focal length
2 marks
which has an f-stop = 2.
(ii)
Find the area of light that will pass through this aperture as described in part (i).
2 marks
Give your answer to the nearest mm2.
(iii)
(iv)
To get the same amount of light through the lens with a 56 mm focal length,
calculate the diameter of this new lens to the nearest mm.
2 marks
Area of aperture (mm 2 )
f stop
1.4
2
2.8
4
5.6
8
11
16
314
154
78.54
38.48
19.63
9.62
5.06
2.41
The f stop column is an example of a “geometric series”, where
each successive number is a fixed multiple of the previous number.
This is called the “common ratio”.
1 mark
Which irrational number do you think this common ratio is for the f stop column?
Question 5. Image formation by Lenses
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Tasks developed by Dave Sedgman, Knox Grammar School, 2013
The thin lens equation relates the object distance and the image distance to the focal length of a lens
as follows:
1
1
1


focal length object distance image distance
or
1 1 1


f d0 di
(i) What focal length must a convex lens have to produce an image distance of 6cm
when an object is placed 6cm from the lens?
2 marks
(ii) Use the lens formula to calculate the image distance when an object is
placed 10 cm from a convex lens with a focal length of 8cm.
2 marks
Question 6. Intensity of Light at a Depth
8
Tasks developed by Dave Sedgman, Knox Grammar School, 2013
The percent I(d) remaining of the original intensity of light that is available at a depth d metres
below the surface of the water in a lake is given by the equation,
I (d ) 
100
(2.72)0.95d
(i) What percentage of the light, to the nearest tenth of a percent, is remaining
two metres below the surface of the lake?
(ii) At what depth, to the nearest tenth of a metre, is the intensity of the light
one-half the intensity at the surface?
Hint: Use I(d) = 50 and then guess and check.
Question 7. Refraction of Light and Snell’s Law.
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2 marks
3 marks
Tasks developed by Dave Sedgman, Knox Grammar School, 2013
The Law of Refraction
When light travels from one medium to another, it generally bends, or refracts. The law of refraction
gives us a way of predicting the amount of bend. This law is more complicated than that for reflection,
but an understanding of refraction is necessary in discussion of lenses and their applications. The law
of refraction is also known as Snell's Law, named for Willobrord Snell, who discovered the law in
1621.
Like with reflection, refraction also involves the angles that the incident ray and the refracted ray make
with the normal to the surface at the point of refraction. Unlike reflection, refraction also depends on
the media through which the light rays are travelling. This relationship gives rise to Snell's Law via
refractive indices, numbers which are constant for given media1.
Snell's Law is given in the following diagram.
We measure the angles to the normal ( i.e. at right angles) to the surface, at the point of contact. The
constants n1 and n2 are the indices of refraction for the corresponding materials through which the
light ray is passing.
Here are a few refractive indices for specific substance through which light is passing.
Substance
Refractive Index, n
Air (1 atmosphere pressure, 0 degrees C)
1.00029
Water (20 degrees C)
1.33
Crown Glass
1.52
Flint Glass
1.66
Question 7 continued.
10
Tasks developed by Dave Sedgman, Knox Grammar School, 2013
(i) Use Snell’s Law to find the refractive index of a substance if a light ray is projected into
water n1 at an angle of 30 0 to the normal and emerges at an angle of 40 0 to the normal in the
new substance n2 .
Snell’s Law n1 sin(i)  n2 sin(r )
Calculate the refractive index of the new substance n2 to 2 decimal places.
2 marks
( Use sin 30 0 = 0.500 and sin 40 0 = 0.643 which are found using the Sine button on your
calculator. )
(ii) Using the table above and the SHIFT SIN key on your calculator, find the angle that a light ray
that is projected initially in air at 30 0 to the normal enters flint glass and leaves at an angle of  to
the normal in the glass.
Find the value of  to the nearest degree.
2 marks
Question 7 continued.
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Tasks developed by Dave Sedgman, Knox Grammar School, 2013
(iii) Using the rearranged Snell's Law, with i and r being the incident and refracted angles,
sin(r ) 
n1
sin(i)
n2
supply the missing words in following sentence:
2 marks
When light is travelling from an area of lower refractive index to an area of higher index,
the ratio
n1
< 1, so that the refracted
n2
than the incident angle i;
angle r will be _______________
i.e. the refracted ray is bent __________________ to the normal.
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Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Question 8. Computer generated colours.
Computers use 256 different values for each of the three primary colours. This includes just
about every possible shade of every colour imaginable, more than you could possibly name. Why do
computers make so many colours? The answer is that, as far as the computer is concerned, every
different combination of Red, Green, and Blue counts as a different colour, even if you couldn't tell
the difference just by looking.
Calculate the total number of colour combinations possible on a computer screen.
13
2 marks
Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Question 9. Colour Blindness
"Colour blindness" is an inability to see certain colours. Most colour blindness is a
result of a lack of one or more of the types of colour cones in the retina. About 10% of
males have some colour perception defect, but it is practically non-existent in females.
The most common colour perception defects are for red or green or both.
A group consists of 200 people with equal number of males and females and one person
is selected at random by a blindfolded person.
Write down the probability P(CBM), that the person selected at random will be a
colour blind male.
14
2 marks
Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Question 10. Colour and Hexadecimal Coding.
Red, Green and Blue
Hexadecimal numbers are used to specify colors.
Any color is defined by its mix of Red, Green and Blue, each of which can be in the range:
0 to 255 (in decimal) , or
00 to FF (in hexadecimal, base 16 )
It is based on the idea that any color can be made by mixing Red,
Green and Blue, and so it is called the "RGB Color System".
It is also called an "Additive" color system, because it starts at
black, and then color is added.
Some Common Colours
Decimal
(Red, Green, Blue)
(0, 0, 0)
(255, 255, 255)
(255, 0, 0)
(0, 255, 0)
(0, 0, 255)
(255, 255, 0)
(0, 255, 255)
(255, 0, 255)
Color
Black
White
Red
Green
Blue
Yellow
Cyan
Magenta
Hexadecimal
(#RRGGBB)
#000000
#FFFFFF
#FF0000
#00FF00
#0000FF
#FFFF00
#00FFFF
#FF00FF
Hexadecimals
The format ("notation") used is #(#RRGGBB), where RR specifies the value of the Red component
of the color (using two hexadecimal digits), GG the Green component and BB the Blue component.
i.e. 3 pairs of hexadecimal characters are used to describe a colour.
Decimal:
Hexadecimal:
0
0
1
1
2 3
2 3
4 5
4 5
Question 10 continued
15
6
6
7 8
7 8
9
9
10
A
11
B
12
C
13
D
14
E
15
F
Tasks developed by Dave Sedgman, Knox Grammar School, 2013
Conversion of hexadecimal numbers (base 16) into decimal form.
Example: LIGHT SKY BLUE is shown below.
LIGHT SKY BLUE
#87CEFA
=((8 x 16+ 7 ), (12 x 16 +14), (15 x 16 +10)),
=( 138, 206, 250 )
Conversion of decimal form into hexadecimal numbers (base 16).
Example: A shade of blue is made up of:
64/255 Red,
48/255 Green
255/255 (ie full) Blue
So this shade of blue is (64,48,255) in decimal, which is equivalent to (40,30,FF) in hexadecimal and
would be coded as #4030FF.
(i) Convert this hexadecimal code for LIGHT PINK into decimal mode
LIGHT PINK
3 marks
#FFB6C1
Answer ( ___ , ____ , ____ )
(ii) Convert this decimal code for MEDIUM SLATE BLUE into hexadecimal form.
MEDIUM SLATE BLUE
Answer #____ ____ ____
End of Challenge
16
(123, 110, 238 )
3 marks