Northern NSW da Vinci Decathlon · 2016-06-01 · 5.6 19.63 8 9.62 11 5.06 ... Use Snell’s Law to find the refractive index of a substance if a light ray is projected into water

$Page 1: Northern NSW da Vinci Decathlon · 2016-06-01 · 5.6 19.63 8 9.62 11 5.06 ... Use Snell’s Law to find the refractive index of a substance if a light ray is projected into water$
Tasks developed by Dave Sedgman, Knox Grammar School, 2013

Northern NSW da Vinci Decathlon

An academic gala day for years 9 and 10

Mathematics Challenge

Session 1

Team Number _____________


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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 = 83 10× metres/second Use 1 year = 365 days.

Please show all working as marks will be awarded for thinking! You can use a calculator where it is necessary.


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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? 2 marks Give your answer to 3 significant figures in scientific notation ( standard form).

(iii) Suppose that a child is born on Earth in the year 2000. 2 marks 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.


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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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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.

02

21

MMvc

=

−

Where M is the mass (kg) of the object travelling at speed v metres per second. 0M 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

047MM =

(ii) Show that this represents an increase in mass, compared to stationary, of 2 marks approximately 51% . (iii) What would be the situation for the mass M kg when its speed actually approached 1 mark

the speed of light ?

(This can be put another way. i.e. evaluate 02

2

limit1

v c

Mvc

→

−

)


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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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Question 4 continued.

(i) Photographers use a formula to find the f-stop.

focal length of lensf stopdiameter 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) To get the same amount of light through the lens with a 56 mm focal length, 2 marks

calculate the diameter of this new lens to the nearest mm.

(iv) f stop Area of aperture (mm 2 )

1.4 314 2 154

2.8 78.54 4 38.48

5.6 19.63 8 9.62 11 5.06 16 2.41

The f stop column is an example of a “geometric series”, where 1 mark

each successive number is a fixed multiple of the previous number. This is called the “common ratio”. Which irrational number do you think this common ratio is for the f stop column?


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Question 5. Image formation by Lenses

The thin lens equation relates the object distance and the image distance to the focal length of a lens as follows:

1 1 1focal length object distance image distance

= + or 0

1 1 1

if d d= +

(i) What focal length must a convex lens have to produce an image distance of 6cm 2 marks when an object is placed 6cm from the lens? (ii) Use the lens formula to calculate the image distance when an object is 2 marks placed 10 cm from a convex lens with a focal length of 8cm.


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Question 6. Intensity of Light at a Depth

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,

0.95100( )

(2.72) dI d×

=

(i) What percentage of the light, to the nearest tenth of a percent, is remaining 2 marks 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 3 marks one-half the intensity at the surface? Hint: Use I(d) = 50 and then guess and check.


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Question 7. Refraction of Light and Snell’s Law.

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 1n and 2n 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


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(i) Use Snell’s Law to find the refractive index of a substance if a light ray is projected into water 1n at an angle of 30 0 to the normal and emerges at an angle of 40 0 to the normal in the new substance 2n .

Snell’s Law 1 2sin( ) sin( )n i n r=

Calculate the refractive index of the new substance 2n 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 θ to the normal in the glass.

Find the value of θ to the nearest degree. 2 marks


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(iii) Using the rearranged Snell's Law, with i and r being the incident and refracted angles,

1

2sin( ) sin( )nr in=

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 1

2

nn < 1, so that the refracted angle r will be _______________

than the incident angle i; i.e. the refracted ray is bent __________________ to the normal.


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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. 2 marks


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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 2 marks colour blind male.


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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

Color Decimal (Red, Green, Blue)

Hexadecimal (#RRGGBB)

Black (0, 0, 0) #000000 White (255, 255, 255) #FFFFFF Red (255, 0, 0) #FF0000 Green (0, 255, 0) #00FF00 Blue (0, 0, 255) #0000FF Yellow (255, 255, 0) #FFFF00 Cyan (0, 255, 255) #00FFFF Magenta (255, 0, 255) #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: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Hexadecimal: 0 1 2 3 4 5 6 7 8 9 A B C D E F


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Question 10 continued

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 3 marks

LIGHT PINK #FFB6C1

Answer ( ___ , ____ , ____ ) (ii) Convert this decimal code for MEDIUM SLATE BLUE into hexadecimal form. 3 marks

MEDIUM SLATE BLUE (123, 110, 238 )

Answer #____ ____ ____

End of Challenge

Documents

Northern NSW da Vinci Decathlon · 2016-06-01 · 5.6 19.63 8 9.62 11 5.06 ... Use Snell’s Law to find the refractive index of a substance if a light ray is projected into water