Chapter 14.1 - Working of Diode

In the previous section we saw semiconductors. In this section we will see some basic electronic components.

We have seen p-type and n-type semiconductors. Consider the silicon crystal shown in fig.14.7 below:

Fig.14.7

We will write an analysis in steps:
1. Left side of the crystal is doped with phosphorus 
♦ So the left side becomes a N-type semiconductor
2. Right side of the crystal is doped with boron
♦ So the right side becomes a P-type semiconductor
■ There is a clear boundary between the two regions
3. Let us connect this new device in a circuit. It is shown in fig.14.8 below:

Fig.14.8

In the fig.14.8:
(i) n-region is connected to the positive terminal of the cell
♦ So the free electrons in the n-region are attracted towards the positive terminal
(ii) p-region is connected to the negative terminal of the cell
♦ The holes in the p-region are attracted towards the negative terminal
(iii) So the result:
♦ The electrons in the left are moving towards the left
♦ The holes in the right are moving towards the right
(iv) So the electrons and holes are moving away from each other. They are not moving towards each other. 
■ Thus no electron will cross the boundary between the two regions. So current will not flow. The bulb will not glow.
4. Let us flip the terminals of the cell. It is shown in fig.14.9 below:

Fig.14.9

In the fig.14.9:
(i) n-region is connected to the negative terminal of the cell
♦ So the free electrons in the n-region are repelled towards the boundary
(ii) p-region is connected to the positive terminal of the cell
♦ The holes in the p-region are repelled towards the boundary
(iii) So the result:
♦ The electrons in the left are moving towards the right
♦ The holes in the right are moving towards the left
(iv) So the electrons and holes are moving towards each other. They are not moving away from each other. 
■ Thus electrons will cross the boundary between the two regions. So current will complete the circuit. The bulb will glow.

---

The semiconductor device that we just saw above is called a diode. We can write:
■ Diode is an electronic component obtained by suitable doping of a semiconductor in such a way that one region is p-type and other region is n-type
• It's symbol is shown in fig.14.10(a) below.
♦ We can see an arrow pointing towards the left. 
♦ The base of this arrow is the p-region of the diode. So it is + ve
♦ The tip of this arrow is the n-region of the diode. So it is - ve
• Using this symbol:
♦ We can simplify fig.14.8 that we saw above into the smaller fig.14.10(b) below. 
♦ We can simplify fig.14.9 that we saw above into the smaller fig.14.10(c) below:

Fig.14.10

• When the switch in fig.14.10(b) above is turned on, the bulb will not glow 
• When the switch in fig.14.10(c) above is turned on, the bulb will glow

---

■ Current flows through a diode when the connection is made in this way:
• The p-region of that diode is connected to the positive terminal of the cell
• The n-region of that diode is connected to the negative terminal of the cell
• Connecting a diode in this manner is forward biasing 
■ Current does not flow through a diode when the connection is made in this way:
• The p-region of that diode is connected to the negative terminal of the cell
• The n-region of that diode is connected to the positive terminal of the cell
• Connecting a diode in this manner is reverse biasing.

---

In the next section, we will see some applications of diode.

PREVIOUS        CONTENTS          NEXT

Copyright©2018 High school Physics lessons. blogspot.in - All Rights Reserved

Chapter 14 - Basic Electronics - Semiconductors

In the previous section we completed a discussion on colours of light. In this chapter we will see some basics about electronics.

Semiconductors

• Substances can be classified as conductors and insulators. 
♦ Conductors readily allow the passage of electric current through them
♦ Insulators do not allow the passage of electric current through them
• There are certain substances which fall in between conductors and insulators
• Silicon is such a substance. It is a common substance found in sand and quartz. 
Let us learn about it in a bit more detail. We will write the steps:
1. Silicon is an element in the periodic table
• It is below carbon and above germanium
2. Carbon, silicon and germanium has a common property:
• They have 4 electrons in the outermost shell
• So each of their atoms need four more electrons to attain octet. 
3. Consider any one silicon atom. It already has 4 atoms in the outermost shell. It is shown in fig.14.1(a) below:

Fig.14.1

• The 'extra 4 electrons required' can be obtained by 'sharing' electrons with other silicon atoms
4. When this 'sharing' takes place, strong bonds are formed between individual atoms. 
• These bonds are called covalent bonds. We have seen it in our chemistry classes. Details here.  
5. So the single silicon atom in fig.14.1(a) will obtain four new silicon neighbours. This is shown in fig.14.1(b) above.
• The structure in fig.b can be shown in a simplified form as in fig.c
6. But each of the four new neighbours are in need of 3 electrons to attain octet. 
• So they enter into bonds with other silicon atoms. 
• This pattern continues, and we get a large number of silicon atoms bonded together to form a 'crystal lattice'. This is shown in fig.14.2 below:

Fig.14.2

• A crystal lattice is a three dimensional symmetric arrangement of atoms.

---

• The same process takes place in carbon also. In chemistry classes, we have seen that the crystalline form of carbon is diamond. Details here.
• Here, we are discussing about silicon. It is a semiconductor. Why do we call it a semi conductor? Let us analyse. We will write the steps:
1. Metals are good conductors of electric current. This is because, metals have free electrons
2. A silicon crystal is shown below:

Source: Wikimedia commons

• It looks metallic. But it is not. 
• This is because, all the 'four outer electrons' of the atoms take part in bonding. There are no free electrons
3. We can change this behaviour of silicon by adding some impurities to it. The commonly used impurities are phosphorus, arsenic, boron, gallium etc., 
4. Let us see what happens when phosphorus is added:
(i) The P atoms takes up position among the Si atoms. This is shown in fig.14.3 below:

Fig.14.3

(ii) But any P atom has 5 outer electrons. We need only 4. So one electron becomes free. 
• The free electrons thus produced are indicated by the independent red dots in the fig.14.3 above
(iii) The newly obtained free electrons can move around. So the silicon crystal can now conduct electricity. 
(iv) But we have to distinguish between this 'modified silicon' from the 'original silicon'.
• For that, the modified silicon is given a special name: n-type silicon
or, in general: n-type semiconductor
• 'n' stands for 'negative'. The conductivity is achieved by electrons, which are 'negatively charged particles'. Hence the name.
5. Let us connect this semiconductor in a circuit and see what happens. It is shown in fig.14.4 below:

Fig.14.4

(i) Actually, there is nothing special. It is just like connecting an ordinary metallic conductor. The bulb will glow.
• We can see that free electrons are available just as in the case of a metallic conductor
(ii) But the conductivity will be less than a metallic conductor

---

• If instead of phosphorus, we use arsenic as the impurity, we will get the same result. 
• This is because, like phosphorus, arsenic also has 5 outer electrons
■ So we can write:
N-type semiconductor is made by using phosphorus or arsenic as the impurity

---

6. Let us see what happens when boron is used as the impurity:
(i) The B atoms takes up position among the Si atoms. This is shown in fig.14.5 below:

Fig.14.5

(ii) But any B atom has only 3 outer electrons. We need 4. So we are one electron short. 
• The 'shortage of electrons' thus produced are indicated by the 'holes' in the fig.14.5 above
7. Let us connect this semiconductor in a circuit and see what happens. It is shown in fig.14.6 below:

Fig.14.6

(i) When the switch is turned on, the electrons, although bonded between atoms, will experience a pull towards the positive terminal of the cell
(ii) So, in the above fig.14.6, electrons will break bonds and move towards the left. Because they want to reach the positive terminal of the cell
(iii) When they move to the left, they fall into the holes. The holes are filled up. 
•  But the position previously occupied by the electrons now become holes  
(iv) This process continues and we find that:
• Electrons move towards the left
• Holes move towards the right
■ This is shown in the animation below:

In general:
• Electrons move towards the positive terminal
• Holes move towards the negative terminal
■ We can say that, the electrons and holes are moving towards each other. Because: 

    ♦ Electrons on the right side are moving towards the left and
    ♦ Holes on the left side are moving towards the right 
• The crystal will conduct electricity only if the electrons and holes move towards each other in this way
• If they move away from each other, no current will pass
(v) Thus, even though there are no free electrons, we get a flow of electrons. The bulb glows.
(vi) Here also, the conductivity will be less than a metallic conductor
8. So the silicon crystal can now conduct electricity. 
•  But we have to distinguish between this 'modified silicon' from the 'original silicon' and also from the N-type.
• For that, this modified silicon is given a special name: P-type silicon
or, in general: P-type semiconductor
• 'P' stands for 'positive'. The conductivity is achieved by 'holes'
♦ A 'hole' indicate a deficiency of electron
♦ That is., a deficiency of negative charge
♦ 'Deficiency of negative' is equivalent to 'gain in positive'. 
• Hence the name.

---

• If instead of boron, we use gallium as the impurity, we will get the same result. 
• This is because, like boron, gallium also has 3 outer electrons
■ So we can write:
P-type semiconductor is made by using boron or gallium as the impurity

---

■ The semiconductors are obtained by the addition of impurities to silicon. The process of adding impurities to silicon to alter it's conductivity is called doping.

---

The invention of semiconductors led to the development of many important components in electronics like diodes, transistors etc., In the next section, we will see some of those components.

PREVIOUS        CONTENTS          NEXT

Copyright©2018 High school Physics lessons. blogspot.in - All Rights Reserved

Chapter 13.7 - Scattering of Light

In the previous two sections we we have seen the following cases:
• Light falling on opaque objects
• Light falling on transparent objects
In this section we will see what happens when light falls on very small particles
Let us do an activity:
In fig.13.22 below, sodium thiosulphate solution (2 g per 50 mL) is taken in a glass container. It is kept on a table in a dark room.

Fig.13.22

■ Trial 1: Allow white light to fall on one side of the container. This white light will pass through the solution and emerge from the opposite side of the container. This emerging light is focused onto a white screen.
• Observation: We can see the white light on the screen
■ Trial 2: With the light still on, add two drops of hydrochloric acid to the solution in the beaker
• Observation 1: The solution appears sky blue in color
• Observation 2: The focused light on the screen shows a range of colors:
♦ First it is a mixture of yellow, orange and red
♦ Then it is a mixture of orange and red
♦ After that red alone
♦ Finally it becomes black. That is., no light
• Observation 3: In the final stage, when there is no light on the screen, the solution in the beaker is light grey in color

---

The trials are complete. We will write an analysis:
1. In trial 1, the light was able to pass through the solution without any interruptions
• This is because, the solution was clear. There were no solid particles in it
2. In trial 2, We added hydrochloric acid
• Then a reaction takes place between sodium thiosulphate and hydrochloric acid.
• As a result, some colloidal particles are formed
3. So the light cannot pass as easily as before through the solution
• The bluish components of the white light have shorter wave lengths.
• They will not be able to pass the colloidal particles. Those components are scattered
4. Those scattered components reach our eyes from the solution. They never reach the screen
• So the solution appears blue in color
5. The reddish components have  greater wave lengths. 
• They are able to bend around the colloidal particles and reach the screen. 
♦ So we see a mixture of yellow, orange and red on the screen
6. But as the reaction continues, the colloidal particles become larger and larger
• Because of the increase in size, yellow also cannot pass
♦ So we see a mixture of orange and red on the screen
7. When the size of particles increase further, orange also cannot pass
♦ So we see only red on the screen
8. Finally the sizes become so large that, even red cannot pass.
• At that time, we see no light on the screen
9. At this stage, all the components are scattered at the solution itself. 
• So all the components reach our eyes at the same time.
• Because of this mixture of all colors, the solution appears to be pale gray to us

Colour of the sky

• When we look up into the sky, we must never look at the sun. 
• If we ever want to look at the sun, we must do so only with the help of  proper scientific instruments and guidance from experts. 
• So when we look at the sky, we must be looking at portions far away from the sun.
• Thus we can say: When we look at the sky, the light that reach our eyes is not the direct light from the sun
• So which light is it?
• Ans: It is the light scattered by the particles in the atmosphere.
• In the above activity, we saw how white light behaves when they pass through small particles 

---

Now consider fig.13.23 below:

Fig.13.23

We will write an analysis in steps:
1. At noon the sun will be directly above the observer. 
• So the light rays have to travel only a short distance. 
2. The bluish components of the light which has smaller wave lengths, get scattered easily. 
• So the sky appears to be blue. 
• This is much like the blue colour (in the initial stage) of the solution in the beaker in the activity that we saw above.
• Note that, in that activity, the size of the particles gradually increases. 
♦ So the bluish colour of the solution gradually changes. 
♦ But in the atmosphere, there is no change to the size of the particles. 
♦ So the blue colour stays
3. The component colours like violet, indigo and blue, which are of shorter wavelength in sunlight, undergo maximum scattering in the atmosphere. 
• These colours spread in the atmosphere and the combined effect of these colours is seen as the blue colour of the sky during day time.
4. At sunrise and sunset, the light has to travel very long distances. 
• This is also shown in fig.13.23. 
• The bluish components of the light gets scattered away. They cannot travel such long distances. So they do not reach the observer. 
• But the reddish components which have greater wave lengths can bend around the particles and reach the observer. 
• So the sky appears to be reddish at sunrise and sunset

---

■ On the moon, the sky is dark even during day time. 
• This is because, there is no atmosphere for moon. So there are no particles to scatter the light. 
• The light from the sun directly reaches the ground surface of the moon. On reaching the surface, the light gets scattered and so we can see the surface of the moon.

---

Tyndall effect

• When rays of light pass through a colloidal fluid or suspension, the tiny particles get illuminated due to scattering. Because of this, the path of light becomes visible. This is called Tyndall effect. 
• The intensity of scattering depends on the size of particles in the colloid. 
• Some times in the early morning hours, we are able to see this effect among the trees. It is shown in the fig. below. The path of sun's rays are visible because, the mist particles get illuminated.

Tyndall effect
 Source: Wikimedia commons

Infrared photography

• Ordinary cameras use ordinary films or ordinary sensors
♦ Such films or sensors capture the visible spectrum of light and we get an image
• Infrared cameras use films or sensors which are sensitive to infrared waves
• To prevent the visible light from reaching those special films/sensors, suitable filters are used
• An image is formed on the film/sensors by the infrared waves.
•  We get more details of the 'subject which is photographed'
• This is because, infrared waves which have greater wavelength, is less scattered than visible light   
• Such detailed photographs are mainly used for scientific research purposes.
Some images can be seen here.

---

We have completed this discussion on colours of light. In the next chapter, we will see some basics about electronis.

PREVIOUS        CONTENTS          NEXT

Copyright©2018 High school Physics lessons. blogspot.in - All Rights Reserved

Chapter 13.6 - The Electromagnetic spectrum

In the previous section we saw colour of transparent objects. In this section, we will see the electromagnetic spectrum. We will write the details in steps:

1. We have seen that, the white light is composed of seven colours:
Violet, Indigo, Blue, Green, Yellow, Orange and Red.
2. All these seven coloured lights, travel in the form of waves
3. We have seen some basics about wave motion (Details here)
• The distance between any to consecutive crests is the wave length
4. Consider the red light. The distance between any two consecutive crests of it's wave is approximately 700 nm. 
♦ 'nm' stands for nanometer
♦ One nanometer is (1× 10^-9) metre
• So we can see that wave length the red light is very small
5. Similarly, wave length of violet is approximately 400 nm
Let us plot these values on a graph. It is shown in fig.13.19(a) below:

Fig.13.19

• 700 nm is marked on the x axis. At that point red colour is shown
• 400 nm is marked on the x axis. At that point violet colour is shown
6. Normally, as we move from left to right on the x axis, the values increase
• But here, the values decreases. Let us see the reason:
• We know that, wavelength is inversely proportional to frequency
• That is., when wave length decreases, frequency increases.
• So if we consider frequency, the values indeed increase when we move from left to right along the x axis in fig.13.19(a) above
7. The other five colours can be arranged in order between red and violet. 
• This is shown in fig.b above 
8. Now two questions arise:
(i) Is there any wave which has 'wave length greater than 700 nm'? 
(ii) Is there any wave which has 'wave length less than 400 nm'? 
9. The answers:
(i) There are indeed waves which have 'wave length greater than 700 nm'
• But human eyes cannot see 'waves with wave lengths greater than 700 nm'
♦ In other words they are 'invisible waves'
• 'Waves with wave lengths ranging from 1 millimetre to 700 nm' are called infrared waves
• Since those wave lengths are greater than 700 nm, we plot them on the left side of red
• This is shown in fig.13.20(a) below. 
(ii) There are indeed waves which have 'wave length less than 400 nm'
• But human eyes cannot see 'waves with wave lengths less than 400 nm'
♦ In other words, just like infra red, these are also 'invisible waves'
• 'Waves with wave lengths ranging from 400 nm to 1 nm' are called ultraviolet waves
• Since those wave lengths are less than 400 nm, we plot them on the right side of violet
• This is shown in fig.13.20(a) below:

Fig.13.20

10. We know that white light comes from the sun
• The two new waves that we saw (infrared and ultraviolet) also come from the sun 
• So we have a group consisting of three items:
(i) infrared waves
(ii) White light waves
(iii) ultraviolet waves 
11. This group is given a special name: solar spectrum
12 If we allow the 'waves from the sun' to pass through a prism, they will undergo refraction
♦ The components will thus get separated from each other
• But we are not able to see the infrared and ultraviolet waves
• We can see only the seven colours
• So these seven colours together is known as visible spectrum
13. In the fig.13.20(a), the seven colours are shown distinct from each other. This is for a better understanding only.  
• In the actual case, there is a 'gradual gradation' from one colour to the next. An example is shown in fig.13.20(b).  
• The red colour at the extreme left gradually merges into the next colour orange. 
• We can see that the red colour is fading as we move to the right.
• As we move to the right, it becomes more and more orange.
• This indicates that, the wavelength of the red light is not the same everywhere.  
• As we move to the right, the wavelength also decreases.
• Indeed experiments indicate that wavelength of red light varies from 700 to 620 nm
■ Some features of infrared waves:
• When objects are heated, the molecules in them vibrate. Those vibrating molecules emit heat in the form of infrared waves. We cannot see those waves. But we can feel it
• The heat from the sun reaches us in the form of infrared waves
• Infrared waves are used in remote controls and night vision cameras 
■ Some features of visible light:
• Causes sense of vision
• Helps to produce energy by photosynthesis
• Used in solar cells 

■ Some features of ultraviolet waves:

• Exposure to UV rays can cause skin cancer. 
• It can also affect our vision.  
• The UV rays from the sun is absorbed by the ozone layer.  So most of the UV rays do not reach the earth.  
• UV rays in suitable doses can help to produce vitamin D in the skin
14. We have seen that the visible spectrum is a part of the solar spectrum. This is clear from fig.13.20(a) above.
• Now two questions arise:
(i) Is there any wave which has 'wave length greater than 1 mm'? 
(ii) Is there any wave which has 'wave length less than 1 nm'? 
15. The answers:
(i) There are indeed waves which have 'wave length greater than 1 mm' 
• 'Waves with wave lengths ranging from 0.1 m to 1 mm' are called microwaves
• 'Waves with wave lengths greater than 0.1 m are called radiowaves
• Since those wave lengths are greater than 1 mm, we plot them on the left side of infrared
• This is shown in fig.13.21 below.
(ii) There are indeed waves which have 'wave length less than 1 nm' 
• 'Waves with wave lengths ranging from 1 nm to 10^-3 nm' are called X-rays
• 'Waves with wave lengths less than 10^-3 nm are called Gamma rays
• Since those wave lengths are less than 1 nm, we plot them on the right side of ultraviolet
• This is shown in fig.13.21 below:

Fig.13.21

■ Some features of radio waves:
• Used for very high frequency radio transmission
• Used for Ultra high frequency television transmission
■ Some features of microwaves:
• Used in radar and mobile phone
• Used in microwave oven
■ Some features of x-rays:
• Penetrate through flesh
•  Helps to detect defects of bones breakage of pipes in industries etc
•  Dissociates DNA . Hence excessive exposure causes cancer
■ Some features of gamma rays:
•  Causes damage to living cells
•  Comes out in large quantities during nuclear fission
•  Used in cancer treatment
•  Useful in sterilizing surgical instruments
■ The waves ranging from radio waves to gamma rays is called electromagnetic spectrum  

---

In the next section, we will see scattering of light.

PREVIOUS        CONTENTS          NEXT

Copyright©2018 High school Physics lessons. blogspot.in - All Rights Reserved

Chapter 13.5 - Colours of Transparent objects

In the previous section we saw colour of opaque objects. In this section, we will see colours of transparent objects.

■ Transparent objects are objects which allow light to pass through them.
• So what will be the colour of a transparent object?
To find the answer, we must first know what a colour filter is:
■ Colour filter is a special transparent object
    ♦ It has the ability to absorb certain colours
    ♦ The other colours will be allowed to pass
Some examples:
Example 1:
• Fig.13.15(a) below shows a 'blue colour filter'.  

Fig.13.15

• All the seven colours are falling on it 
• But it allows only the blue light to pass through.  
• The other 6 colours will be absorbed.
Example 2:
• Yellow filter will allow only yellow light to pass through it
• But yellow is a secondary colour 
♦ It is composed of two primary colours: red and green
• So an yellow filter will allow yellow, red and green to pass through
• All other colours will be absorbed. This is shown in fig.13.15(b) above 
Example 3:
• Fig.13.6(c) above, shows a 'red colour filter'.  
• All the seven colours are falling on it 
• But it allows only the red light to pass through.  
• The other 6 colours will be absorbed.
■ Images of some actual colour filters can be seen here.

---

■ Colour filters are transparent materials which allow the passage of certain colours alone through them.  
• Primary colour filters green, blue and red allow only their respective colours to pass through them. 
• Secondary colour filters yellow, magenta and cyan allow their component colours also through them.

---

■ The colour of a transparent object is the colour of the light which it transmits. 
• If a material allows all the colours to pass through it, that material will appear colourless.  
• Water is colourless because it allows all the lights to pass through it.

---

Now we will see some solved examples:
Solved example 13.4
What is the colour of a red flower when looked through a yellow filter?
Solution: 
1. Consider fig.13.16 below.  

Fig.13.16

• White light is falling on a red flower.  
2. Flower is an opaque object.  So it will absorb all colours except red.
• The red colour will be reflected
• This reflected red colour should reach our eye.  Then only we can see the flower.
3. But there is a filter in between.  It is a yellow filter.  
• Yellow filter is a secondary filter.  It will allow yellow, red and green to pass through.  
4. So the red light will easily reach our eye.  We will see the red flower as red itself.

Solved example 13.5
In the above problem, if a green filter is used instead of yellow filter, what will be the colour of the flower?
Solution:
1. Consider fig.13.17 below.  

Fig.13.17

• White light is falling on a red flower.  
2. Flower is an opaque object.  So it will absorb all colours except red.
• The red colour will be reflected
• This reflected red colour should reach our eye.  Then only we can see the flower.
3. But there is a filter in between.  It is a green filter.  
Green filter is a primary filter.  It will allow only green to pass through.  
4. So the red light will not reach our eye. 
• But red is the only colour coming from the flower. Now, it is blocked by the filter
• So the flower will appear as black.

Solved example 13.6
White light is passed through green and red filters and allowed to fall on a white paper what will be the colour of the paper?
Solution: 
1. Consider the fig.13.18 below.

Fig.13.18

• The white light can be represented by the three colours: red, blue and green. 
2. The white light first falls on the green filter 
• It will absorb the blue and red lights
• Only green will pass. 
3. This green will fall on the red filter
• The red filter absorbs all the light except red
• So it will absorb the green light
4. In effect, no light will pass the red filter
• So the paper will appear black

Solved example 13.7
The card shown below is kept in white light. It is viewed through a cyan glass filter. In what colors will it appear?

Solution:
(a) White flower
• White light is falling on the white flower
• This white flower reflect all the seven colors of the white light
• These reflected colors should reach our eyes. Only then we can see the flower
• But a cyan filter is kept between our eyes and the flower
• Cyan filter is a secondary filter. it will allow only cyan, blue and green to pass through
• So, out of the seven colors coming from the flower, we will get blue and green
• So the flower will appear cyan in color
(a) Green leaves
• White light is falling on the green leaves
• These green leaves reflect green color.  All the other six colors of the white light will be absorbed
• These reflected green light should reach our eyes. Only then we can see those leaves
• But a cyan filter is kept between our eyes and the flowers
• Cyan filter is a secondary filter. It will allow only cyan, blue and green to pass through
• So, the green light will reach our eyes
• So the leaves will appear green in color
(c) Blue letters
• White light is falling on the blue letters
• These blue letters reflect blue color.  All the other six colors of the white light will be absorbed
• These reflected blue light should reach our eyes. Only then we can see those letters
• But a cyan filter is kept between our eyes and the flowers
• Cyan filter is a secondary filter. It will allow only cyan, blue and green to pass through
• So, the blue light will reach our eyes

• So the letters will appear green in color
(d) Red signature
• White light is falling on the red letters
• These red letters reflect red color.  All the other six colors of the white light will be absorbed
• These reflected red light should reach our eyes. Only then we can see those letters
• But a cyan filter is kept between our eyes and the flowers
• Cyan filter is a secondary filter. It will allow only cyan, blue and green to pass through
• So, the red light will not reach our eyes
• So the letters of the signature will not be visible

---

In the next section, we will see the electromagnetic spectrum.

PREVIOUS        CONTENTS          NEXT

Copyright©2018 High school Physics lessons. blogspot.in - All Rights Reserved

Chapter 13.4 - Colour of Opaque objects

In the previous section we saw primary, secondary and complementary colours. In this section, we will see colours of opaque objects.

Let us do an activity:
1. A sheet of red paper is placed flat on top of a table in a dark room. See fig.13.12(a) below:

Fig.13.12

2. Arrange a sheet of white paper in such a way that, it is near to the red paper and perpendicular to the table.  
3. Now consider fig.13.12(b)
• A dashed line is drawn perpendicular to the red paper.  
• This perpendicular line is the 'normal to the surface of the red paper'
4. Allow a ray of white light from a torch to fall on the red paper 
• This ray should be at an angle with the normal.
5. A portion of the white light will get reflected from the red paper
• This reflected light will make the same angle with the normal.
• Note that, there will not be any reflection if the ray of light falls along the normal. 
6. Now look at the white paper. Its colour has become red.  
• This is because, red light is reflected from the red paper 
7. We know that white light is composed of seven colours. 
• But now, after falling on the red paper, six colours are not available.
• It is obvious that, all those six colours are absorbed by the red paper. In short:
■ The red paper reflects only red light. All other colours will be absorbed

---

The same activity can be repeated after replacing the red paper with a blue one. It is shown in fig.13.13 below:

Fig.13.13

We can write:
■ The blue paper reflects only blue light. All other colours will be absorbed

---

The same activity can be repeated after replacing the blue paper with a green one. It is shown in fig.13.14 below:

Fig.13.14

We can write:
■ The green paper reflects only green light. All other colours will be absorbed

---

We can write a summary:
1. White light falls on an opaque object
2. That object reflects its own colour. All other colours are absorbed
3. So we can write:
■ The 'colour of an opaque object' is the 'colour of the light that it reflects'

---

Colour pigments

1. We find many uses for colour pigments (Some images of pigments can be seen here.)
Some of their uses are:
• For making drawings and paintings
• For making art works
• For painting walls and furniture of our houses 
2. Any colour can be made using three basic pigments. They are: cyan, magenta and yellow 
• So these three pigments are called primary colour pigments 
3. Let us see how they work:
• Consider some cyan pigment.  Why does it appear cyan in colour?
Ans: Cyan pigment is an opaque object
• So it is reflecting cyan colour and absorbing all other colours
4. But which are the components of cyan?
Ans: Green and blue
5. So we can write:
The cyan pigment is reflecting only two colours: green and blue
6. That means: cyan pigment is absorbing red. 
7. So we can write in simple terms:
■ The cyan pigment:
• Absorbs one colour: red
• Reflects two colour: blue and green

---

8. Consider some magenta pigment.  Why does it appear magenta in colour?
Ans: Magenta pigment is an opaque object
• So it is reflecting magenta colour and absorbing all other colours
9. But which are the components of magenta?
Ans: Red and blue
10. So we can write:
The magenta pigment is reflecting only two colours: Red and blue
11. That means: magenta pigment is absorbing green. 
12. So we can write in simple terms:
■ The magenta pigment:
• Absorbs one colour: green
• Reflects two colour: Red and blue

---

13. Consider some yellow pigment.  Why does it appear yellow in colour?
Ans: Yellow pigment is an opaque object
• So it is reflecting yellow colour and absorbing all other colours
14. But which are the components of yellow?
Ans: Red and green
15. So we can write:
The yellow pigment is reflecting only two colours: Red and green
16. That means: yellow pigment is absorbing blue. 
17. So we can write in simple terms:
■ The yellow pigment:
• Absorbs one colour: blue
• Reflects two colour: Red and green

---

■ Now we will see what happens when two primary pigments are mixed together
1. Take equal amounts of cyan and yellow pigments
2. Mix them well.  
• Since they are taken in equal amounts, the particles of both the pigments will be distributed uniformly in the mixture.  
3. So what happens when white light falls on that mixture?
Ans: The cyan particles will absorb all the red lights
• The yellow particles will absorb all the blue lights
4. So the only remaining light is green and it will be reflected
• So the mixture will appear green
■ Next we will see what happens when all the three primary pigments are mixed together
1. Take equal amounts of cyan, magenta and yellow pigments
2. Mix them well.  
• Since they are taken in equal amounts, the particles of all the pigments will be distributed uniformly in the mixture.  
3. So what happens when white light falls on that mixture?
Ans: The cyan particles will absorb all the red lights
• The magenta particles will absorb all the green lights
• The yellow particles will absorb all the blue lights
4. So there is nothing left to reflect
• So the mixture will appear black

---

Now we will see some solved examples

Solved example 13.1
Green and red color lights are allowed to fall on a white cricket ball simultaneously. What will be the color of the ball?
Solution:
1. Green and red are two primary colors. 
2. They combine to give the secondary color yellow. 
3. So the cricket ball will appear yellow in color

Solved example 13.2
A 'plant with green leaves and red flowers' is kept in a dark room. Light with following colors are allowed to fall on it. What will be the observed colors of leaves and flowers?
1. Green 
2. Yellow 
3. Red 
4. Blue 
Solution:
1. Green light
(a) Leaves:
• All the seven colors can fall on the leaves. It will reflect the green. And absorb the other six
• If green alone falls on the leaves, then also, green will be reflected. 
• So the leaves appear green
(b) Flowers:
• All the seven colors can fall on the flowers. It will reflect the red. And absorb the other six
• If green alone falls on the flowers, that green will be absorbed. Then there is nothing left to reflect. 
• So the flowers will appear black
2. Yellow light
(a) Leaves:
• All the seven colors can fall on the leaves. It will reflect the green. And absorb the other six
• Yellow is a secondary color. If yellow alone falls on the leaves, it means that, (Red + Green) is falling on the leaves
• So the red will be absorbed and green will be reflected
• So the leaves appear green
(b) Flowers:
• All the seven colors can fall on the flowers. It will reflect the red. And absorb the other six
• Yellow is a secondary color. If yellow alone falls on the flowers, it means that, (Red + Green) is falling on the flowers
• So the green will be absorbed and red will be reflected
• So the flowers appear red
3. Red light
(a) Leaves:
• All the seven colors can fall on the leaves. It will reflect the green. And absorb the other six
• If red alone falls on the leaves, that red will be absorbed. Then there is nothing left to reflect
• So the leaves appear black
(b) Flowers:
• All the seven colors can fall on the flowers. It will reflect the red. And absorb the other six
• If red alone falls on the flowers, then also that red will be reflected
• So the flowers will red
4. Blue light
(a) Leaves:
• All the seven colors can fall on the leaves. It will reflect the green. And absorb the other six
• If blue alone falls on the leaves, that blue will be absorbed. Then there is nothing to reflect
• So the leaves appear black
(b) Flowers:
• All the seven colors can fall on the flowers. It will reflect the red. And absorb the other six
• If blue alone falls on the flowers, that blue will be absorbed. Then there is nothing to reflect. 
• So the flowers will appear black

Solved example 13.3
The greeting card shown below is kept in a dark room and red light is allowed to fall on it. In which all colors will it appear? 

Solution:
(a) White flower:
• All the seven colors can fall on the white flower. It will reflect all those colors
• In our present case, only red light is falling on it. That red will be reflected.
• So the flower will appear red in color
(b) Green leaves
• All the seven colors can fall on the leaves. It will reflect the green. And absorb the other six
• If red alone falls on the leaves, that red will be absorbed. Then there is nothing to reflect
• So the leaves appear black
(c) Blue letters
• All the seven colors can fall on the leaves. It will reflect the blue. And absorb the other six
• If red alone falls on the leaves, that red will be absorbed. Then there is nothing to reflect
• So the letters appear black
(d) Red Signature
• All the seven colors can fall on the signature. It will reflect the red. And absorb the other six
• In our present case, only red light is falling on it. That red will be reflected.
• So the signature will appear red in color

---

In the next section, we will see colour of transparent objects.

PREVIOUS        CONTENTS          NEXT

Copyright©2018 High school Physics lessons. blogspot.in - All Rights Reserved