### REFRACTION OF LIGHT AT PLANE SURFACES

I. CONCEPTUAL UNDERSTANDING

1. Why is it difficult to shoot a fish swimming in water?   (2 Marks)

A: * It is difficult to shoot a fish swimming in water because one can not judge the actual position of fish in water.

* The fish appears to be raised high in its position due to refraction of light.

2. The speed of the light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s.  (2 Marks) Speed of light in air

3. Refractive index of glass relative to water is 9/8 . What is the refractive index of water relative to glass?  (1 Marks)

A: * Refractive index of glass relative to water = 9/8

Refractive index of water relative to glass = 8/9

4. The absolute refractive index of water is = 4/3. What is the critical angle?  (2 Marks)

A: Absolute Refractive index of water n = 4/3

* We know n = 1/sin c  or sin c = 3/4  = 0.75

* From natural sin c tables sin 48.5o = 0.75

... Critical angle c = 48.5o

5. Determine the refractive index of Benzene if the critical angle is 42o(2 Marks)

A: * Refractive index of Benzene (n) = 1/sin c

6. Explain the formation of a mirage?   (4 Marks)

A:

* During a hot summer day, air just above the road is very hot and the air at higher altitudes is cool.

* It means that the temperature decreases with height. As a result the density of air increases with height.

* We know that the refractive index of air increases with density. Thus the refractive index of air increases with height.

* So the coller air at the top has greater refractive index than the hotter air just above the road.

* Light travels faster through the thinner hot air than through the denser cool air above it.

* When light from a tree passes through a medium just above the road, whose refractive index decreases towards the ground, it suffers, refraction and takes a curved path because of total internal reflection.

* This appears to the observer as if the ray is reflected from the ground. Hence we feel the illusion of water being present on the road.

* The inverted virtual image of the tree is seen by the observer. This is a mirage.

7. How do you verify experimentally that  sin i/sin r is a constant.  (4 Marks)

A: Aim: To obtain a relation between angle of incidence and angle of refraction.

Materials required: A plank, white chart, protractor, scale, small black painted plank, a semi circular glass disc of thickness nearly 2 cm, pencil and laser light.

Procedure:

* Take a wooden plank which is covered with white chart. Draw two perpendicular lines, passing through the middle of the paper as shown in the figure.

* Let the point of intersection be O. Mark one line as NN which is normal to the another line marked as MM.

* Here MM represents the line drawn along the interface of two media and NN represents the normal drawn to this line at 'O'.

* Take a protractor and place it along NN in such way that its centre coincides with O as shown in figure (a).

* Then mark the angles from 0o to 90o on both sides of the line NN as shown in figure (a).

* Repeat the same on the other side of the line NN. The angles should be indicated on the curved line.

* Now place a semi - circular glass disc so that its diameter coincides with the interface line (MM) and its center coincides with the point O.

* Point a laser light along NN in such a way that the light propagates from air to glass through the interface at point O and observe the path of laser light coming from other side of disc as shown in figure (b).

* There is no deviation.

* Send laser light along a line which makes 15o (angle of incidence with NN and see that it passes through point O.

* Measure its corresponding angle of refraction, by observing laser light coming from the other side (Circular side) of the glass slab.

* Note these values in the table below. Do the same for the angles of incidence such as 20o, 30o, 40o, 50o and 60o and note the corresponding angles of refraction.

* Find  sin i/sin r for every 'i' and 'r' and evaluate sin i/sin r for every incident angle.

* We get the ratio sin i/sin r as a constant. This ratio gives the value of refractive index of glass.

8. Explain the phenomenon of total internal reflection with one or two activities.

A: Total Internal Reflection:

* Place the semi circular glass disc in such a way that its diameter coincides with interface line MM and its centre coincides with point "O".

* Now send light from the curved side of the semicircular glass disc.

* This means that we are making the light travel from denser medium to rarer medium. Start with angle of incidence (i) equal to 0° i.e., along the normal and look for the refracted ray on the other side of the disc.

* We observe that it does not deviate its path when entering into rarer medium.

* Send laser light along angles of incidence 5°, 10°, 15° etc and measure the angle of refraction.

* Tabulate the results in table as shown below and note the values 'i' and 'r' in table.

* We observe that at a certain angle of incidence the refracted ray does not come out but grazes the interface separating air and glass. This angle of incidence is known as critical angle.
* Now send light at an angle greater than the critical angle. We find that the light is reflected into the denser medium at interface.
* It means that light never enters into the rarer medium. This phenomenon is called total internal reflection.

9. How do you verify experimentally that the angle of refraction is more than the angle of incidence when light ray travels from denser to rarer medium.       (4 Marks)
A:

* Take a metal disk. Use a protractor and mark angles along its edge as shown in the figure (a).

* Arrange two straws at the centre of the disc in such a way that they can be rotated freely about the centre of the disc.

* Adjust one of the straws to make an angle 10°.

* Immerse half of the disc vertically into the water filled in a transparent vessel.

* While dipping, verify that the straw at 10º is inside the water.

* From the top of the vessel try to view the straw which is inside the water as shown in figure (b).

* Then adjust the other straw which is outside the water until both straws appear to be in a single straight line.

* Then take the disc out of the water and observe the two straws on it we will find that they are not in a single straight line.

* Measure the angle between the normal and second straw.

* Note the value in the table given below.

* Do the same activity for various angles and find the corresponding angles of refraction.

* Note all these values in the Table. Use the data in the table and find refractive index of water.

(Note that the incident angle should not be greater than 48°)

Observation:

We note that the angle of refraction (r) is more than the angle of incidence (i).

Conclusion: When light travels from denser (water) to rarer (air) medium it bends away from the normal.

II. ASKING QUESTIONS & MAKING HYPOTHESIS

10. Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water. How does it appear? And why? (Make hypothesis and do the above experiment.) (2 Marks)
A:

* As the bright metal ball is made black with soot air enters in the in between space of the soot and the metal ball.

* This air acts as a rarer medium and water as a denser medium.

* The ray of light travels from water to the air around the ball.

* When the angle of incidence exceeds the critical angle then total internal reflection takes place and the ball with a soot coating appears bright.

* In addition to this due to the refraction of light the ball to be raised high in its position.

11. Take a glass vessel and pour some glycerine into it and then pour water upto the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel. (2 Marks)

a) What changes do you notice?

b) What could be the reasons for these changes?

A:

a) When the glass rod is observed from the sides of the glass vessel the part of the rod immersed in water is visible and that immersed in glycerine is not visible.
b) The refractive index of glass is the same as the refractive index of glass. As such the speed of light in both is the same. So they do not appear separately.

III. EXPERIMENTATION AND FIELD INVESTIGATION

12. Do activity 7 again. How can you find critical angle of water? Explain your steps briefly.  (4 Marks)

* Take a cylindrical transparent Vessel.
* Place a coin at the bottom of the vessel.
* Now pour water until we get the image of the coin on the water surface.
* When we try to see the coin from the side of the vessel the coin is not visible in the absence of water in it.
* The coin is visible when water is poured in the vessel.
* This is because of the total internal reflection.
The refractive index of water = 1.33

= 48.7°
* For water the critical angle is c = 48.7°.

IV. INFORMATION SKILLS & PROJECTS

13. Collect the values of refractive index of the following media.
Water, coconut oil, flint glass, crown glass, diamond, benzene and hydrogen gas.
A:

14. Collect the information on working of optical fibres. Prepare a report about various uses of optical fibres in our daily life.     (4 Marks)
A:

Optical Fibre: An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer (10-6 m). A bunch of such thin fibres form a light pipe.

Working: Because of the small radius of the fibre, light going into it makes a nearly glancing incidence on the wall. The angle of incidence is greater than the critical angle and hence total internal reflection takes place. The light is thus transmitted along the fibre.

Uses:

* Optical fibres are used by doctors in 'Endoscope' to see all organs inside the human body. For example throat, intestines... etc.

* Optical fibres are used to transmit communication signals through light pipes.

* Optical fibres are used in International telephone cables laid down under the sea.

15. Take a thin thermocol sheet, cut it in circular discs of different radii like 2 cm, 3 cm, 4 cm, 4.5 cm, 5 cm etc., and mark centres with a sketch pen. Now take needles of length nearly 6 cm. Pin a needle to each disc at its centre vertically. Take water in a large opaque tray and place the disc with 2 cm. radius in such a way that the needle is inside the water as shown in the figure. Now try to view the free end (head) of the needle from the surface of the water.

15a. Are you able to see the head of the needle? Now do the same with other discs of different radii. Try to see the head of the needle, each time.

Note: The position of your eye and the position of the disc on water surface should not be changed while repeating the activity with other discs.

15b. At what maximum radius of the disc, where you not able to see the free end of the needle.

15c. Why were you not able to view the head of the nail for certain radii of these discs.

15d. Does this activity help you to find the critical angle of the medium (water)? Draw a diagram to show the passage of light ray from the head of the nail in different situations. (4 Marks)

A: * Yes, I am able to see the head of the needle.

* At a radius 6 cm of the disc, I am not able to see the free end of the needle.

* This is because the light rays coming from the head of the nail for certain radii of the discs are undergoing total internal reflection when these rays are incident on the surface of water.

* We can find the critical angle of the medium water as follows

a) Refractive index of air (n2) = 1.003

Refractive index of water (n1) = 1.33

d) From natural sines
sin 48.7° = 0.7521
= Critical angle of water = 48.7°

V. COMMUNICATION THROUGH DRAWING AND MODEL MAKING

16. Explain the refraction of light through a glass slab with a neat diagram.         (4 Marks)
A: Aim: To study the refraction of light through a glass slab.
Materials required: Plank, Chart Paper, Clamps, Scale, Pencil, Thin glass slab and Pins.
Procedure:
* Place a Piece of chart (paper) on a Plank. Clamp it.
* Place a glass slab in the middle of the paper and draw border line along the edges of the slab by using a pencil. Remove the slab and name the vertices of the rectangle as A, B, C and D.
* Draw a perpendicular at a point X on the longer side (AB) of the rectangle.
* Now draw a line PX such that it makes an angle 30° with the normal at X.
* Now put the glass slab on the paper in such a way that it fits in the rectangle ABCD.
* Now fix two identical pins on the line PX at P1 and P2
* By looking at the two pins P1 and P2 from the other side (CD) of the slab, fix two pins P3 and P4 in such a way that all the four pins appear to be along a straight line.

* Remove the slab and the pins.
* Draw a straight line joining Q, P3 and P4. This line represents the emergent ray. This meets the line CD at Y.
* PX is the incident ray.
* Join XY which represents the refracted ray in the glass slab.

17. Place an object on the table. Look at the object through the transparent glass slab. You will observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray in this situation.               (2 Marks)
A:

* ABCD is the trace of the glass slab.
* O is the object placed at the point 'O' on the side CD of the glass slab.
* OPQ is the normal ray emerged out of the slab undeviated.
* OR is the incident ray in the glass slab refracted along RS. This ray RS appears to have been coming from the point I which is the image of the object O.
* Thus the image appears closer to the viewer.

VI. APPRECIATION AND AESTHETIC SENSE, VALUES

18. What is the reason behind the shining of diamonds and how do you appreciate it.  (2 Marks)
A: * The reason behind the shining of diamonds is total internal reflection of light in it.
* The refractive index of diamond is 2.42. So its critical angle is 24.4°.
* As such a light ray enters the diamond will have always the angle of incidence more than 24.4°.
* So light suffers total internal reflection when it enters the diamond.
* The diamond cutters skill is highly appreciable, because he could cut it in such a way that light enters suffers total internal reflection.

19. How do you appreciate the role of Fermat Principle in drawing ray diagrams.
A: * According to Fermat's Principle that light selects the shortest path to travel.
* This is the reason for the propogation of light in straight lines.
* We use this principle, when we draw ray diagrams to find the image of an object formed by mirrors to illustrate the laws of reflection, refraction... etc.

VII: APPLICATION TO DAILY LIFE, CONCERN TO BIODIVERSITY
20. A light ray is incident on air - liquid interface at 45° and refracted at 30°. What is the refractive index of the liquid? For what angle of incidence will be the angle between reflected and refracted ray be 90°?    (2 Marks)
A: * Angle of incidence i = 45°
Angle of refraction r = 30°

* Given the angle between the reflected and refracted ray = 90°
* It means angle of reflection (r) + angle of incidence (i) = 90°
∴ Angle of refraction (r) = (90° - i)

* But refractive index n =

* From Natural tangent tables tan 54.7° = 1.414 = tan i
* The angle of incidence i = 54.7°

21. Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface for a certain viewing position?
A: * When rays of light from water enter the rarer medium they get refracted.
* Some of these refracted rays may enter the eye of the viewer and the viewer can observe the brightness of water depending upon the quantity of refracted rays received by his eye.
* When a test tube is immersed in water, rays of light coming from water may also fall on the walls of the test tube.
* The rays that fall on the walls of the test tube at angles greater than the critical angle of water suffers total internal reflection.
* Some of such rays when they enter the eye of the viewer, he receives more light than the light without the immersion of the test tube.
* Now the extra light perceived by the viewer makes him feel the test tube much brighter than the surrounding water.
* This is the reason as why the viewer feels that the test tube immersed in water appears to have a mirror surface for certain viewing position.

22. What is the angle of deviation produced by a glass slab? Explain with ray diagram. (2 Marks)
A: * Deviation is the angle between the incident ray and the emergent ray.
* In the case of a glass slab the incident ray and the emergent ray are parallel to each other as shown in the figure.
* So the angle of deviation produced by a glass slab is zero.
* In the figure

P1 P2 X is the incident ray and Y P3 P4 is the emergent ray which are parallel to each other.

23. In what cases does a light ray not deviate at the interface of the two media.      (1 Mark)
A: The light ray which incidents along the normal to the interface does not deviate at the interface of the two media.

24. A ray of light travels from an optically denser to rarer medium. The critical angle for the two media is 'c'. What is the maximum possible deviation of the ray?   (2 Marks)
A: * 'c' is the critical angle of the two media.
* When the incident ray incidents at critical angle in the denser medium, the refracted ray grazes at the interface of the two media.
* So the maximum possible deviation of the ray must be less than 'c'.
* As such the maximum possible deviation of the ray is Π-2c.

25. When we sit at a camp fire, objects beyond the fire are seen swaying. Give reasons for it.   (2 Marks)
A: * When we sit at a camp fire, heat is transmitted to the surroundings.
* As a result of this transmitted heat the apparent density of air changes and its refractive index changes accordingly.
* Due to this changing density and refractive index of the surrounding air, the angles of reflection and deviation change.
* This makes the objects behind the fire appear as if they are swaying.

26. Why do stars appear twinkling?       (2 Marks)
A: * We view the stars from earth through thick layers of turbulent air of the earth's atmosphere.
* The light from the stars reach the viewer after passing through many layers of atmosphere.
* As the light from the stars pass through these layers of air it is refracted many times in random directions.
* These random refractions makes the star to appear twinkling to the viewer.
* A viewer can not experience this type of vision of twinkling if he could see the stars beyond the atmosphere of the earth.

27. Why does a diamond shine more than a glass piece cut to the same shape?      (2 Marks)
A: * The refractive index of diamond is 2.42 and its critical angle is 24.4° which is very less.
* The faces of the diamond are cut in such a way the light that enters the diamond undergoes total internal reflection and emerge out of the diamond.
* As a result of these total internal reflections a diamond shines more.
* The glass piece cut to the same shape of the diamond is not suitable for total internal reflections of the light rays that enter in to it.
* So diamond shines more than a glass piece cut to the same shape, because the eye receives more light from the diamond than it does from the glass.

QUESTIONS AND ANSWERS GIVEN IN THE LESSON
1. What difference do you notice in fig (a) and fig (b) with respect to refracted rays. (AS - 5)

A: * As light ray enters from a rarer medium to a denser medium as shown in fig (a), the refracted ray bends towards the normal.
* As light ray emerges from a denser medium to a rarer medium as shown in fig (b), the refracted ray bends away from the normal.

2. Is there any relation between behaviour of refracted rays and speeds of light.       (AS - 1)
A: The speed of light changes with the change of medium. So the ray of light travelling from one medium to another medium bends away or bends towards the normal depending upon the nature of the medium.

3. Why do different material media possess different values of refractive indices?     (AS - 1)
A: * The speed of light is different in different media.
* Further refractive index depends upon the nature of the material.
* So different media have different refractive index.

4. On what factors does the refractive index of a medium depend?        (AS - 1)
A: Refractive index depends upon
a) Nature of the material
b) Wave length of light used

5. Can we derive the relation between the angle of incidence and the angle of refraction theoretically?        (AS - 1)
A: * We can derive the relation between the angle of incidence i and angle of refraction r theoretically.
* The relation is n1 sin i = n2 sin r.

6. Is there any chance that angle of refraction is equal to 90°? When does this happen? (AS - 2)
A: Yes. The angle of refraction is equal to 90° when the angle of incidence is equal to critical angle.

7. At what angle of incidence do you notice that the refracted ray grazes the interface separating the two media. (air and glass) (AS - 2)
A: At the critical angle of incidence, the refracted ray grazes the interface separating the two media.

8. What happens to light when the angle of incidence is greater than critical angle. (AS - 2)
A: When the angle of incidence is greater than critical angle the light ray gets reflected in to the denser medium at the interface i.e. light never enters the rarer medium. This phenomenon is called total internal reflection.

9. How does light behave when a glass slab is introduced in its path.      (AS - 2)
A: The light ray under goes refraction Twice.

ACTIVITIES

Activity - 1
1. What is your observation when you see a pencil kept in a tumbler of water? (AS - 3) (1 Mark)
A: * Take some water in a glass tumbler.
* Keep a pencil in it.
* When we look at the pencil from one side of the glass we find that the pencil in air and water appear separately is a disjointed manner due to refraction.
* When we look the pencil from the top of the tumbler, the pencil appears thicker than its thickness.

Activity - 2
2. Explain the formation of image of an object on a long wall which is facing the Sun.
A: * Go to a long wall (of length about 30 feet) facing the Sun.
* Go to one end of a wall and ask your friend to bring a bright metal object near the other end of the wall.
* When the object is few inches from the wall, it appears distorted and you will see a reflected image in the wall as though the wall was mirror.
* This is due to the refraction of light.

Activity - 3
3. Show that the speed of light changes when light propogates from one medium to another medium.     (AS - 3) (4 Marks)

* Take a shallow vessel with opaque walls such as a mug.
* Place a coin at the bottom of the vessel.
* Move away from the vessel until you can not see the coin. See figure (a)
* Stand there and ask your friend to fill the vessel with water.
* When the vessel is filled with water the coin comes back into view. See figure (b)
Explanation:
* Draw a ray diagram from the coin to the eye.
* By observing the path of the ray, it is clear that light ray changes its direction at the interface separating the two media. i.e. water and air.
* This path is chosen by the light ray so as to minimize time of travel between coin and eye.
* This is possible only if the speed of the light changes at interface of the two media.
* Thus we can conclude that the speed of light changes when light propogates from one medium to another medium.

Lab Activity - 1
4. Derive the relation between the angle of incidence and angle of refraction through an activity. What is your conclusion.   (4 Marks)
A: Aim: To obtain a relation between angle of incidence and angle of refraction.
Materials required: A plank, white chart, protractor, scale, small black painted plank, a semi circular glass disc of thickness nearly 2 cm, pencil and laser light.

Procedure
* Take a wooden plank which is covered with white chart.
* Draw two perpendicular lines, passing through the middle of the paper as shown in the figure.
* Let the point of intersection be O. Mark one line as NN which is normal to the another line marked as MM.
* Here MM represents the line drawn along the interface of two media and NN represents the normal drawn to this line at 'O'.
* Take a protractor and place it along NN in such a way that its centre coincides with O as shown in figure (a).
* Then mark the angles from 0° to 90° on both sides of the line NN as shown in figure (a).
* Repeat the same on the other side of the line NN. The angles should be indicated on the curved line.
* Now place a semi-circular glass disc so that its diameter coincides with the interface line (MM) and its center coincides with the point O.
* Point a laser light along NN in such a way that the light propagates from air to glass through the interface at point O and observe the path of laser light coming from other side of disc as shown in figure (b).
* Send Laser light along a line which makes 15° (angle of incidence) with NN and see that it passes through point O. Measure its corresponding angle of refraction, by observing laser light coming from he other side (Circular side) of the glass slab.
* Note these values in the table below.
* Do the same for the angles of incidence such as 20°, 30°, 40°, 50° and 60° and note the corresponding angles of refraction.

* Find  in each case. We get this ratio as constant.
* This ratio gives the refractive index of glass. In this experiment we notice that 'r' is less than 'i' in all the cases and the refracted ray bends towards the normal in each case.
* From this experiment, we conclude that when light travels from a rarer medium (air) to a densar medium (glass), the value of 'r' is less than the value of 'i'. and the refracted ray bends towards the normal.

Activity - 4
5. Prove that when light ray travels from denser to rarer medium, it bends away from the normal and the angle of refraction is greater than the angle of incidence.    (4 Marks)

* Take a metal disc. Use a protractor and mark angles along its edge as shown in the figure (a).
* Arrange two straws at the centre of the disc in such a way that they can be rotated freely about the centre of the disc.
* Adjust one of the straws to make an angle 10°.
* Immerse half of the disc vertically into the water, filled in a transparent vessel.
* While dipping, verify that the straw at 10º is inside the water.
* From the top of the vessel try to view the straw which is inside the water as shown in figure (b).
* Then adjust the other straw which is outside the water until both straws appear to be in a single straight line.
* Then take the disc out of the water and observe the two straws on it. We will find that they are not in a single straight line.
* Measure the angle between the normal and second straw.
* Note the value in the table given below.
* Do the same activity for various angles and find the corresponding angles of refraction.
* Note all these values in the Table. Use the data in the table and find refractive index of water.
(Note that the incident angle should not be greater than 48°)
Observation:
We note that the angle of refraction (r) is more than the angle of incidence (i).
Conclusion: When light travels from denser (water) to rarer (air) medium it bends away from the normal.

 S.no. Angle of incidence(i) Angle of refraction (r)

6. Derive Snell's Law with an example.           (4 Marks)

* Let a person has fallen out of a boat and is seeking for help in the water at the point b as shown in fig (a).
* Let OX be the shore. A person at A saw the accident and to save the person at B person at A has to travel some distance on land and some distance in water.
* To save the person at B, person at A should reach the point B in the shortest possible time.
* It is advantageous, if the person at A travels more distance on land in order to decrease the distance in water as he goes slow in water.
* Whatever be the paths, the find path person at A has to reach the person at B is ACB.
* This path takes the shortest time of all possible paths. Any other route becomes longer.

* If we plot a graph for the time taken to reach the person against the position of any point when the person crosses the shore line we get a curve shown in fig (b).
* 'C' is the point on the shore line, corresponds to the shortest of all possible times.
* Consider a point 'D' on the shore line which is very close to point 'C' such that there is essentially no change in time between path ACB and ADB.
* Let us find how long it would take from A to B by the two paths ADB and ACB (Fig c) assuming no change in time between these two paths.
* Draw perpendiculars ED at D and CF at C.
* We can understand that the path is shorted by an amount EC. But the person has to go an extra distance DF in water.
* It means we gain a time that is equal to go through the distance EC on land, but we lose the time that is equal to go extra distance DF in water.
* These times must be equal as we assumed there is no change in time between the two paths.
Derivation: (Snell's Law) n1 sin i = n2 sin r
* Let time of travel from E to C and D to F be Δt and v1 and v2 be the speeds of running and swimming.
From fig(c): We get
EC = v1 Δt and DF = v2 Δt

Let i and r be the angles measured between the path ACB and normal NN drawn perpendicular to OX.

* To save the person at B, one should take such a path to satisfy the eq. (3). The same thing can be applied to light ray also.

* n1 sin i = n2 sin r
This is called Snell's Law.

Activity - 5
7. Explain total internal reflection by an activity.
A: Total Internal Reflection:
* Place the semi circular glass disc in such a way that its diameter coincides with interface line MM and its centre coincides with point "O".
* Now send light from the curved side of the semicircular glass disc.
* This means that we are making the light travel from denser medium to rarer medium. Start with angle of incidence (i) equal to 0° i.e., along the normal and look for the refracted ray on the other side of the disc.
* We observe that it does not deviate its path when entering into rarer medium.
* Send laser light along angles of incidence 5°, 10°, 15° etc and measure the angle of refraction.
* Tabulate the results in table as shown below and note the values 'i' and 'r' in table.

* We observe that at a certain angle of incidence the refracted ray does not come out but grazes the interface separating air and glass. This angle of incidence is known as critical angle.
* Now send light at an angle greater than the critical angle. We find that the light is reflected into the denser medium at interface.
* It means that light never enters into the rarer medium. This phenomenon is called total internal reflection.

8. Derive an expression for the critical angle.         (AS - 3) (4 Marks)
A: From Snell's Law
n1 sin i = n2 sin r

* Let us consider the light ray from the denser medium (refractive index: n1) is travelling to the rarer medium (refractive index: n2)
* We know that the angle of refraction is more than the angle of incidence when light ray travels from dencer to rarer medium

* It means that the angle of refraction (r) is greater than the angle of incidence (i).
* The angle of incidence at which the light ray travelling from denser to rarer medium, grazes the interface is called critical angle for denser medium.
* Let 'c' be the critical angle. Then 'r' becomes 90°.

* We get sin c =  . We know that  i.e. n12 is called refractive index of denser medium with respect to the rarer medium.
∴  sin c =

Activity - 6
8. Place a coin on a table and place a transparent glass tumbler on the coin. When the glass tumbler is filled with water explain why the coin disappears from view.   (AS - 3) (2 Marks)
A: * When a coin is placed on a table and a transparent glass tumbler is placed on the coin.
* If we see the coin from the side of the glass it is visible to some extent.
* When the glass is filled with water and if the coin is viewed from the side of the glass it is not visible.
* The cause for the disappearance of the coin is total internal reflection.

Activity - 7
9. Why does a coin placed in a transparent vessel appear above the water level when the vessel is sufficiently filled with water.     (AS - 3) (2 Marks)
A: * Take a transparent cylindrical vessel.
* Place a coin at the bottom of the vessel.
* Now pour water until we get the image of the coin on the water surface.
* When we try to see the coin from the side of the vessel the coin is not visible in the absence of water in it.
* The coin is visible when water is poured in the vessel.
* This is because of total internal reflection.

Activity - 8
10. Derive the relation between thickness of a glass slab, vertical shift and refractive index of a glass slab by an activity.      (AS - 3) (4 Marks)

* Measure the thickness of the slab and note it in your note book.
* Take a white chart and fix it on the table.
* Take the slab and place it in the middle of the chart. Draw its boundary. Remove the slab from its place. The lines form a rectangle.
* Name the vertices at A, B, C and D. Draw a perpendicular to the longer line AB of the rectangle at any point on it.
* Place slab again in the rectangle ABCD.
* Take a pin. Place at a point P in such a way that its length is parallel to the AB on the perpendicular line at a distance of 15 cm from the slab.
* Now take another pin and by looking at the first pin from the other side of the slab try to place the pin so that it forms a straight line with the first pin.
* Remove the slab and observe the positions of the pins.
* Draw a perpendicular line from the second pin to the line on which first pin is placed. Call the intersection point Q. Find the distance between P and Q.
* We may call it vertical shift.
* Do the same activity for another distance of the pin from the slab. We will get the same vertical shift.
* We could use a formal to find out refractive index of the glass.

I. Conceptual Understandings. (AS - 1)
1. How do you identify the denser medium in the given two media.     (2 Marks)
A: Denser medium is the medium which has more refractive index and rarer medium has less refractive index in the given two media.

2. What are optically denser and rarer media.           (2 Marks)
A: * Let v1 and v2 be the speed of light in the two given media respectively.
* If v1 > v2 then the first medium is denser medium and the second medium is rarer medium.
* If v2 > v1 then the second medium is denser medium and the first medium is rarer medium.

3. Define absolute refractive index.          (1 Mark)

4. What do you understand if you are told that the refractive index of glass is 1.5.  (2 Marks)

5. Write Snell's Law and explain the terms in it.     (2 Marks)
A: * Snell's law is: n1.sin i = n2 sin r
* n1 and n2 are the refractive indices of the two media.
* Angle of incidence = i and angle of refraction = r

6. What is a mirage?               (1 Mark)
A: * Mirage is an optical illusion where it appears that water has collected on the road at a distant place but when we get there, we do not find any water.

7. Write the laws of refraction.       (2 Marks)
A: Laws of refraction.
* The incident ray, the refracted ray and the normal to interface of two transparent media at the point of incidence all lie in the same plane.
* During refraction, light follows Snell's law.
n1 sin i = n2 sin r (or)  = Constant.

8. What are optical fibres. How do they work.      (4 Marks)
A: Optical fibres: An optical fibre is a very thin fibre made of glass or plastic having radius about a micrometer (10-6 m). A bunch of such thin fibres form a light pipe.
Working: Because of the small radius of the fibre, light going into it makes a nearly glancing incidence on the wall.
* The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
* The light is thus transmitted along the fibre.

9. How does a doctor examine the stomach of a patient using an optical fibre.    (4 Marks)
A: * All organs of human body are not accessible to the naked eye of the doctor. For example intestines.
* The doctor inserts an optical fibre into the stomach through the mouth. Light is sent down through one set of fibres in the pipe.
* This illuminates the inside of the stomach. The light from the inside travels back through another set of fibres in the pipe and the viewer gets the image of the outer end.
* This is generally fed to the computer screen.

10. Distinguish between reflection and total internal reflection.      (4 Marks)
A:

 Reflection Total internal reflection Reflection takes place for all angles of incidence. Total internal reflection takes place when the angle of incidence exceeds critical angle. Reflection takes place on smooth polished surfaces No smooth polished surface is required for total internal reflection. Some part of the incident light is absorbed by the reflecting surface. No part of the incident light is absorbed by reflecting surface.

11. A rectangular glass wedge (prism) is immersed in water as shown in figure (a). For what value of angle α, will the beam of light, which is normally incident on AB, reach AC entirely as shown in figure (b). Take the refractive index of water as  and the refractive index of glass as .

A: From the geometry of figure b it is clear that, the angle of incidence on the side BC is equal to (dotted line is a normal drawn at the point of incidence.)
* The ray should undergo total internal reflection to reach A.C.
* For occurrence of total internal reflection, the value of α must be greater than the critical angle at interface of water and glass.
* Let 'C' be the critical; angle for the interface of glass and water. From the given condition,
α > C ........ (1)
* We know, sin C =  ........ (2)

From equation 2, We get
sin c =   ⇒ c = 62° 30'
Hence α is greater than C = 62° 30'

II. Asking questions and making Hypothesis

12. If a ray of light travels from one medium to another medium having same refractive indices. What is your observation?                (AS - 2) (2 Marks)
A: As the two media has the same refractive indices light travels in these two media along a straight line without any deviation or bending.

13. Why planets do not twinkle?          (AS - 2) (2 Marks)
A: * Planets are relatively closer to earth compared to stars.
* A planet behaves as an extended source of light. It means it appears as a collection of a number of Point-sized sources of light.
* The - total variation in the amount of light entering our eye from all individual point sized sources of light will average out to be zero.
* So planets do not twinkle.

14. Why are the bubbles rising up the fish tank appear Silvery?     (AS - 2) (2 Marks)
A: * Light rays travelling through water, strike the water air interface of the bubbles at an angle greater than the critical angle of water.
* This results in the total internal reflection of these light rays.
* These internal reflected rays on reaching the viewers eye appear to be coming from air bubbles.
* So the air bubbles rising up the fish tank appear silvery.

15. Why does a crack in a window pane appear silvery?      (AS - 2) (2 Marks)
A: * In a crack of a window pane some air is always present.
* The rays of light striking the glass, the glass air interface incident at an angle greater than the critical angle of glass.
* So these rays of light are totally internally reflected.
* When these rays enter the viewer eye, they appear to come from the crack, which intern appear silvery.

16. A straight stick dipped in water appears to the bent. Why?       (AS - 2) (2 Marks)
A: * Rays of light coming from the end of the stick dipped in water bend away from the normal when they are refracted in to air.
* These refracted rays in air when extended backwards, they meet at a point higher than the end of the stick.
* This point gives the apparent position of the stick. So the stick dipped in water appears to be bent.

17. A pond appears to be shallower than it is really is when viewed of obliquely. Why? (AS - 2) (2 Marks)
A: * Rays of light coming from the bottom of the pond bend away from the normal when they are refracted in to air.
* These refracted rays in air when extended backwards, they meet at a point higher than the bottom of the pond.
* This point gives the apparent position of the bottom. So the pond appears to be shallower than it is really when viewed obliquely.

IV. Information Skills and Projects

18. The following table gives the refractive indices of different materials.
a) Identify the medium of highest optical density and the medium of lowest optical density.
b) In which of the media does the light travel fast.
1) Kerosene      2) Turpentine oil        3) Water
c) What do you understand by the statement that the refractive index of diamond is 2.42
d) What happens to the light rays when they travel from water to crown glass.
e) How a ray of light effects when it enters diamond from air.     (AS - 4) (4 Marks)

A: a) 1) Diamond is having the highest optical density as its refractive index is 2.42.
2) Air is having the lowest optical density as its refractive index is 1.0003
b) speed of light in the medium v =
1) So light travels faster in the medium of low refractive index.
2) The refractive indices of Kerosene, Turpentine oil and Water are 1.44, 1.47, 1.33.
3) So light travels faster in water and slower in turpentine oil. Light travels with a speed in between its speed in turpentine oil and water when travelling in kerosene.
c) We understand that as the refractive index of diamond is 2.42, its critical angle will be small and light rays entering the diamond easily undergo total internal reflection.
d) When light travels from water to crown glass its speed decreases and the light ray bends towards the normal.
e) When light travels from diamond to air, the speed of light rays increases and the light rays bends away from the normal.

V. Communication Through Drawing, Model Making
19. In the figure the refraction and emergence of a ray of light incident on a glass slab is shown. Re draw the figure and show the lateral displacement of the incident ray. Mention two factors on which the lateral displacement of the incident ray depends.
PQ: incident ray
QR: refracted ray
RS: Emergent ray
A: The lateral displacement of the incident ray depends on
a) The angle of incidence of the incident ray on the face of the glass slab.
b) The thickness of the glass slab.
c) In the fig (b) D is the lateral displacement of the incident ray.

VII: Application to daily life, Concern to Biodiversity

20. What is total internal reflection? What are the applications of total internal reflection. (AS - 7) (2 Marks)
A: Total internal reflection:
When the angle of the incidence is grater than critical angle, the light ray gets reflected into the denser medium at the interface of the denser medium enters the rarer medium. This phenomenon is called total internal reflection.
Applications of total internal reflections:
1) Optical fibres: Total internal reflection is the basic principle of working of optical fibres.
2) Brilliance of diamonds: Light ray incident on the face of a diamond undergoes internal reflections and the diamond appear brilliant. The critical angle for diamond is 24.4° which is very low.

21. What are the uses of optical fibres in daily life.        (AS - 7) (2 Marks)
A: Uses of optical fibres:
* Optical fibres are used to see the internal parts of the human body by the doctors.
* Fibre optics is used to transmit communication signals through light pipes. The clarity of signals transmitted in this way is much better than the other conventional methods.

22. Write the applications of optical fibres in communications.         (AS - 7) (2 Marks)
A: The other important applications of fibre optics is to transmit communication signals through light pipes.
* For example, about 2000 telephone signals, approximately mixed with light waves, may be simultaneously transmitted through a typical optical fibre.
* The clarity of the signals transmitted in this way is much better than other conventional methods.

Writer: C.V. Sarveswara Sarma

Posted Date : 13-12-2021

గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.