Refraction Of Light At Plane Surfaces Class 10th Physics & Chemistry AP Board Solution

Class 10th Physics & Chemistry AP Board Solution
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  1. Why is it difficult to shoot a fish swimming in water? (AS1)
  2. The speed of the light in a diamond is 1,24,000 km/s. Find the refractive index of diamond…
  3. Refractive index of glass relative to water is 9/8. What is the refractive index of water…
  4. The absolute refractive index of water is 4/3. What is the critical angle? (AS1)…
  5. Determine the refractive index of benzene if the critical angels 42°. (AS1)…
  6. Explain the formation of mirage? (AS1)
  7. How do you verify experimentally that sin i/sin r is a constant? (AS1)…
  8. Explain the phenomenon of total internal reflection with one or two activities. (AS1)…
  9. How do you verify experimentally that the angle of refraction is more than angle of…
  10. Take a blight metal ball and make it black with soot in a candle flame. Immerse it in…
  11. Take a glass vessel and pour some glycerine into it and then power water upto the brim.…
  12. Do activity-7again. How can you find critical angle of water? Explain your steps briefly.…
  13. Collect the values of refractive index of the following media. (AS4) Water, coconut oil,…
  14. Coiled information a working of optical fibers. Prepare a report about various uses of…
  15. • Are you able to see the head of the needle? Take a thin thermal sheet. Cut it in…
  16. • At what maximum radius of disc, were you not able to see the free end of the needle?…
  17. • Why were you not able to view the head of the nail fix certain radii of the discs? Take…
  18. • Does this activity help you to limit he ethical age of the medium (water)? Take a thin…
  19. • Draw a diagram to show the passage of light ray from the head of the nail in different…
  20. Explain the refraction of light through a glass slab with a neat ray diagram. (AS5)…
  21. Place an object on the table. Look at the object through the transparent glass slab. You…
  22. What is the reason behind the shining of diamonds and how do you appreciate it? (AS6)…
  23. How do you appreciate the role of Fermat principle in drawing ray diagrams. (AS6)…
  24. A light ray is incident on air-liquid interface at 45o and is refracted at 30°. What is…
  25. Explain why a tube immersed at a certain angle in a tumbler of water appears to have a…
  26. In what cases does a light ray not deviate at the interface of two media? (AS7)…
  27. When we sit at a camp fire, objects beyond the fire are seen swaying. Give the reason for…
  28. Why do stars appear twinkling? (AS7)
  29. Why does a diamond shine more than a glass piece cut to the same shape? (AS7)…

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Question 1.

Why is it difficult to shoot a fish swimming in water? (AS1)


Answer:

Due to the phenomenon of refraction the depth of pond appears to rise upwards. Hence, we cannot estimate the actual depth of fish in water and when we shoot the fish it appears to be raised therefore it becomes difficult to shoot the fish swimming in water. The figure below illustrates the phenomenon.




Question 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 3.00,000 km/s. is (AS I)


Answer:

Refractive index of diamond (n) = 

Given;


Speed of light in air = 3.00,000 km/s.


Speed of light in diamond = 1,24,000 km/s.


⇒ n = 


⇒ n = 2.42


Hence refractive index of diamond is 2.42.



Question 3.

Refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass? (AS1)


Answer:

Refractive index of water relative to glass = 

⇒ Refractive index of water relative to glass = 


⇒ Refractive index of water relative to glass = 


Refractive index of water relative to glass is = 



Question 4.

The absolute refractive index of water is 4/3. What is the critical angle? (AS1)


Answer:

Given;

Absolute refractive index of water (n) = 


By Snell’s Law critical angle(c) is given by =  (sin


inverse of  )


⇒ c = 


⇒ c = 


∴ c = 48.5o. (y looking in sine table).


Hence the critical angle is  or 48.5o.



Question 5.

Determine the refractive index of benzene if the critical angels 42°. (AS1)


Answer:

Given;

Critical angle (c) = 42o .


By Snell’s Law;


Refractive index(n) = 


⇒ n = 


⇒ n =  (By looking in sine table sin42 = 0.6691).


Hence refractive index of benzene is 1.51



Question 6.

Explain the formation of mirage? (AS1)


Answer:

Definition: Mirage is the optical illusion; the water appears at the distance but when we go there we don’t find any water. It occurs due to Total Internal Reflection.



Formation:


1. During hot summer days air just above road is very hot and air at higher altitudes is cold.


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


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


4. So cooler air has higher refractive index than the hotter air just above the road.


5. Therefore light travels faster through thinner hot air than dense cold air.


6. When light from tree passes through the medium just above ground whose refractive index decreases towards ground it suffers total internal reflection and takes the curved path.


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


8. The inverted virtual image seen by observer is called as the mirage.


The figure below depicts the formation.






Question 7.

How do you verify experimentally that sin i/sin r is a constant? (AS1)


Answer:

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

Materials Required: A plank, white chart paper, protector, semicircular glass disc, pencil, laser light.


Procedure:


1. Take a wooden plank covered with white chart paper and draw two perpendicular lines passing through center as shown in figure.


2. Mark one line as NN which is perpendicular to the other line MM. Let O be the point of intersection of both lines.


3. Take protector and place it along NN and then mark the angles from 0o to 90o on both sides of NN as shown in figure.


4. Now place the semicircular glass disc such that the diameter coincides with the MM and center coincides with the O.



5. Send the laser light along line which makes an angle of 20o with NN. Measure its corresponding angle of refraction.



6. Repeat the procedure for 30o, 40o, 50o and 60o.


Observation: If we calculate the ratio of the sine of angle of incidence and sine of angle of reflection than it will come out to be constant.


Result: Ratio of sine of angle of incidence and angle of refraction is constant.



Question 8.

Explain the phenomenon of total internal reflection with one or two activities. (AS1)


Answer:

When the angle of incidence is greater than the particular angle (called as critical angle) than the ray of light instead of refracting gets reflected into denser medium at the interface. This phenomenon is called as the total internal reflection.


Conditions Of Total internal Reflection: Total internal reflection takes place when both of the following two conditions are met:


• The light is in the denser medium and approaching the less dense medium.


• The angle of incidence is greater than the so-called critical angle.


Activity 1:


1. Place the semicircular disc in such a way that diameter coincides with the interface and centre coincides with the center of the interface.


2. Now send the light along the curved side of the disc and. Start with angle of incidence 0o .


3. Send the laser along the angles 5o, 10o, 15o,… and measure the angle of refraction.


Observation: We will observe that at certain angle of incidence the refracted ray does not come out but gazes the interface. This angle of incidence is called as the critical angle. Figure below depicts the same.




Question 9.

How do you verify experimentally that the angle of refraction is more than angle of incidence when light rays travel from denser to rarer medium. (AS1)


Answer:

Aim: Verify that the angle of refraction is more than angle of incidence when light rays travel from denser to rarer medium.

Procedure:


1. Take a metal disc and mark angle along its edges using protractor as shown in the figure.



2. Arrange two straws at center of disc such that they can be rotated freely about center of disc.


3. Adjust one of the straw to make the angle of 10o and immerse half of the disc vertically into the water verify that straw is at 10o inside the water.


4. From the top of the vessel view the straw as shown in the figure.



5. Adjust straw until both straw are in single straight line.


6. Take disc out of water and we will observe that two straws are not in straight line.


7. Measure the angle between the normal and second straw and note it.


8. Repeat for various angles and find their corresponding angle of refraction.


Observation: we will note that angle of refraction(r) is more than angle of incidence (i).


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



Question 10.

Take a blight metal ball and make it black with soot in a candle flame. Immerse it in water. How does it appear and why? (Make hypothesis mid do the above experiment). (A52)


Answer:

When the bright metal ball is made black with soot in candle flame the air enters the space between the soot and metal ball.

• Light passes from the denser medium (water) to the rarer medium (air).


• When the angle of incidence exceeds the critical angle, the total internal reflection takes place.