CBSE Class 10 Science Lab Manual

CBSE Class 10 Science Lab Manual – Refraction Through Glass Slab

Board	CBSE
Textbook	NCERT Books
Class	Class 10
Subject	Science (Physics)
Experiment Name	Refraction Through Glass Slab
Category	Class 10 Science Lab Manual

The experiment to determine Refraction Through Glass Slab are part of Science Lab Manual for Class 10 CBSE Experiments is designed to help students bridge the gap between theoretical concepts and practical applications through hands-on experiments in Physics.

Science Lab Manual Class 10 CBSE Refraction Through Glass Slab Experiment

Determine Refraction Through Glass Slab Class 10 Practical

Class 10 Physics Practical Glass Slab Introduction

• Refraction: When light travels from one medium to another, it generally bends,this bending is called refraction.
• Snell’s law: The law of refraction is also known as Snell’s law. Snell’s law gives the relationship between angles of incidence and angle of refraction.
  

Laws of refraction of light

1. The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
2. The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media.
   This law is also known as Snell’s law of refraction. If i is the angle of incidence and r is the angle of refraction, then:
   

• The speed of light in a vacuum is 3.00 x 10 s m/s.
• When light travels through something else, such as glass, water or oil it travels at a different speed.
• Refractive index: The speed of light in a given material is related to a quantity called the refractive index, n, which is defined as the ratio of the speed of light in vacuum to the speed of light in the medium.
• Index of refraction: n = c/v, where the speed of light in a medium is v and c is the speed of light in vacuum.
• The ability of a medium to refract light is also expressed in terms of its optical density.
• The medium with larger refractive index is optically denser medium than the medium with comparatively smaller refractive index.
• The speed of light is higher in a rarer medium than a denser medium.
• Convex lens converges a beam of light whereas concave lens diverges a beam of light incident on them.
• Rules of refraction:
  Rule 1: When a light ray travels from a rarer medium to a denser medium, the light ray bends towards the normal (RDTN).
  Rule 2: When a light ray travels from a denser medium to a rarer medium, the light ray bends away from the normal (DRAN).
• The diagram obtained is a linear diagram of refraction of light through glass slab.
  
• Refraction in daily life:
  (i) Bottom of a tank or a pond containing water appears to be raised.
  (ii) Printed matter through a thick glass slab placed on a table appears to be raised.
  (iii) Pencil immersed in a glass containing water appears to be displaced at the interface of air and water.

Science Lab Manual Class 10 Glass Slab Experiment – 5

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.

Theory

• Refraction of Light: When light passes from one medium to other it deviates/changes its path, this property of light is called refraction of light.
• Normal Ray: A ray of light which forms an angle of 90° with the refracting surface is said to be normal. When a ray of light travels along the normal, it does not suffer any refraction.
• Incident Ray: A ray of light that travels towards the refracting surface is called incident ray.
• Refracted Ray: A ray of light that changes its path when passes through a refracting surface is said to be refracted ray.
  
  
• Emergent Ray: A ray of light which emerges out into the original medium after refraction is said to be an emergent ray.
• Lateral Displacement: The perpendicular shift in the path of light, seen when it emerges out from the refracting medium is called lateral displacement.
• Angle of Incidence (i): The angle formed between the normal and incident ray is called angle of incidence.
• Angle of Refraction (r): The angle formed between the refracted and normal ray is called angle of refraction.
• Angle of Emergence (e): The angle formed between the normal and emergent ray is called angle of emergence.
• DRAN: When a ray of light travels from denser medium to rarer medium it bends away from the normal.
• RDTN: When a ray of light travels from rarer medium to denser medium, it bends towards the normal.
• During Refraction:
  (i) Angle of incidence = Angle of emergence.
  (ii) Incident ray and emergent ray are parallel.
• Laws of Refraction:
  (i) The incident ray, the normal ray and the refracted ray, all lie in the same plane.
  (ii) The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant quantity for the two given media. This law is also known as Snell’s law. sin i/ sin r
  This constant value is called the refractive index of the second medium with respect to the first.

Materials Required
A drawing board, 4-6 all pins, white sheet of paper, rectangular glass slab, a protractor, a scale, a pencil and thumb pins.

Procedure

1. Take a soft drawing board. Fix a white sheet on it with the help of thumb pins.
2. Place the rectangular glass slab in the centre of the white paper and draw its outline boundary with pencil.
3. Mark this rectangular figure obtained as ABCD.
4. On one side of this figure, i.e., AB take one point E, draw a perpendicular EN and label it as normal ray.
5. With the help of a protractor draw one angle of 30° with the EN. Fix two pins P and Q on the ray of this angle, the distance between the pins should be more than 4-5 cm.
6. Put the glass slab on the rectangular figure ABCD.
7. See through the glass slab from side CD and fix pin R and S such that when seen through the glass slab all
   the pins lie in straight line, [i.e., Pins P, Q, R and S should lie in straight line when seen through the glass slab], ‘
8. Now, remove the pins P, Q, R and S one by one and draw small circles around the pin points.
9. Remove the glass slab.
10. Join points R and S such that it meets CD at point F.
    Draw perpendicular to CD at point F as N’M’.
11. Join points E and F with the pencil.
12. Measure the angles formed at AB and CD, i.e., the incident angle, refracted angle and emergent angle.
13. Extend ray PQ with scale and pencil in dotted line. It will be parallel to ray FRS. The distance between these two parallel rays is called lateral displacement (d).
14. Measure the lateral displacement.
15. Repeat the above procedure for angles 45° and 60°.
    Diagram
    
    ABCD = Glass slab
    EN and FM’ = Normal rays
    P, Q, R, S = All pins ∠PEN = ∠i = incident angle = 30°
    ∠MEF = ∠r = refracted angle
    ∠SFM’= ∠e = emergent angle = 30° ~ 31°
    d = lateral displacement.

Glass Slab Experiment Class 10 Observations Table

S.No.	Angle of incidence ∠i = ∠PEN	Angle of refraction ∠r = ∠MEF	Angle of emergence ∠e = ∠SFM’	∠i – ∠e ∠PEN – ∠SFM’
1.	30°	28°	30°	0°
2.	45°	43°	44.8°	0.2°
3.	60°	56°	59.8°	0.2°

During performing this experiment, ∠i – ∠e may not be zero at times as shown above due to human error.

Conclusion

1. The angle of incidence is nearly equal to the angle of emergence.
2. The angle of refraction is less than angle of incidence because light is travelling from rarer to denser optical medium.
3. The lateral displacement remains the same for different angles of incidences.
4. When the light ray travels from optically rarer medium (air) to optically denser medium (glass) the light bends towards the normal.

Precautions

1. The glass slab should be perfectly rectangular with all its faces smooth.
2. The drawing board should be soft so that pins can be easily fixed on it.
3. The angle of incidence should lie between 30° and 60°.
4. All pins base should lie in straight line.
5. While fixing the pins P and Q or the pins R and S, care should be taken to maintain a distance of about 5 cm between the two pihs.
6. Draw thin lines using a sharp pencil.
7. Use a good quality protractor having clear markings.
8. Place the protractor correctly to measure the angles.
9. Perpendiculars should be drawn correctly.

Sources of Error

1. The glass slab should not have any air-bubbles.
2. All measurement of angles using protractor should be done accurately.

NCERT Class 10 Science Lab Manual Viva Voce

Question 1:
What is refraction of light?
Answer:
When light travels from one optical medium to another, it (changes its path) deviates from its path. This phenomenon is called refraction of light.

Question 2: 
During this experiment of tracking the path of light through glass slab, what should be the angle of incidence?
Answer:
The angle of incidence should be taken between 30° to 60°.

Question 3:
What is lateral displacement?
Answer:
The perpendicular shift in the path of incident ray when it travels from one medium to another is called lateral displacement.

Question 4:
How is angle of incidence and emergence related?
Answer:
The angle of incidence and the angle of emergence are always equal or the difference may be of 1°.

CBSE Class 10 Science Lab Manual Practical Based Questions

Question 1:
Out of VIBGYOR, which light bends the least on refraction and which light bends the most?
Answer:
Red light bends the least and violet light bends the most.

Question 2:
State one condition during refraction of light where light does not deviate.
Answer:
When a ray of light is perpendicular to the refracting surface it will not show any deviation.

Question 3:
If a ray of light makes an angle of 30° with the refracting surface, then what will be the angle of incidence?
Answer:
The angle of incidence will be 90° – 30° = 60°.

Question 4:
A ray of light travels from optically denser medium to rarer medium. What will happen to its path?
Answer:
(DRAN) When light travels from optically denser medium to optically rarer medium it bends away from the normal.

Question 5:
Which property of light causes the rainbow formation?
Answer:
The refraction of light due to tiny water droplets in the atmosphere causes rainbow in the sky.

Question 6:
If a ray of light travels from water to oil, in which direction will the light bend?
Answer:
Oil is optically denser than water hence, it will bend towards the normal.

Question 7:
A ray of light travels in the path of normal ray. What will be the angle of incidence?
Answer:
As the ray of light, i.e., incident ray is forming 0° angle with normal ray hence, the angle of incidence is 0°.

Question 8:
Under what condition the emergent ray and incident ray will be parallel?
Answer:
When the angle of incidence is equal to the angle of emergence then both the rays are parallel.

Question 9:
Why does a ray of light bend towards the normal when it enters from air to glass slab?
Answer:
When light travels from rarer medium, i.e., air to the denser medium, i.e., glass slab, its speed decreases and hence it bends to take the shortest path.

Question 10:
On what factor is the lateral displacement of the glass slab dependent?
Answer:
Lateral displacement is dependent on the thickness of glass slab. When the thickness increases the lateral displacement also increases.

Question 11:
Why should we preferably take angle of incidence between 30° and 60°?
Answer:
For angles beyond 30° and 60° the refracted ray may not appear on the opposite face of the glass slab.

Question 12:
Why is emergent ray parallel to the incident ray, after the refraction of incident ray through a glass slab?
Answer:
∠i = ∠e in this case, and these are alternate angles. So, the incident ray is parallel to the emergent ray.

Question 13:
When a ray of light passes through a glass slab, then how many times does it change its path and why?
Answer:
The ray of light bends twice.
First time when it enters from air to the glass slab, it bends towards the normal, i.e., from rarer medium to denser medium.
Second time, when the ray moves out from the glass slab to air, it bends away from the normal, i.e., it moves from denser medium to rarer medium.

CBSE Class 10 Physics Lab Manual Questions

Question 1:
Why are incident and emergent rays parallel to each other in case of a rectangular glass slab?
Answer:
The incident ray and the emergent ray are parallel to each other in case of a rectangular glass slab because the angle of incidence and angle of emergence are same at opposite parallel surfaces of slab.

Question 2:
Why does a ray of light bend towards the normal when it enters from air in a glass slab and bends away from the normal when it emerges out into air?
Answer:
When light enters from air to glass slab its speed decreases and bends towards the normal. But when the light travels from glass slab to air its speed increases and 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:
The path of light when it enters perpendicular to the surface of glass slab would be a straight line passing through
the glass slab.

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:
This is because for any angle greater or smaller than this range (30 degrees to 60 degrees) the emergent ray would not appear on the opposite side of the slab.

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

Lab Manual Class 10 Science Multiple Choice Questions

Questions based on Procedural and Manipulative Skills
Question 1:
In the experiment to trace the path of a ray of light through a rectangular glass slab using pins P₁, P₂, P₃ and P₄, four students did the following:
A. Looked at heads of P₁ and P₂, while placing P₃, and heads of P₁, P₂ and P₃ while placing P₄ .
B. Looked at feet of P₁, P₂ while placing P₃ and feet of P₁, P₂ and P₃ while placing P₄ .
C. Looked at heads of P₁ and P₂, while placing P₃ and feet of all the pins while placing P₄.
D. Looked at feet of P₁ and P₂ while placing P₃ and heads of all the pins while placing P₄.
The correct procedure is that of student
(a) A               (b) B               (c) C                 (d) D.
Answer:
(b)
Explanation:
It is the correct procedure.

Question 2:
Lateral displacement depends on:
(a) angle of incidence
(b) angle of refraction
(c) thickness of glass slab
(d) angle of emergence.
Answer:
(c)
Explanation:
The bending of light depends on the medium.

Question 3:
If the angle of incidence is 0°, the angle of refraction in the same medium will be:
(a) 90°              (b) 0°
(c) 180°            (d) less than 0°.
Answer:
(b)
Explanation:
As per formula.

Question 4:
The relation between ∠i, ∠r and n (refractive index) is:
(a) n = sin i/sin r
(b) n = sin r/sin i
(c) n sin i = n sin r
(d) sin i = n sin r
Answer:
(a)
Explanation:
Formula of refraction.

Question 5:
A student suggested the following ‘guidelines’ to his friend for doing the experiment on tracing the path of a ray of light passing through rectangular glass slab for three different angles of incidence.
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 the directions making angles of 30°, 45°, 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 he showed these ‘guidelines’ to his teacher, the teacher corrected and modified the ‘guidelines’ labelled as: (The information was insufficient or incorrect)
(a) B, C, E           (b) B, D, E
(c) B, C, D           (d) C, D, E.
Answer:
(b)
Explanation:
It is the correct option.

Question 6:
The incident ray and the emergent ray in the glass slab are:
(a) always parallel               (b) converging always
(c) sometimes parallel        (d) sometimes diverging.
Answer:
(a)
Explanation:
These two rays do not meet.

Question 7:
The speed of light is maximum in:
(a) oil               (b) water
(c) glass           (d) air.
Answer:
(d)
Explanation:
The speed of light is higher in a rarer medium than a denser medium.

Question 8:
In refraction of light, the angle of incidence and angle of refraction is same when:
(a) ∠i = 90°            (b) ∠i = 180°
(c) ∠i = 0°               (d) none of these
Answer:
(c)
Explanation:
As per the formula and law.

Question 9:
When ray of light travels from air to glass slab its wavelength:
(a) increases
(b) no change
(c) decreases
(d) depends on glass slab thickness.
Answer:
(c)
Explanation:
Speed decreases hence wavelength decreases.

Question 10:
The bending of light when it passes from one medium to the other is called:
(a) dispersion          (b) diffraction
(c) refraction            (d) reflection.
Answer:
(c)
Explanation:
Refraction is changing the paths of light when it passes through different media.

Question 11:
In an experiment to trace the path of a ray of light passing through a rectangular glass slab, four students tabulated their observation as given below.

The student most likely to have done the experiment properly is
(a) A               (b) B              (c) C              (d) D
Answer:
(a)
Explanation:
We must not only have the angle of emergence
(nearly) equal to the angle of incidence but also have an idea of the magnitude of the angle of refraction (for a glass slab) for the three most often used values (30°, 45°, 60°) of the angle of incidence.

Questions based on Observational Skills
Question 12:
A student traces the path of a ray of light passing through a rectangular glass slab. For measuring the angle of incidence, he must position the protractor in the manner shown in the figure:

(a) A          (b) B            (c) C               (d) D
Answer:
(b)
Explanation:
The angle is measured between the normal ray and the incident ray.

Question 13:
In an experiment on tracing the path of a ray of light passing through a rectangular glass slab, the correct setting of the protractor (D) for measuring the angle of incidence ∠i and the angle of emergence ∠e corresponds, respectively to diagram is:

(a) K and M          (b) K and N
(c) L and M          (d) L and N.
Answer:
(a)
Explanation:
The angles need to be measured from the normal ray.

Question 14:
Four students traced the path of a ray of light from glass to air as shown below. The correct path of refracted ray is:

Answer:
(b)
Explanation:
Denser to rarer and away from the normal.

Question 15:
When light passes from rarer medium to denser medium, the light will bend:
(a) towards the normal
(b) away from normal
(c) depends on thickness of glass slab
(d) depends on the angle of incidence.
Answer:
(a)
Explanation:
(RDTN) The ray from rarer to denser medium bends towards the normal.

Question 16:
In these diagrams, the angle of refraction r is correctly marked in which diagram?

(a) I               (b) II                (c) III                (d) IV
Answer:
(d)
Explanation:
Angle r is between the normal and the refracted ray.

Question 17:
A student traces the path of a ray of white light through a rectangular glass slab and marks the angles of incidence (∠i), refraction (∠r) and emergence (∠e) as shown in the figure. Which angle or angles has he NOT marked correctly?
(a) ∠i only               (b) ∠i and ∠r
(c) ∠i and ∠e           (d) ∠r and ∠e
Answer:
(c)
Explanation:
Angle should be between the normal ray and the incident ray, the normal and the emergent rays.

Question 18:
Four students showed the following traces of the path of a ray of light passing through a rectangular glass slab.

The trace most likely to be correct is that of student:
(a) I                (b) II                (c) III                 (d) IV
Answer:
(c)
Explanation:
The shift of I and E rays is appropriate.

Question 19:
Select from the following the best set-up for tracing the path of a ray of light through a rectangular glass slab:

(a) I                (b) II                 (c) III                  (d) IV
Answer:
(a)
Explanation:
The angle of incidence and distance between two points is appropriate.

Questions based on Reporting and Interpretation Skills
Question 20:
In an experiment to trace the path of a ray of light passing through a rectangular glass slab, four students tabulated their observations as given below:

Students	A	B	C	D
∠i	30°	30°	30°	30°
∠r	18°	20°	17°	21.5°
∠e	32°	32.5°	30°	34.5°

Which student performed the experiment correctly?
(a) A             (b) B               (c) C                (d) D
Answer:
(c)
Explanation:
∠i and ∠e are same and angle r is less than angle i.

Question 21:
A student does the experiment on tracing the path of a ray of light through a rectangular glass slab for different angles of incidence. He can get a correct measure of the angle of incidence and the angle of emergence by following the labelling indicated in figure:

(a) A               (b) B                (c) C                  (d) D
Answer:
(d)
Explanation:
The angle is measured between the normal ray and the incident ray.

Question 22:
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
(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
Answer:
(a)
Explanation:
∠i and ∠e are same and angle r is less than angle i.

Question 23:
The two dots Pt and P2 shown in each of the following diagrams I, II, III and IV denote the position of two pins in respect of distance and direction for performing an experiment on tracking the path of a ray of light passing through a rectangular glass slab. In which one of the four cases, one is likely to get the best result?

(a) I             (b) II                 (c) III                 (d) IV
Answer:
(c)
Explanation:
Dots P and P, should be on incident ray making an angle of 30 to 60 degree with the normal.

Question 24:
Out of the four set-ups shown for carrying out the experiment to trace the path of a ray of light through a rectangular glass slab, the best set-up is

(a) A              (b) B                 (c) C                  (d) D
Answer:
(b)
Explanation:
Dots should be on incident ray making an angle of 30 to 60 degree with the normal.

Question 25:
In an experiment to trace the path of a ray of light passing through a rectangular glass slab, the correct measurement of angles of incidence (i), refraction (r) and emergence (e) is shown in the diagrams.

(a) A              (b) B                (c) C                (d) D.
Answer:
(a)
Explanation:
∠i = ∠e , ∠i > ∠r.

Question 26:
The path of a ray of light passing through a rectangular glass slab was traced and angles measured. 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:

(a) I            (b) II                 (c) III               (d) IV
Answer:
(d)
Explanation:
The angle is always formed between the normal and the ray (incident/refracted/emergent).

Question 27:
An experiment to trace the path of a ray of light through a glass was performed by four students A, B, C and D. They reported the following measurements of angle of incidence i, angle of refraction r and angle of emergence e.

Student	∠i	∠r	∠e
A	30°	30°	20°
B	40°	50°	40°
C	40°	30°	48°
D	40°	30°	40°

student performed the experiment correctly?
(a) A           (b) B             (c) C             (d) D
Answer:
(d)
Explanation:
angle i = angle e, angle i > angle r.

Question 28:
The correct path of a ray of light passing from air to kerosene oil and from kerosene oil to water is

(a) A               (b) B              (c) C               (d) D
Answer:
(a)
Explanation:
Ray from rarer to denser medium is towards the normal and from denser to rarer is away from the normal.

Question 29:
A ray of light enters air from water and experiences refraction, then
(a) ∠i = ∠r               (b) ∠i < ∠r
(c) ∠i > ∠r               (d) ∠i / ∠r = 0°.
Answer:
(b)
Explanation:
Light is passing from denser to rarer medium.

Question 30:
Four students A, B, C and D traced the paths of incident ray and the emergent ray by fixing pins P and Q for incident ray and pins R and S for emergent ray for a ray of light passing through a glass slab.

The correct emergent ray was traced by the student:
(a) A              (b) B              (c) C                (d) D
Answer:
(b)
Explanation:
The bending of light is correct (DRAN) and (RDTN).

Question 31:
A ray of light is incident normally, the angle of incidence is:
(a) 90°              (b) 60°             (c) 180°                 (d) 0°.
Answer:
(d)
Explanation:
No angle is made as the incident and the normal rays are straight and same lines.

Question 32:
A student traces the path of a ray of light through a rectangular glass slab for four different angles of incidence. He very cautiously measures the angle i, angle r and the angle e. On analyzing his measurements, he is likely to draw the following conclusion:
(a) ∠i = ∠e < ∠r                (b) ∠i  >  ∠r  > ∠e
(c) ∠i = ∠ r < ∠e               (d) ∠i = ∠e > ∠r
Answer:
(d)
Explanation:
Angle i = angle e, but angle of refraction is always less than both the other angles.

Question 33:
Study the following four experimental set-ups I, II, III and IV for the experiment, “To trace the path of a ray of light through a rectangular glass slab”.

Which of the marked set-ups is likely to give best results (P₁ and P₂ are the positions of pins fixed on the incident ray)?
(a) I              (b) II             (c) III              (d) IV
Answer:
(d)
Explanation:
The distance between the two pins should be more than 4-5 cm.

Question 34:
After tracing the path of a ray of light passing through a rectangular glass slab for four different values of the angle of incidence, a student reported his observations in tabular form as given below:

S.No.	∠i	∠r	∠e
I	30°	19°	29°
II	40°	28°	40°
III	50°	36°	50°
IV	36°	40°	59°

The best observation is:
(a) I               (b) II              (c) III              (d) IV
Answer:
(c)
Explanation:
∠i = ∠e , ∠i > ∠r.

More Resources CBSE Class 10 Lab Manual Practical Skills:

• CBSE Class 10 Science Reflection through Glass Slab Practical Skills
• CBSE Class 10 Maths Lab Manual
• CBSE Class 10 Math Labs with Activity

NCERT Class 10 Science Lab Manual 

• Determine Ohm’s Law Class 10 Experiment
• Determine Resistors in Series Class 10 Experiment
• Determine Resistors in Parallel Class 10 Experiment
• Determine Focal Length of Concave Mirror and Convex Lens Class 10 Experiment
• Determine Refraction Through Glass Slab Class 10 Experiment
• Determine Refraction Through Prism Class 10 Experiment
• Determine Image Formation by a Convex Lens Class 10 Experiment

Board	CBSE
Textbook	NCERT Books
Class	Class 10
Subject	Science (Physics)
Experiment Name	Ohm’s Law
Category	Class 10 Science Lab Manual

The experiment to determine Ohm’s Law are part of Science Lab Manual for Class 10 CBSE Experiments is designed to help students bridge the gap between theoretical concepts and practical applications through hands-on experiments in Physics.

Science Lab Manual Class 10 CBSE Ohm’s Law Experiment

Determine Ohm’s Law Class 10 Practical

Experiment To Verify Ohm’s Law Class 10 CBSE

Ohm’s Law Experiment Class 10 Pdf Introduction

• Charge: There are two charges in nature i.e., positive and negative. The negative charge is due to electron. Its value is 1.6 x 10^-19C. It is measured in coulombs.
• Coulomb: One coulomb is the amount of charge present on 6.25 x 10¹⁸ electrons.
• Electric Current: It is the rate of flow of charge through a conductor. If a net charge Q flows across any cross-section of conductor in time t, then the current I, through the cross-section is
  \(I=\frac { Q }{ t } \)
• The unit of current is ampere.
• One Ampere: One ampere is constituted by the flow of one coulomb of charge per second.
  \(IA=\frac { 1C }{ 1s }\)
• It is measured by a device called ammeter which is always connected in series in a circuit.
• Potential difference In an electric circuit carrying current, the work done to move a unit charge from one point to the other is called potential difference.
  \(V=\frac { W }{ Q } =\frac { Work\quad done }{ Charge } \)
• The SI unit of potential difference is volt (V).
• One Volt: When 1 joule of work is done to move a charge of 1 Coulomb from one point to the other then potential difference is of 1 volt.
  \(1Volt=\frac { 1 Joule }{ 1 Coulomb }\)
• It is measured by an instrument called the voltmeter. The voltmeter is always connected in parallel in a circuit.
• e.m.f.: Electro motive force, is the force which disturbs the equilibrium of free electrons flowing in the metal wire. The source of e.m.f. like cell or battery can develop a potential difference across the ends of the wire and the electrons can flow through the wire.
• Ammeter: The number of electrons flowing through a wire can be measured using ammeter. It is always connected in series in the circuit. The positive electrode of battery/cell is connected to the positive electrode of the ammeter and the negative end to the negative electrode of the battery.
• Cell: A cell is a device which produces potential difference in the wire and supplies the electrons to flow through the closed circuit. A primary cell produces 1.5 volts of potential difference.

Types of cells:

(a) Primary cell like dry cell, Lechlanche cell is used in torch, transistors etc. It produces 1.5 volts of p.d. (potential difference). For effective use should be used intermittently.
(b) Secondary cells can be recharged using a charger. The cell is connected to the charger and the electrons are stored in it which can be used later. Such cells are also called accumulators or storage cells.

• Battery: A combination of two or more cells is called battery. It is commonly used in cars, invertors.
  
• Conventional flow of current: In the circuit diagram as shown below, the flow of electric current is always shown from positive end to a negative end of the battery. But we know that electrons in the circuit flow from negative terminal to positive terminal.
• When the flow of current was studied it was assumed that the positive electricity is flowing from higher potential to lower potential. The electrons were discovered much later.
• Symbols used in electric circuit:
  
• Circuit diagram:
  
  
• Key: It is used to pass the current through circuit when it is closed. It is also used to stop the current through circuit when it is open.
• Ohm’s law: The current flowing through a metallic conductor held at constant temperature is directly proportional to the potential difference between the ends.
  V = IR
• Resistor: It is an electric wire which offers resistance to the flow of current through it. Its unit is Ohm (Ω).
• Resistance: When the electrons flow through a wire, they collide with the atoms of the wire, due to this collision the speed of electrons flowing gets disturbed and they also lose the energy in the form of heat energy. This obstruction for the flow of electrons is called resistance.
• Each and every wire that may be a very good conductor of electricity will certainly offer some resistance to the flow of electrons.
• Factors affecting resistance:
  (a) If the wire is long then the collision of electrons flowing through the wire will be more and hence it will offer more resistance. But if the wire is thick the collisions would be less and the resistance offer would be less. Hence, the resistance of the wire depends on the thickness, length material and the temperature of the wire.
  (b) If the wire is made up of same material and is thick the resistance will be less as compared the long wire of
  the same material at the constant temperature.
  (c) This law is not valid for semiconductors like diode, thermistor, diode, filament of lamp, light dependent resistor, LED etc. Therefore, all semiconductors are called non-ohmic materials.
• Validity of Ohm’s Law: Ohm’s law is valid only under the condition when temperature is kept constant. This is reasonable because when temperature is constant only then resistance will be constant.
• Galvanometer: It is an instrument used to measure very less current.
• Ammeter: It is an instrument used for measuring the magnitude of current flowing through a circuit. It is always connected in series, it offers very low resistance. It’s unit is ampere (A).
• Voltmeter: It is an instrument used for measuring the potential difference between any two points of a given conductor. It is always connected in parallel in the circuit. The unit is volts (V).
• Rheostat: A component used to regulate the electric current flowing through a circuit without changing the voltage is called Rheostat. It has variable resistance. It is used to change the resistance in the circuit.

Science Lab Manual Class 10 Experiment – 1

Aim
To study the dependence of potential difference (V) across a resistor on the current (I) passing through it and determine its resistance. Also plot a graph between V and I.
Theory

• Ohm’s Law: The potential difference, V across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature is the same. This is Ohm’s law.
  V ∝ I
  .’.    V = IR, (Here R = Constant for the given metallic wire)
• The SI unit of resistance is Ohm (Ω).
  \(R=\frac { V }{ I }\)
• One Ohm: If the potential difference across the ends of a conductor is 1 volt and the current flowing through it is 1 ampere, then the resistance of the conductor R is 1 ohm.
  
• Factors affecting resistance:
• The nature of resistor (a conductor having some resistance.)
• The length of the resistance. (R ∝ l)
  (Resistance increases as the length of the wire is increased)
• The area of cross-section of the resistor. \(R\propto \frac { 1 }{ A } \)
  (Resistance decreases with the increase in the cross-section area of the wire)
• Circuit Diagram:
  
• In a circuit ammeter is always connected in series and voltmeter is connected in parallel across the points between which potential difference is to be measured.
• A straight line graph obtained between V and I verifies the Ohm’s law.
• Least Count: It is very important to find the least count of ammeter and voltmeter before using them.

If in the ammeter, there are 10 divisions from 0 to 0.1 A then each division indicates 0.01 A.
A.  To calculate the least count of ammeter.
Range of ammeter = A_{R…………………….}
Number of divisions in ammeter = A_{N…………………………..}
.’. Least count of ammeter = \(\frac { { A }_{ R } }{ { A }_{ N } }\)= …………….. ampere.
B. To calculate the least count of voltmeter.
Range of voltmeter = V_R……………………
Number of divisions in voltmeter = V_N
.’. Least count of voltmeter = \(\frac { { V }_{ R } }{ { V }_{ N } }\) =  ………………. volt.

Materials Required
A battery, an insulated copper wire (cut into 10 pieces), a key, an ammeter, a voltmeter, a rheostat, a resistor and a piece of sand paper.
Procedure

1. Keep the devices as shown in the circuit diagram.
2. Connect them with the connecting wires and keep the key open.
3. Positive terminal of the battery is connected to the positive terminal of the ammeter.
4. Check the +ve and -ve terminals of voltmeter before connecting it in the circuit.
5. Once the circuit is connected, insert the key and check the rheostat, adjust its slider and see whether the ammeter and voltmeter readings are shown.
6. By using the slider of rheostat take three different readings of current 1 and voltmeter V.
7. Record your observations in the observation table.
8. Calculate resistance of a given resistor by formula \(R=\frac { V }{ I }\).
9. Plot a graph of voltmeter reading and current reading. On x axis take V and on y axis take I.
10. Resistance increases with increase in temperature of pure metals.

Observation Table
A. Least count of ammeter and voltmeter

S. No.		Ammeter (A)	Voltmeter (V)
1.	Range	0 – 0.5 A	0-0.1 V
2.	Least Count	0.01 A	0.01 V
3.	Zero Error (e)	0	0
4.	Zero Correction	0	0

B. For reading of ammeter and voltmeter

S. No.	Current in Ampere (I) (Ammeter Reading)		Potential difference in Volts (V) (Voltmeter Reading)		Resistance in Ohms R = V/I(Ω)
	Observed	Corrected	Observed	Corrected
1.	0	0.02	0	0.04	R1 =2 Ω
2.	0	0.03	0	0.06	R2 = 2 Ω
3.	0	0.04	0	0.08	R3 = 2Ω

Conclusions

1. The value of R is found to be same and constant in all three readings.
2. The resistance of a resistor is ratio of potential difference V and current I.
3. The graph of V and I is a straight line. This shows that V∝I. This verifies Ohm’s law.

Precautions

1. The connecting wires should be thick copper wires and the insulation of their ends should be removed using the sand paper.
2. Connections should be tight otherwise some external resistance may introduce in the circuit.
3. Connections should be made as per the circuit. Before closing the circuit show the connections to the teacher to take the readings.
4. The ammeter should be connected in series with the resister such that the current enters at the positive terminal and leaves at the negative terminal of the ammeter.
5. Voltmeter should always be connected in parallel to resistor.
6. Calculate the least count of voltmeter and ammeter correctly.
7. The pointers of the ammeter and voltmeter should be at zero mark when no current flows through the circuit.
8. Current should be passed through the circuit for a short time while taking observations; otherwise current would cause unnecessary heating in the circuit. Heating may change the resistance of resisters.

Ohms Law Practical Class 10 Viva Voce

Question 1:
Define electric current.
Аnswer:
The number of charges flowing through a given wire in unit time is called electric current.
OR
The rate of flow of charge in a conductor is called electric current.
\(I=\frac { Q }{ t }\)

Question 2:
What is the value of charge in 1 electron?
Аnswer:
1 electron = 1.6 x 10^-19C

Question 3:
What is coulomb?
Аnswer:
One coulomb is the amount of charge present on 6.25 x 10¹⁸ electrons.

Question 4:
Define potential difference.
Аnswer:
The work done in moving a unit charge from one point to the other is called potential difference.

Question 5:
Define 1 volt.
Аnswer:
When 1 joule of work is done to move a charge of 1 coulomb from one point to the other, then potential difference
is of 1 volt.

Question 6:
What is resistance? Give its SI unit.
Аnswer:
It is the property of a conductor to resist the flow of charges through it.
Its SI units is ohm(Ω). \(R=\frac { V }{ I }\)

Question 7:
Define 1 ohm.
Аnswer:
If the potential difference across the ends of a conductor is 1 volt and the current flowing through it is 1 ampere, then resistance of the conductor is 1 ohm.

Question 8:
What are the factors that affect resistance?
Аnswer:

1. The nature of resistor, i.e., material of a conductor.
2. The length of the resistor (R∝ l).
3. The area of cross-section of the resistor\(\left( R\propto \frac { 1 }{ A } \right)\)
4. Temperature of wire.

Question 9:
What is Ohm’s law?
Аnswer:
The potential difference V across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains same.

Question 10:
How many electrons are present in 1 coulomb?
Аnswer:
In 1 coulomb 6.25 x 10¹⁸ electrons are present.

CBSE Class 10 Science Lab Manual Practical Based Questions

Question 1:
What is the unit of current and how do we measure current flowing through a wire?
Аnswer:
The unit of current is ampere. It is measured by the device called ammeter.

Question 2:
What is the unit of potential difference and how do we measure potential difference?
Аnswer:
Its unit is volt and is measured by a device called voltmeter.

Question 3:
In an electric circuit containing resistance, ammeter, key and battery, where will you connect voltmeter to verify Ohm’s law?
Аnswer:
Voltmeter will be connected parallel to the resistance.

Question 4:
If the length of a given resistor is increased, what will happen to the overall resistance?
Аnswer:
On increasing the length of the resistor its resistance increases.

Question 5:
To make electric heater what type of wire should be used?
Аnswer:
To make electric heater the wire used should be offering high resistance.

Question 6:
What is the nature of graph obtained for V and I?
Аnswer:
The graph is a straight line.

Question 7:
What does the straight line of a graph indicate?
Аnswer:
The straight line of the graph indicates that current I is directly proportional to voltage V.

Question 8:
What is the formula used to calculate the resistivity of a given wire?
Аnswer:
To calculate resistivity

Question 9:
What is meant by least count of an instrument?
Аnswer:
The least value that an instrument can measure is called its least count. It should be non-zero number.

Science lab manual class 10 NCERT Lab Manual Questions

Question 1:
In this experiment it is advised to take out the key from the plug when the observations are not being taken. Why?
Аnswer:
This helps in taking accurate readings. The unnecessary current flows through the circuit causes the heating effect and changes the resistance.

Question 2:
Suppose the ammeter (or voltmeter) you are using in this experiment do not have positive (+) and negative (-) terminal markings. How will you use such ammeter (or voltmeter) in the circuit?
Аnswer:
We need to identify the positive and negative terminal of the device by connecting it to the battery. Connect the ammeter in series and check for the deflection and connect voltmeter in parallel to the resistor in the circuit and check for the deflection.

Question 3:
If the resistor of a known resistance value is replaced with a nichrome wire of 10 cm length (say). How do the values of current through the nichrome wire and potential difference across the two ends of it may change? How the values will change if the replaced wire is of manganin in place of nichrome?
Аnswer:
The resistance of nichrome is more as compared to manganin. If the known resistor is replaced by the nichrome wire than the current will decrease and the potential difference will decrease. On replacing nichrome wire with manganin the current will increase and the potential difference will increase.

Question 4:
Suppose in this experiment you see that the deflection on ammeter (or voltmeter) scale goes beyond the full scale. What will you infer from such an observation? What will you infer if the deflection takes place in opposite direction?
Аnswer:
In the experiment if the deflection on ammeter or voltmeter scale goes beyond the full scale than the device needs to be replaced with the one which can measure higher current and voltage.
If the deflection takes place in opposite direction than the connections need to be checked and the terminals need to be interchanged.

Question 5:
Why is it advised to clean the ends of connecting wires before connecting them?
Аnswer:
The ends of the wire may get corroded or some impurities may be deposited on it hence to remove the same and get correct readings the ends of the wire should be cleaned.

Science lab activities for class 10 CBSE Multiple Choice Questions (MCQs)

Questions based on Procedural and Manipulative Skills
Question 1:
The instrument used to measure electric current is
(a) voltmeter
(b) ammeter
(c) resistor
(d) none of these

Question 2:
The instrument used to measure the potential difference is
(a) voltmeter
(b) ammeter
(c) rheostat
(d) galvanometer

Question 3:
The unit of electric current is
(a) volt
(b) ampere
(c) ohm
(d) joule

Question 4:
The SI unit of resistance of a wire is
(a) volt
(b) ampere
(c) ohm
(d) joule

Question 5:
The unit of charge is
(a) volt
(b) ampere
(c) joule
(d) coulomb

Question 6:
The resistance of a wire depends on
(a) material of wire
(b) length of wire
(c) cross-sectional area of wire
(d) all of these

Question 7:
According to Ohm’s law, the relationship between V, I and R is

Question 8:
An ammeter has 20 divisions between mark 0 and mark 2 on its scale. The least count of the ammeter is
(a) 0.02 A
(b) 0.01 A
(c) 0.2 A
(d) 0.1 A

Question 9:
In a voltmeter there are 20 divisions between the 0 mark and 0.5 V mark. The least count of the voltmeter is
(a) 0.020 V
(b) 0.025 V
(c) 0.050 V
(d) 0.250 V.

Question 10:
An ammeter has a range of (0-3) ampere and there are 30 divisions on its scale. What is the least count?
(a) 10
(b) 27
(c) 0.1
(d) 0.01

Question 11:
The resistance of an alloy
(a) increases with temperature
(b) decreases with temperature
(c) is constant with rise in temperature
(d) is zero.

Question 12:
What will happen to current passing through a conductor if potential difference across it is doubled and the resistance is halved?
(a) remains unchanged
(b) becomes double
(c) becomes halved
(d) becomes four times

Question 13:
For the experiment “to find the equivalent resistance of the two given resistors connected in parallel” the following circuit was drawn by a student.

The teacher pointed out the possibility of the following faults:
A. the ammeter was not correctly connected in the circuit
B. the voltmeter was not correctly connected in the circuit
C. the resistors IT and R2 were not correctly connected in parallel.
D. the rheostat and the key were not correctly connected in the circuit
The two faults pointed out correctly by the teacher, are
(a) A and B
(b) B and C
(c) C and D
(d) D and A

Question 14:
A voltmeter should have:
(a) high resistance
(b) low resistance
(c) moderate resistance
(d) variable resistance.

Question 15:
The ammeter connected in a circuit reads 0.01 A when battery is switched off. It means there is:
(a) wrong connections
(b) zero error
(c) positive error
(d) negative error.

Question 16:
In an electric circuit the key should be kept off to avoid:
(a) damage of instrument
(b) damage of resistor
(c) incorrect readings
(d) none of these.

Question 17:
In Ohm’s circuit which of the following does not have © and © terminals?
(a) Voltmeter
(b) Ammeter
(c) Battery
(d) Resistor

Question 18:
The resistance of a wire depends on:
(a) material of the wire
(b) length of the wire
(c) temperature of the wire
(d) all of these

Questions based on Observational Skills
Question 19:
The current flowing through a resistor connected in an electric circuit and potential difference developed across its ends are shown in the diagram.
Find the value of the resistance of the resistor is

Question 20:
The current flowing through a resistor connected in a circuit and the potential difference developed across its ends are as shown in the diagram. The approximate value of the resistor is:

Question 21:
The current flowing through a resistor connected in an electrical circuit and the potential difference developed across its ends are shown in the given diagram.

The value of resistance of the resistor in Ohm is
(a) 25 (b) 20 (c) 15 (d) 10

Question 22:
The current flowing through a conductor and the potential difference across its two ends are as per reading of the ammeter and the voltmeter shown below. The resistance of the conductor would be:

Question 23:
For the circuit diagram shown below, the student would observe

(a) same reading in both the ammeter and the voltmeter
(b) no reading in either the ammeter or the voltmeter
(c) some reading in the ammeter but no reading in the voltmeter.
(d) some reading in the voltmeter but no reading in ammeter.

Question 24:
The following circuit diagram shows the experimental set-up for the study of dependence of current on potential-difference. Which two components are connected in the series?

(a) battery and Voltmeter
(b) ammeter and voltmeter
(c) ammeter and rheostat
(d) resistor and voltmeter

Question 25:
A student arranged an electric circuit as shown below:

He would observe
(a) no reading in either the ammeter or the voltmeter.
(b) no reading in the voltmeter but a finite reading in the ammeter.
(c) no reading in the ammeter but a finite reading in the voltmeter.
(d) a finite reading in both the ammeter and the voltmeter.

Question 26:
To verify Ohm’s law the given circuit diagram was drawn by a student. What does X, Y and Z in the circuit stand for respectively?

(a) Ammeter, Voltmeter and Resistance
(b) Voltmeter, Ammeter and Resistance
(c) Ammeter, Voltmeter and Rheostat
(d) Voltmeter, Ammeter and Rheostat.

Question 27:
In the given circuit diagram, the components connected in series are:

(a) battery and ammeter
(b) ammeter and resistor
(c) ammeter and rheostat
(d) all of the above.

Question 28:
In the above circuit diagram, the components connected in parallel are:
(a) battery and ammeter
(b) resistor and voltmeter
(c) rheostat and ammeter
(d) ammeter and voltmeter.

Questions based on Reporting and Interpretation Skills
Question 29:
A voltmeter has a least count of 0.05 volt. While performing Ohm’s law experiment a student observed that the pointer of the voltmeter coincides with 15th division. The observed reading is:
(a) 0.75 V
(b) 0.075 V
(c) 7.5 V
(d) 75 V

Question 30:
The only correct statement for the following electric circuit is

(a) The voltmeter has been correctly connected in the circuit.
(b) The ammeter has been correctly connected in the circuit.
(c) The resistors R₁ and R₂ have been correctly connected in series.
(d) The resistors R₁ and R₂ have been correctly connected in parallel.

Question 31:
The correct set-up for studying the dependence of the current on the potential difference across a resistor is

Question 32:
To study the dependence of the current (I) on the potential difference (V), across a resistor, two students used two set-ups shown in figures (A) and (B) respectively. They kept the contact J in four different positions, marked (a), (b), (c) and (d) in the two figures.
For the two students, the value of the emf used by student (A) and the resistance due to the rheostat

used by student (B), will each be minimum when the contact J is in the position.
(a) (d) in both the set-ups.
(b) (a) in both the set-ups.
(c) (d) in set-up (A) and (a) in set-up (B).
(d) (a) in set-up (A) and (d) in set-up (B).

Question 33:
In the circuit given below the voltmeter and ammeter readings are respectively

(a) 2 V and 2 A
(b) 1 V and 2 A
(c) 2 V and 1 A
(d) 1 V and 1 A.

Question 34:
Which of the following set-up is correct for the verification of Ohm’s law.

Question 35:
Identify the circuit in which the electrical components have been properly connected.

Question 36:
The plot correctly showing the dependence of the current 1 on the potential difference V across a resistor R is

Question 37:
The graph of V-I is a straight line. The slope of this straight line graph gives:
(a) potential difference
(b) power
(c) resistance
(d) rheostat.

Question 38:
The best graph plotted by a student for Ohm’s experiment is:

Question 39:
The given graph, is plotted for V-I to verify Ohm’s law.
The resistance of the conductor used in the experiment is:
(a) 1 Ω
(b) 1.5 Ω
(c) 3 Ω
(d) 2 Ω

Question 40:
A student wanted to make a battery of 6 V of cells with e.m.f 1.5 V each. The correct arrangement is:

Question 41:
For the circuits shown in figures I and II, the ammeter readings would be:

(a) 0 A in circuit I and 1 A in circuit II
(b) 0 A in both the cases
(c) 1 A in both the cases
(d) 1 A in circuit I and 1 A in circuit II.

Question 42:
For the circuits shown in figures I and II given below, the ammeter reading is 1A so the voltmeter reading would be:

(a) 0 V in both the circuits
(b) 2 V in both the circuits
(c) 2 V in circuit I and 0 V in circuit II
(d) 0 V in circuit I and 2 V in circuit II.

Question 43:
In an experiment to study dependence of current on the potential difference across a given resistor, four students P, Q, R and S kept the plug key in the circuit closed for time t{ and then open for time t2 as given in the table below:

Students	Closed time t1 seconds	Open time t2 seconds
P	30	60
Q	60	30
R	60	15
S	45	15

The best choice of open and closed time is that of student
(a) P (b) Q (c) R (d) S

CBSE Class 10 Science Practicals Lab Manual MCQ Аnswers:

Physics Lab Manual CBSE Class 10 Scoring Key With Explanation

1. (b) Ammeter measures current.
2. (a) Potential difference is measured by voltmeter. S.I. unit of current is ampere.
3. (b) S.I. unit of resistance is ohm.
4. (c) S.I. unit of charge is coulomb.
5. (d) S.I. unit of charge is coulomb.
6. (d) R depends on all the given factors.
7. (b) Ohm’s Law.
8. (d) Least count is range divided by number of divisions.
9. (b) Least count is range divided by number of divisions.
10. (c) Least count is range divided by number of divisions.
11. (a) R is proportional to temperature.
12. (d) As per Ohm’s law of formula calculation.
13. (b) The -ve of voltmeter should be connected to +ve of ammeter and R( and R, should be connected in parallel.
14. (a) The voltmeter should have very high resistance so that it doesn’t allow current to flow through it and thus change the currents in the rest of the circuit.
15. (c) The reading is shown on switched off circuit.
16. (c) The ON switch will interfere in correct readings as some current will keep flowing.
17. (d) Resistor is just a piece of wire.
18. (d) Resistance depends on all the given factors.
19. 
20. 
21. 
22. 
23. (c) Current will not flow through voltmeter as the key is open.
24. (c) They are in series.
25. (b) Voltmeter has open key.
26. (b) In series is ammeter, in parallel is voltmeter and Z is symbol of resistance.
27. (d) Through series, the current flows the same.
28. (b) Resistor and voltmeter are the two components connected in parallel.
29. (a) Reading = Least count x Division of reading.
30. (b) It is the correct statement for the circuit.
31. (a) Voltmeter is connected in parallel while ammeter is connected in series.
32. (b) In (A) as only cell is in circuit, therefore, voltmeter reads minimum emf. In (B) as, the initial point at rheostat is in circuit therefore, minimum resistance is in circuit.
33. (c) Current (I) =V/R= 1 A and potential difference (V) = RI = 2V.
34. (b) Circuit B can help in verification of ohm’s law.
35. (b) Ammeter is in series and voltmeter is in parallel.
36. (a) I is directly proportional to V. Hence, straight line graph.
37. (c) As per ohm’s law, V = IR.
38. (c) Straight line is obtained as V and I are directly proportional to each other.
39. (d) It is the slope value of the graph.
40. (b) 1.5 V x 4 = 6 V
41. (d) In circuit I, the current 5V/5 ohm = 1 A.
    In circuit II, the key is open.
    Hence, no current will flow in circuit II.
42. (d) In circuit I, the key is open, so 0V. In circuit II, the current flows and hence 2V.
43. (a) We must keep the circuit closed for a relatively shorter time and open for a relatively longer time.

More Resources CBSE Class 10 Lab Manual Practical Skills:

• CBSE Class 10 Science Practical Skills
• CBSE Class 10 Maths Lab Manual
• CBSE Class 10 Math Labs with Activity

NCERT Class 10 Science Lab Manual 

• Determine Ohm’s Law Class 10 Experiment
• Determine Resistors in Series Class 10 Experiment
• Determine Resistors in Parallel Class 10 Experiment
• Determine Focal Length of Concave Mirror and Convex Lens Class 10 Experiment
• Determine Refraction Through Glass Slab Class 10 Experiment
• Determine Refraction Through Prism Class 10 Experiment
• Determine Image Formation by a Convex Lens Class 10 Experiment

CBSE Class 10 Science Lab Manual – Image Formation By Convex Lens Experiment

Board	CBSE
Textbook	NCERT Books
Class	Class 10
Subject	Science
Experiment Name	Image Formation by a Convex Lens
Category	Class 10 Science Lab Manual

The experiment to determine Image Formation by a Convex Lens are part of Science Lab Manual for Class 10 CBSE Experiments is designed to help students bridge the gap between theoretical concepts and practical applications through hands-on experiments in Physics.

Science Lab Manual Class 10 CBSE Image Formation by a Convex Lens Experiment

Determine Image Formation by a Convex Lens Class 10 Practical

Convex Lens Practical Class 10 Introduction

• Lens: A transparent material bound by two surfaces, of which one or both surfaces are spherical, forms a lens.
  Convex lens: Convex lens is a transparent curved device that is used to refract light. A lens is usually made from glass. There are two different shapes for lenses. They are called convex and concave lens. A convex lens is thicker in the middle and thinner at the edges, it is also called a converging lens. A convex lens will focus light and make a real and inverted image. The size of the image will depend on the position of the object and lens.
• Concave lens: A concave lens is thinner in the middle and thicker at the edges. A concave lens is also called a diverging lens. A concave lens will disperse light and make an image that is always virtual, upright and smaller than the object.
• Principal axis: The principal axis is a line that is perpendicular to and passes through the centre of the lens.
• Principal focus: The principal focus, F is the point through which all incident rays travelling parallel to the principal axis after refraction appear to meet.
• The focal length “ƒ’ is the distance from the center of the lens, O to the principal focus, F.
• Centre of curvature of lens: A convex lens is made up of two spherical surfaces. These surfaces are the parts of a sphere. The centre of these spheres are called centres of curvature of the lens. It is represented by
  

• Optical centre: The central point of a lens is its optical centre (O).
• Aperture of lens: The diameter of the circular outline of a spherical lens is called its aperture. Lenses with thin apertures are called thin lenses.
• Real image: When the light rays after refraction actually meet at a point to form the image is said to be real image, it is always inverted and can be obtained on the screen.
• Virtual image: When the light rays after i-efraction through a lens do not actually meet but appears to meet, then the image formed is virtual. It cannot be obtained on the screen and is always erect.
• Magnification of image: When an image is formed after refraction, then its size may be same, enlarged or diminished. The ratio of size of image to size of object is called magnification.
• Power of a lens: The power of a lens is the reciprocal of the focal length (ƒin metres). The S.I. unit of power is dioptre (D).
• Ray diagram through the convex lens:
  
  Concave Lens Ray Diagram Class 10

CBSE Class 10 Science Lab Manual Experiment 7

Convex Lens Experiment Class 10

Aim
To find the image distance for varying object distances in case of a convex lens and draw corresponding ray diagrams to show the nature of image formed.
Theory

• Lens: It is a piece of a transparent medium. It may be curved at both the surfaces or may be plane at one surface and curved at the other surface. Two main types of lenses are: (a) convex lens or converging lens and (b) concave lens or diverging lens.
• Convex lens and its types: It is also called converging lens because it converges a beam of light incident on it. Double convex lens is bulging in the centre, i.e., it is thicker in the middle and thinner at the edges.
  

• Convex lens: It is a converging lens with real focus. These lenses are thick in the middle and thin at the edges.
  

•  The distance of object from the centre of lens is represented by letter ‘u’. The distance of image from the centre of lens is represented by letter V.
• Lens formula: This formula gives the relationship between object distance (u), image-distance (v) and the focal length (ƒ). The lens formula is expressed as
  \(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u }\)
• New Cartesian sign conventions for u, v and ƒ
  

1. The object is always placed to the left of the lens.
2. Incident rays are drawn from left towards the right of the lens.
3. All distances are measured from the optical centre as origin.
4. Distances measured in the direction of the incident rays are taken as positive.
5. Distances measured opposite to the direction of the incident rays are taken as negative.
6. Distances measured perpendicular to and above the principal axis are taken as positive.
7. Distances measured perpendicular to and below the principal axis are taken a negative.
8. Focal length of a convex lens is taken as positive while that of concave lens is taken as negative.

•  For the sake of clarity of ray diagrams, only two rays are considered.
• The intersection of atleast two refracted rays gives the position of image of the point object. Any two of the following rays can be considered for locating the image:
  

Materials Required
A convex lens of a short focal length (12-20 cm), measuring scale, optical bench and a needle or a candle.

Procedure

1. Fix a thin convex lens on a lens holder and place the screen on the other side of the lens.
2. Focus a sharp, clear and inverted image of the distant object on the screen. This is the rough focal length, measure it with the help of a metre scale.
3. Mark the position of lens on optical bench or on a table. Fix the lens at this point, label it as ‘O’.
4. Mark a point ‘F’ at both the sides of the lens as focus of the lens by knowing the focal length as calculated in first step.
5. Mark a point 2F at both the sides of the lens, the distance of 2F from the lens is double the focal length of the lens.
6. Place a candle on the table or needle on optical bench at distance beyond 2F and adjust the height of the centre of lens nearly equal to the height of the flame of the candle.
7. To locate a sharp image of the candle flame in the convex lens from the other side of the lens, adjust the position of the screen and record your observations.
8. Now, place the object, e., the lighted candle or the needle at 2F and record your observations.
9. Now, shift the object between F and 2F and record the observations.
10. Now, place the object at F and record the observations.
11. Place the object between O and F of the lens and record your observations.
12. Draw ray diagrams for all the positions of the object.
    

Observation Table

Mathematical Calculation of ƒ:

> Hence, the focal length of the given lens is 10 cm.

Ray Diagrams

Result
The focal length of the given lens is 10 cm.

Precautions

1. The focal length of the convex lens must be between 15 to 20 cm.
2. Use thin convex lens of small aperture.
3. Perform this experiment in calm air to avoid the flickering of the candle flame.
4. To obtain distinct and sharp image of the candle flame, perform this experiment in a dark room.
5. The optical bench or the bench holding the lens, object and image screen should not be shaky.

CBSE Class 10 Science Lab Manual Viva Voce 

Question 1:
What is a lens?
Answer:
It is a transparent medium bounded by two surfaces, one or both may be curved.

Question 2:
How many types of convex lens do you know?
Answer:
The three types of lens are double convex lens, plano-convex and concavo-convex lens.

Question 3:
What is meant by principal axis of a lens?
Answer:
A straight ray of light passing through the centre of a lens and centre of curvatures is called principal axis

Question 4:
If half of the convex lens is covered by a paper, will you get an image?
Answer:
Yes, the image will be obtained but little blurred.

Question 5:
Define focal length of a lens.
Answer:
It is the distance between the optical centre and its principal focus. Its unit is metre.

Question 6:
Define power of a lens.
Answer:
Power of lens is defined as the reciprocal of its focal length in metre.
\(p=\frac { 1 }{ f }\)(in metres)
Unit of power of a lens is dioptre.

Question 7:
Define one dioptre.
Answer:
One dioptre is the power of a lens when focal length is 1 metre.

Question 8:
Give one use of concave lens.
Answer:
Concave lens is used in spectacles.

Question 9:
Which lens is used to correct myopia, i.e., short sightedness?
Answer:
A concave lens.

Question 10:
Which lens is used to correct hypermetropia, i.e., long sightedness?
Answer:
A convex lens.

CBSE Class 10 Science Lab Manual Practical Based Questions

Question 1:
How is convex lens different from concave lens?
Answer:
Convex lens is thicker in the middle and thinner at the edges. Concave lens is thinner in the middle than at the edges

Question 2:
Without touching the lenses how can you distinguish between the convex and concave lens.
Answer:
If the image formed by the given lens is inverted then it is convex lens otherwise it is a concave lens.

Question 3:
Which lens is called converging lens and why?
Answer:
Convex lens is called converging lens because it converges the parallel beam of light passing through it at a point.

Question 4:
Which lens is called diverging lens and why?
Answer:
Concave lens is called diverging lens because it diverges all the rays incident on it.

Question 5:
What do you mean by optical centre of a lens?
Answer:
A point in the centre of the lens lying on the principal axis is said to be optical centre. A ray of light passing through optical centre does not deviate.

Question 6:
Define principal focus of convex lens.
Answer:
A beam of light parallel to the principal axis of convex lens passes through the lens and converges at a point called principal focus.

Question 7:
Give uses of convex lens.
Answer:
Convex lens is used in spectacles, telescopes and microscopes (simple and compound).

Question 8:
What is the nature of image formed by convex lens when object is placed at 2F.
Answer:
The image formed is inverted, real and of same size as that of object, obtained on the screen at 2F.

Question 9:
What is the nature of image formed by convex lens if the object is placed beyond 2F?
Answer:
The image formed is real, inverted and diminished, between F and 2F.

Question 10:
What is magnification of image?
Answer:
The ratio of height of image to height of object is said to be the magnification. It is given by m =\(\frac { -v }{ u } =\frac { h’ }{ h }\).

Question 11:
What type of image is formed by convex lens?
Answer:
Convex lens forms real image. It can be magnified or diminished.

Question 12:
How can you use convex lens as a magnifying glass?
Answer:
When the object is placed between focus of the lens and the aperture of the lens then the image formed will be magnified and lens can be used as magnifying lens.

Question 13:
What type of image is formed by concave lens?
Answer:
A concave lens always form virtual, erect and diminished image.

Question 14:
If an object is placed at infinity where is the image formed in case of concave lens?
Answer:
If the object is placed at infinity the image is formed at the focus, virtual, erect and diminished.

Question 15:
The power of the lens is -ID. What is the nature of the lens?
Answer:
If the power of the lens is negative then it is concave lens.

Question 16:
The power of the lens is +1.5D. What is the nature of the lens?
Answer:
When the power of the lens is positive, then it is a convex lens.

Question 17:
Where will you keep an object to obtain real and enlarged image?
Answer:
To obtain real and enlarged image we will use convex lens and place the object between F and 2F.

NCERT Class 10 Science Lab Manual Questions 

Question 1:
What is the nature of an image formed by a thin convex lens for a distant object? What change do you expect if the lens were rather thick?
Answer:
The image formed by thin convex lens for distant object is real, inverted, highly diminished and at the focus.
If the convex lens is thick the image formed will be real, inverted, highly diminished and the focal length will be smaller as compared to the thin lens.

Question 2:
You are provided with two convex lenses of same aperture and different thickness. Which one of them will be of shorter focal length?
Answer:
Thick convex lens will have shorter focal length.

Question 3:
If we cover one half of the convex lens while focusing a distant object, in what way will it affect the image formed?
Answer:
The image formed will be blurred and less clear but complete image will be formed of the object.

Question 4:
Which type of lens is used by the watch-makers while repairing fine parts of a wrist watch?
Answer:
Convex lens.

Question 5:
Sometimes, the image formed by a convex lens, of an object placed at 2F₁ is not of the same size and at location 2F₂ on the other side of the convex lens. What could be the possible reason(s) for such a situation?
Answer:
The convex lens generally produces the image of the object placed at 2F₁ of same size at 2F₂ as that of the object. But this is true only for thin convex lenses with small apertures. For thick convex lens with larger aperture does not show the image of same size.

Question 6:
A ray of light is passing through the principal focus of a convex lens. How will it emerge after refraction through the lens?
Answer:
A ray of light passing through the principal focus of a convex lens will be parallel to the principal axis after refraction through the lens.

Question 7:
An object is placed on the left side of a lens (having 10 cm focal length) at a distance of 20 cm. What will be the sign of object distance?
Answer:
The distance of the object is measured against the direction of incident ray of light and hence will be negative.

Question 8:
How will you distinguish between a convex lens and a concave lens by holding in hand and looking the printed page through them?
Answer:
If the print on the page is magnified than the lens is convex and if the print appears to be diminished then the lens is concave.

Question 9:
In what way will image of the lighted candle be affected when the experiment is performed in a bright light area and on a windy day?
Answer:
In a bright light area the clarity of the image will be less and on the windy day the image will be shaky.

Question 10:
A distinct image of the lighted candle has been obtained on screen with fixed position, using a thin convex lens. Why does the image of the candle get blurred if the position of any one of them is slightly disturbed?
Answer:
The sharp and clear image of the object is formed only when all the rays after refraction meet sharply at one place and if the position of the object or screen is changed then all the refracted rays don’t meet on the screen and hence the image formed is not clear, sharp and is blurred.

Question 11:
What effect do you expect if the lens is thick?
Answer:
The focal length of the lens will change thereby affecting the position of the image formed.

Question 12:
Why do we require a calm atmosphere to perform this experiment?
Answer:
The calm atmosphere allows the accuracy in measuring the distances, focal length and position of the image formed.

Question 13:
Why is it preferred to perform this experiment in dark or in shade?
Answer:
To obtain the clear and sharp image of the candle flame.

CBSE Class 10 Science Lab Manual Multiple Choice Questions ( MCQ’s ) 

Questions based on Procedural and Manipulative Skills 

1. The part of the lens through which the ray of light passes without suffering any deviation is
(a) focus
(b) centre of curvature
(c) optical centre
(d) pole

2. Double convex lens is used in
(a) spectacles
(b) microscope
(c) telescope
(d) all of these

3. The unit of power of a lens is
(a) metre
(b) dioptre
(c) centimetre
(d) none of these

4. The lens formula is:

5. A lens that always produces virtual image is
(a) convex lens
(b) concave lens
(c) double convex lens
(d) concavo-convex lens

6. The lens used by watch maker is:
(a) concave lens
(b) convex lens
(c) both (a) and (b)
(d) none of these

7. The imaginary straight line passing through the centre of the lens is called:
(a) optical centre
(b) pole
(c) principal axis
(d) principal focus

8. The power of the lens is -2.5 D. The nature of the lens is:
(a) concave
(b) convex
(c) concavo-convex
(d) convexo-concave

9. The diameter of the convex lens is called:
(a) curvature
(b) aperture
(c) focal point
(d) focal length

Questions based on Observational Skills

10. Convex lens always gives real image only if the object is placed
(a) beyond optical centre ‘O’
(b) beyond centre of curvature
(c) beyond focus F
(d) beyond radius of curvature

11. Incident rays parallel to the principal axis on passing through convex lens meet at
(a) focus
(b) optical centre
(c) at centre of curvature
(d) radius of curvatures

12. Convex lens produces an image of same size as that of object. The position of object is
(a) at F
(b) at 2F
(c) between F and O
(d) between F and 2F

13. Convex lens is used as a magnifying glass. To see the size of object magnified, the object must be placed at a distance
(a) less than double focal length
(b) more than double focal length
(c) less than focal length
(d) more than focal length

14. An object is placed beyond 2F of a convex lens. The image formed will be
(a) diminished
(b) real and inverted
(c) between F and 2F
(d) all of these

15. The image formed by convex lens is virtual, erect and enlarged only when the object is placed
(a) at F
(b) between F and lens
(c) at 2F
(d) at infinity

16. A child covered more than half of the convex lens to get an image of a distant object. The correct observation would be
(a) no image is
(b) virtual image is formed
(c) real image is formed
(d) real image is formed but blurred in nature

17.

In the above figure, AB is object and A’ B’ is image formed. The correct sign convention for u, v and/is
(a) -u, -v, -f
(b) – u, -f, + v
(c) –u, +f, + v
(d) +u,+f, +v

18.

When the object is placed between F and O of the convex lens the correct sign convention for u, v and/is
(a) +u, + v, +f
(b) – u, – v, +f
(c) – u, – v, -f
(d) – u, + v, +f

19. To obtain the image at infinity in a convex lens the object should be placed
(a) at F
(b) at 2F
(c) between F and 2F
(d) beyond 2F

Questions based on Reporting and Interpretation Skills

20. A person is suffering from an eye defect in which his eye lens is not able to focus the image on retina. But the image is formed in front of retina.

To correct this eye defect a patient should be advised to use
(a) convex lens which is converging in nature
(b) concave lens which is diverging in nature
(c) combination of both the lenses
(d) none of the above

21. When the object is at infinity, the rays coming from it are parallel to each other

The image is formed
(a) at F
(b) at 2F
(c) between F and 2F
(d) beyond 2F.

22. In the given figure of Q.21, the nature of image formed is
(a) real, erect, magnified
(b) real, inverted, highly diminished
(c) real, erect, diminished
(d) virtual, erect, diminished

23. To obtain a magnified image on screen of an object for convex lens the object should be placed at
(a) F
(b) between F and 2F
(c) beyond 2F
(d) between F and O

24. To get a real, inverted and enlarged image and beyond 2F, the object should be placed:
(a) at 2F
(b) between F and 2F
(c) beyond 2F
(d) between F and O

25. A convex lens has focal length of 20 cm. To get an image of same size, real and inverted the object should be placed from the lens:
(a) at 30 cm
(b) 20 cm
(c) at 10 cm
(d) 40 cm

26. A candle with flame is kept at the focus (F) of the convex lens. The image is formed:
(a) between F and 2F
(b) at 2F
(c) at F
(d) at infinity

27. The power of the lens is one dioptre when its focal length is:
(a) 1 cm
(b) 1000 cm .
(c) 100 cm
(d) 10 cm

28. A convex lens of power +5 dioptre is dipped in a beaker containing water. The power will:
(a) decrease
(b) increase
(c) no change
(d) none of these

29. A convex lens has focal length of 10 cm. What is its power?
(a) 10 D
(b) ID
(c) 0.1 D
(d) 100 D

30. A student obtained an image of a distant object on a screen to determine the focal length f₁ of the given lens. His teacher after checking the image, gave him another lens of focal length f₂ and asked to focus the same object on the same screen. The student found that to obtain a sharp image he has to move the lens away from the screen. From this finding we may conclude that both the lenses given to the student were: [Delhi 2014]
(a) Concave and f₁ < f₂
(b) Convex and f₁ < f₂
(c) Convex and f₁ > f₂
(d) Concave and f₁ > f₂

31. A concave lens is dipped in a beaker containing water it will behave like:
(a) convex lens
(b) plane glass
(c) concave lens
(d) none of these

Class 10 Science Lab Practicals MCQ Scoring Key With Explanation 

1. (c) Optical center is the point through which any ray passes straight.

2. (d) The double convex lens,helps in magnifying the object.

3. (b) When / is in metre, the unit of power is in dioptre.

4. (c) Lens formula,\(\frac { 1 }{ v } =\frac { 1 }{ u } =\frac { 1 }{ f }\), gives the relationship between the object-distance (u), image-distance (v), and the focal length (f) of a spherical lens.

5. (b) Concave lens produces virtual focus.

6. (b) It gives magnified image.

7. (c) Principal axis is an imaginary straight line passing through the centre of the lens.

8. (a) Concave lens has -ve power.

9. (b) Aperture of lens = diameter of lens.

10. (c) Rays coming from beyond F meet after refraction.

11. (a) All rays parallel to principal axis meet at F after refraction through lens.

12. (b) The object placed at 2F will give same size image at 2F on the other side of the lens.

13. (a) For convex lens, if object is placed between F and 2F, the image is magnified.

14. (d) For convex lens when the object is placed beyond 2F, the image formed is diminished, real and between F and 2F.

15. (b) For convex lens when object is placed between focus F and optical centre O, the image is virtual and erect.

16. (d) On covering half or more lens the image is formed in the same position but it is blurred.

17. (c) Distances measured in the direction of the incident rays are positive and the distances measured in the direction opposite to that of the incident rays are negative.

18. (b) Distances measured in the direction of the incident rays are positive and the distances measured in the direction opposite to that of the incident rays are negative.

19. (a) If the object is at F, the image formed will be at infinity.

20. (b) Concave lens will diverge the rays to make them meet at retina.

21. (a) The parallel rays converge at F of convex lens.

22. (b) For object at infinity the image formed at F is highly diminished, real and inverted.

23. (b) If object is placed between F and 2F, the image is magnified.

24. (b) If object is placed between F and 2F, the image is magnified.

25. (d) f = 20 cm, then 2f= 40 cm, the object at 2F gives the image at 2F.

26. (d) Object at F gives image at infinity.

27. (c) 100 cm = 1 m; P = \(\frac { 1 }{ f } =\frac { 1 }{ 1 }\) = 1 D

28. (a) Focal length will increase.

29. p=\(\frac { 1 }{ f } =\frac { 1 }{ 0.1 }\)D=10 D

30. (b) Convex lens gives image on the screen and the lens and screen distance = f which is more in second case.

31. (a) The refractive index of water is added to the refraction of light.

More Resources CBSE Class 10 Lab Manual Practical Skills:

• CBSE Class 10 Science Image formation by a convex lens Practical Skills
• CBSE Class 10 Maths Lab Manual
• CBSE Class 10 Math Labs with Activity

NCERT Class 10 Science Lab Manual 

• Determine Ohm’s Law Class 10 Experiment
• Determine Resistors in Series Class 10 Experiment
• Determine Resistors in Parallel Class 10 Experiment
• Determine Focal Length of Concave Mirror and Convex Lens Class 10 Experiment
• Determine Refraction Through Glass Slab Class 10 Experiment
• Determine Refraction Through Prism Class 10 Experiment
• Determine Image Formation by a Convex Lens Class 10 Experiment