CBSE Class 10 Science PrIndirect Speech Exercises for Class 7 CBSE With Answersactical Skills – Refraction Through Glass Slab

CBSE Class 10 Science Practical Skills – Refraction Through Glass Slab

BASIC BUILDING CONCEPTS
Refraction of Light

1. When a beam of light falls obliquely at the interface of the two optical media, direction of its path changes when it enters into the other medium. This phenomenon exhibited by the light rays is called refraction of light. This is due to the change in speed of light while going from one medium to the other.
   
2. During the refraction, when light travels from one medium to another transparent medium, the speed and wavelength of light changes, whereas frequency remains the same.
3. When a light ray travels from rarer to denser medium, its speed gets slow down and the ray bends towards normal (∠i >∠r).
4. When the ray of light travels from denser to rarer medium, its speed goes up and the ray bends away from the normal (∠i < ∠r).
   From the above two points, we can conclude that the light bends on undergoing refraction.
5. Laws of Refraction:
   • At the point of incidence the incident ray, the refracted ray and the normal to the interface, all lie in the same plane.
   • The ratio of the sine of angle of incidence to the sine of angle of refraction is constant for the same colour of light and for the same pair of transparent media. Thus,
     sin i/sin r = constant = n₂₁
     Where n21 is called the refractive index of the second medium with respect to first medium.
     This law is also known as Snell’s law of refraction.
6. Refraction through rectangular glass slab with parallel faces:
   The refraction takes place at both air-glass interface and glass-air interface has the following characteristics:
   • When a light ray travels from air to glass, the angle of incidence is greater than angle of refraction as ray bends towards normal.
   • When a light ray travels from glass to air, the angle of refraction (also called angle of emergent in case of glass) is greater than the angle of incidence of glass-air interface as ray of light bends away from the normal.
   • If the angle of incidence is zero, i.e. the incident ray is normal to the interface, the ray of light continues to travel in the same direction after refraction.
   • The angle of emergence and angle of incidence will be equal.
     Emergent ray is parallel to the incident ray along with original direction but it will be laterally displaced to the left of the incident ray.
   • For the same angle of incidence, lateral displacement is proportional to the thickness of the glass slab.
   • For the same thickness of glass slab, the lateral displacement is proportional to the angle of incidence.

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AIM
To trace the path of a ray of light passing through a rectangular glass slab for different angles of incidence. Measure the angle of incidence, angle of refraction, angle of emergence and interpret the result.

MATERIALS REQUIRED
Drawing board, drawing pins, three plane sheets of white paper, a rectangular glass slab, geometry instruments, and sharp pointed pins.

THEORY
The refraction takes place at both air-glass interface and glass-air interface of a rectangular glass slab.

• When the light ray incident on air-glass interface (DC) obliquely, it bends towards the normal.
• The refracted ray is incident obliquely on the second parallel surface inside the rectangular glass slab, i. e. glass-air interface (AB) and after refraction, it moves away from the normal.
  These refractions at both the surfaces obey the laws of refraction (Refer to basic building concept).
  

PROCEDURE

1. Fix a white paper on a drawing board with the help of drawing pins and divide the sheet in two parts by a vertical line.
2. Place a rectangular glass slab in the first part. Draw its boundary. Remove the glass slab and label the boundary as A₁, B₁, C₁, D₁ as shown below.
   
3. Draw a normal (perpendicular) MN on the side A₁B₁ at a point O₁, slightly away from the centre towards A₁
4. Draw an oblique line P₁Q₁ (incident ray) such that ∠P₁O₁M = 30° (Angle of incidence). Fix two sharp pointed pin P₁ and Q₁ vertically erected on the line P₁O₁ at a distance of 4 to 6 cm apart.
5. Place the glass slab again within its boundary. Look at the feet of pins (not their heads) P₁ and Q₁ from the other parallel opposite face of the slab, i.e. from C₁D₁ along the plane of paper. Fix other two pins R₁ and S₁ in such a way that R₁ , S₁ and the image of P₁ and Q₁ lie on a same straight line.
6. Remove the glass slab and all the four pins. Encircle all the prick of the four pins. Join the points R₁ and S₁ within the encircle and produce upto the edge C₁D₁. Let R₁S₁ meet C₁D₁ at O₁. This will act as an emergent ray.
7. Draw a normal M₁N₁ at O₂. Join O₁ and O₂. It will represent the path of ray inside the glass slab, i.e. refracted ray.
8. Measure the angle of emergence, i.e. ∠e = ∠N₁O₁S₁ and angle of refraction, i.e. ∠r = ∠NO₁O₂
9. Repeat the experiment by taking the different angles of incidence such as 45° and 60° on the other part of paper and measure the angle of refraction and emergence accordingly and tabulate them.

OBSERVATION TABLE

RESULT

1. At the point of incidence the incident ray, refracted ray and the normal to the air-glass interface, all lie in the plane of paper.
2. Within experimental limits, the angle of emergence and angle of incidence are equal.
3. The emergent ray is parallel to the incident ray.
4. Emergent ray is laterally displaced.
5. When the light ray travels from optically rarer medium (air) to optically denser medium (glass), the angle of refraction is less than the angle of incidence.
6. The refracted angle at the air-glass interface and the incident angle at the glass-air interface are found to be equal.
7. From the observation table, it is clear that with the increase in angle of incidence, angle of refraction also increases.

PRECAUTIONS

1. Drawing board should be made of soft wood.
2. Glass slab must be clean and should be free from air bubble.
3. The pins should be fixed in vertical position.
4. The base of all the pins (on the paper) should be placed in a straight line.
5. Keep your eyes along the plane of paper and in line while observing the image and fixing the pins.
6. For incident ray, the angle of incidence must lie between 30° and 60°.

INTERACTIVE SESSION

Question 1.
What is the refraction?
Answer:
The phenomenon in which a beam of light falls obliquely at the interface of the two optical media, i.e. air-glass interface etc., and direction of its path changes when it enters into the second medium is called refraction.

Question 2.
Is there any relation between two velocities in two different media?
Answer:
The ratio of velocity of light in vacuum/air to the velocity of light in optically denser medium is constant and it is called refractive index of the medium, i.e. n₂₁ = c/ν

Question 3.
On which factors does the refractive index of a medium depend?
Answer:
It depends upon

• density of the medium
• nature of the medium

Question 4.
What is the change in frequency during refraction of light?
Answer:
During the journey of a light ray from one medium to another medium, frequency remains the same.

Question 5.
For which colour the refractive index of glass is maximum and minimum?
Answer:
Maximum for violet colour and minimum for red colour.

Question 6.
What is the unit of refractive index?
Answer:
It has no unit.

Question 7.
Why?
Answer:
Because it is the ratio of sine of the angle of incidence to the sine of the angle of refraction.

Question 8.
In this experiment, which one is greater: angle of incidence or angle of refraction?
Answer:
When the ray travels from air to glass, ∠ i > ∠ r.
When the ray travels from glass to air, ∠ i < ∠ r.

Question 9.
What is the relation made by the refracted ray with both the normals?
Answer:
Both the normals at both the interfaces are parallel so refracted ray will act as a transversal. Hence, refracted ray will make equal angles with the normal.

Question 10.
What happens to the emergent angle on increasing the incident angle at air-glass interface?
Answer:
Angle of emergence also increases.

Question 11.
What do you mean by the lateral displacement?
Answer:
The lateral displacement is the perpendicular distance between the incident ray and emergent ray.

Question 12.
Why do you measure the perpendicular distance?
Answer:
Because incident ray and emergent ray both are parallel to each other.

Question 13.
With the help of this experiment, would you be able to find the refractive index of the glass slab?
Answer:
Yes, by finding the values of sin i and sin r, we can calculate the refractive index of the glass slab.

Question 14.
State the conditions when no refraction occurs.
Answer:

1. The light ray falls along the normal and
2. The refractive index of the two optical media are equal.

Question 15.
Does the glass slab appear to us of the same thickness as actually it is?
Answer:
No, it appears to be less thicker than actually it is.

Question 16.
Why?
Answer:
Due to refraction.

Question 17.
Name some natural phenomena which are based on the refraction.
Answer:

1. Bottom of the tank or pond filled with water appears to be raised.
2. Spoon or pencil partly immersed in water looks tilted.
3. Twinkling of stars.
4. Words on paper appear to be raised when viewed by keeping glass slab over it.

Question 18.
Why should the pins be fixed at least 4 to 6 cm apart?
Answer:

1. To get more light.
2. It will be more convenient to fix other pins, if the two pins on incident ray are far apart.

Question 19.
How is reflection of light different from refraction of light?
Answer:
In reflection, ray of light turns back in the same medium after reflection, while in refraction, ray of light passes from rarer to denser medium or denser to rarer medium.

NCERT LAB MANUAL QUESTIONS

Question 1.
Why are the incident and emergent rays parallel to each other in case of refraction through a rectangular glass slab?
Answer:
In case of refraction through a rectangular glass slab, the incident and emergent rays are parallel to each other because the angle of emergence ∠e is equal to the angle of incidence ∠i, i.e. ∠e = ∠i.

Question 2.
Why does a ray of light bend towards the normal when it enters from air into glass slab and bends away from the normal when it emerges out into air?
Answer:
When a ray of light enters from air into the glass slab, it bends towards the normal due to decrease in its speed and when it emerges out into air, its speed increases so it bends away from the normal.

Question 3.
Draw the path of a ray of light when it enters perpendicular to the surface of a glass slab.
Answer:

Question 4.
While tracing the path of ray of light through a glass slab, the angle of incidence is generally taken between 30° and 60°. Explain the reason on the basis of your performing this experiment for different angles of incidence.
Answer:
On the basis of experiment, it is observed that

• the angle of emergence also increases with the increase in angle of incidence or vice versa.
• if the angle of incidence is less than 30s, very less bending occurs at the emergent glass-air interface.
• if the angle of incidence is greater than 60s, the emergent ray may emerge from the side surface of the rectangular glass slab instead of opposite parallel surface.

Hence, angle of incidence should be taken between 30° and 60° for tracing the path of ray of light through a glass slab.

Question 5.
How does the lateral displacement of the emergent ray depend on the width of the glass slab and angle of incident?
Answer:
Lateral displacement of emergent ray is proportional to

• the thickness of the glass slab for a given angle of incidence and a pair of media.
• the angle of incidence for the given thickness of the glass slab and a pair of media.

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA (1 Mark)

Question 1.
A student traces the path of a ray of light passing through a rectangular glass slab. [Delhi 2009]

For measuring the angle of incidence, he must position the protractor in the manner as shown in figure:
(a) A
(b) B
(c) C
(d) D

Question 2.
While performing the experiment on tracing the path of a ray of light passing through a glass slab as shown in the given diagram, four students interpreted the results as given below. Which one of the four interpretations is correct? [CBSE 2012]

(a) ∠r > ∠e
(b) ∠r = ∠e
(c) ∠i = ∠r
(d) ∠i > ∠r

Question 3.
A student performed the experiment of tracing the path of light through a glass slab as shown. Which is the angle of refraction out of the following? [CBSE 2012]

(a) PQN₁
(b) N₂QR
(c) DQR
(d) M₂RE

Question 4.
After tracing the path of ray of light through a glass slab for different angles of incidence, a student measured the corresponding values of angle of refraction r and angle of emergence e and recorded them in the table given below: [Delhi 2013]

The correct observations are:
(a) I and II
(b) II and III
(c) I and III
(d) I, II and III

Question 5.
For tracing the path of ray of light through a rectangular glass slab, the range of angle of incidence should be between [AI 2013]
(a) 15° and 40°
(b) 20° and 45°
(c) 30° and 60°
(d) 50° and 75°

Question 6.
A student, while tracing the path of ray of light passing through the glass slab, fixes two pins vertically on an inclined straight line and looks at the images of these pins from the other side. He should fix two other pins such that:
(a) his eye and feet of the pins are in a straight line.
(b) his eye and head of pins are in a straight line.
(c) his eye and middle point of pins are in a straight line.
(d) his eye and feet of all pins are in different line.

Question 7.
While performing the experiment to -trace the path of ray through glass slab, the teacher instructed her students to ensure that during the experiment, glass slab must not displace its boundary. This instruction was given because if glass slab gets displaced out of its boundary then
(i) the angle of incidence of ray will change.
(ii) the diagram will not look nice.
(iii) the refracted ray will not be traceable.
(iv) the emergent ray will not be seen.
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)

Question 8.
Four students traced the path of a ray of light passing through a glass slab. The correct trace is that of student

(a) A
(b) B
(c) C
(d) D

Question 9.
The path of a ray of light passing through a rectangular glass slab was traced and angles were marked. Which one out of the following is the correct representation of an angle of incidence ‘i’, angle of refraction ‘r’ and angle of emergence ‘e’ as shown in the diagrams? [Delhi 2007]

(a) I
(b) II
(c) III
(d) IV

Question 10.
Which one of the following is the best set-up for tracing the path of a ray of light through a rectangular glass slab? [Delhi 2013, AI 2007]

(a) I
(b) II
(c) III
(d) IV

Question 11.
In which of the following figures, the correct placement of the protractor (or ‘D’), for measuring the angles of incidence and emergence, is shown in the glass slab experiment? [Delhi 2007C]

(a) A
(b) B
(c) C
(d) D

Question 12.
An experiment to trace the path of a ray of light through a glass slab was performed by four students I, II, III and IV. They reported the following measurements of angle of incidence ‘i’ , angle of refraction ‘r’ and angle of emergence ‘e’ Which one of the students has performed the experiment correctly? [AI 2007C]

(a) I
(b) II
(c) III
(d) IV

Question 13.
A student performs the experiment on tracing the path of a ray of light passing through a rectangular glass slab for different angles of incidence. He measures the angle of incidence ∠i, angle of refraction ∠r and angle of emergence ∠e for all his observations. He would find that in all cases [AI 2013, Delhi 2008]
(a) ∠i is more than ∠r but (nearly) equal to ∠e
(b) ∠i is less than ∠r but (nearly) equal to ∠e
(c) ∠i is more than ∠e but (nearly) equal to ∠r
(d) ∠i is less than ∠e but (nearly) equal to ∠r

Question 14.
A student performed the experiment of tracing the path of light through a rectangular glass slab as shown. From the given figure, the angle of incidence is: [CBSE 2012]

(a) 40°
(b) 30°
(c) 50°
(d) 90°

Question 15.
A teacher advises his student that he should take the angle of incidence not more than 60°. He says this because for higher angle of incidence, the emergent ray
(a) moves along the same path
(b) tends to graze along the side surface of the slab
(c) tends to move along the normal
(d) none of the above

Question 16.
A student carries out the experiment of tracing the path of a ray of light through a rectangular glass slab, for two different values of angle of incidence: ∠i = 30° and ∠i = 45°. The set of values of the angle of refraction (∠r), and the angle of emergence (∠e), she is likely to observe in the two cases are [Delhi 2007C]
(a) [∠r = 30°, ∠e = 20°] and [∠r = 45°, ∠e = 28°]
(b) [∠r = 20°, ∠e = 30°] and [∠r = 45°, ∠e = 28°]
(c) [∠r = 20°, ∠e = 30°] and [∠r = 25°, ∠e = 45°]
(d) [∠r = 30°, ∠e = 20°] and [∠r = 25°, ∠e = 45°]

Question 17.
A student suggested the following ‘guidelines’ to his friend for doing the experiment on tracing the path of a ray of light passing through a rectangular glass slab for three different angles of incidence: [CBSE 2012, AI 2008]
A. Draw the ‘outline’ of the glass slab at three positions on the drawing sheet.
B. Draw ‘normals’ on the top side of these ‘outlines’ near their left end.
C. Draw the incident rays on the three ‘outlines’ in directions making angles of 30°, 45° and 60° with the normals drawn.
D. Fix two pins vertically on each of these incident rays at two points nearly 1 cm apart.
E. Look for the images of the ‘heads’ of these pins while fixing two pins from the other side, to get the refracted rays.
When these ‘guidelines’ were shown to his teacher, the only ones that were left uncorrected were the ‘guidelines’ labelled as
(a) (A), (B)
(b) (A), (C)
(c) (B), (C)
(d) (B), (E)

Question 18.
In the given figure, MN represents the lateral displacement of a ray of light emerging out from a rectangular glass slab of thickness AB. If the thickness of the glass slab is increased to AC, the lateral displacement will be

(a) the same
(b) less than MN
(c) more than MN
(d) zero as the ray emerges out.

ANSWER KEY

1. (b)
2. (d)
3. (b)
4. (d)
5. (c)
6. (a)
7. (a )
8. (c)
9. (d)
10. (b)
11. (d)
12. (a)
13. (a)
14. (c)
15. (b)
16. (c)
17. (b)
18. (c)

Short Answer Questions (2 Marks)

Question 1.
In the experiment to trace the path of a ray of light through a glass slab,

1. if the angle of incidence is increased, how does angle of refraction change?
2. What relationship students work out when they measure the angle of incidence and angle of emergence?

Answer:

1. From Snell’s law,
   sin i / sin r = constant, so if angle of incidence increases, angle of refraction also increases.
2. Both are equal. This is because when light gets refracted through a glass slab, the incident ray and emergent ray are parallel to each other, so ∠i = ∠e.
   

Question 2.
In an experiment with a rectangular glass slab, a student observed that a ray of light incident at an angle of 55° with the normal on one face of the slab, after refraction, strikes the opposite face of the slab before emerging out into air making an angle of 40° with the normal. Draw a labelled diagram to show the path of this ray. What value would you assign to the angle of refraction and angle of emergence?
Answer:
OA- incident ray
∠i is angle of incidence = 55°
Given ∠r₂ = 40°
∠r₁ and ∠r₂ are alternate interior angles
∠r₁ = ∠r₂ = 40°
So, angle of refraction = 40°
Since, the emergent ray is parallel to the incident ray, the angle of emergent must be equal to angle of incidence, i.e. ∠e = ∠i = 55°

Question 3.
A ray of light, incident obliquely on a face of a rectangular glass slab placed in air, emerges out from the opposite face.
(a) Draw a ray diagram to show the path of the ray of light and identify two points where refraction takes place.
(b) If this experiment is repeated with the different angles of incidence, what relation would you observe among the angle of incidence, the angle of refraction and the angle of emergence?
Answer:

In the figure, the two points are O and O’ where refraction takes place as light ray has changed its direction at these two points.
(b) The relation among the angle of incidence, the angle of refraction and the angle of emergence remains the same, i.e. ∠i = ∠e > ∠r

Question 4.
Why the glass slab must not be displaced from its boundary during the experiment to trace the path of a ray of light through a glass slab?
Answer:
If the glass slab gets displaced from its. original position/boundary, the angle of incidence will change accordingly and experiment will not be performed correctly. Boundary helps us to keep the glass slab in a proper position.

Question 5.
What would be the observations of a student who performed the experiment to trace the path of a ray of light through a glass slab, whose two opposite faces are not parallel to each other?
Answer:

1. The emergent ray is not parallel to the incident ray. The direction of emergent ray depends upon the inclination of non- parallel faces.
2. Angles of refraction at both the faces,
   i.e. ∠r₁ and ∠r₂ are not equal.

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