High school Physics Lessons: principal axis

In the previous section we saw that the rays emerging from the object, passes through the lens and then converge to give points on the image. To draw the exact paths taken by the rays, we must be familiar with some rules. They are given below:
Rule 1:
• When a light ray passes through the optic centre of a thin lens, it does not undergo any deviation. This is shown in figs.7.16 (a) and (b) below:

Fig.7.16

• We can see that the rays are passing through P. The lens is not capable of causing any deviation to such rays
Rule 2:
A ray of light falling parallel to the principal axis of a convex lens passes through the principal focus after refraction. This is shown in fig.7.17 below:

Fig.7.17

Rule 3:
A ray of light falling parallel to the principal axis of a concave lens appears to emerge from the focus on the same side of the lens. This is shown in fig.7.18 below:

Fig.7.18

• Note that, after refraction, the ray diverges away. We can only trace it's path backwards by drawing a dashed line on paper. When we draw such a dashed line, we will reach the focus F
Rule 4:
A ray of light passing through the principal focus of a convex lens passes parallel to the principal axis after refraction. This is shown in fig.7.19 below:

Fig.7.19

This can be explained using the following steps:
1. Light rays emerge in all directions from an object
2. Among those rays in (1) abive, some will pass through F
3. Among those rays in (2) above, some will fall on the lens
4. All rays in (3) will undergo deviation.
• After the deviation, they will become parallel to the principal axis
• This can be considered as the reverse of Rule 2
Rule 5:
• All the light rays coming from an 'object at infinity' (or a 'distant object') are considered to be parallel to each other.
• We have seen that such parallel rays will converge at F. See fig.7.13 of the previous section

Now we are ready to draw ray diagrams. The ray diagrams will help us to predict three items:
(i) Position of the image
(ii) Nature of the image
(iii) Size of the image

Case 1: Object beyond 2F
This is shown in fig.7.20 below:

Fig.7.20

It is convenient to draw the ray diagram on a graph paper. Let us write the steps:
1. The optic centre P must be placed at the 'origin of the graph'
• 'Origin of the graph' is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 9.5 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the object from P is denoted as 'u'
■ So we can write: u = -9.5 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 6.9 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = +6.9 cm or simply 7 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is +2.25 cm or simply 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 1.63 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -1.63 cm
10. Considering the numeric values only, we have:
• Height of the object = 2.25 cm
• Height of the image = 1.63 cm
• So the height of the image has decreased.
• Not only the height, the thickness also will decrease. So a thinner yellow line and a smaller square are shown as the image
11. We can write the conclusion:
■ When the object is placed beyond 2F:
(i) Position of the image: Between F and 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Diminished

Case 2: Object at 2F
This is shown in fig.7.21 below:

As before, we will draw the ray diagram on a graph paper.
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 2F
■ So we can write: u = -8 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 8 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = 8 cm
• Note that this is the distance of 2F (on the other side) from P
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is also 2.25 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -2.25 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = -2.25 cm
• So the height of the image has not changed.
• Not only the height, the thickness also will not change.
11. We can write the conclusion:
■ When the object is placed at 2F:
(i) Position of the image: At 2F on the other side
(ii) Nature of the image: Real and inverted

(iii) Size of the image: Same size

Case 3: Object between F and 2F
This is shown in fig.7.22 below:

Fig.7.22

As before, we will draw the ray diagram on a graph paper.
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 6.25 units to the left of P
■ So we can write: u = -6.25 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 11.11 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = 11.11 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 4 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -4 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 4 cm
• So the height of the image has increased.
• Not only the height, the thickness also will increase. So a thicker yellow line and a bigger square are shown as the image
11. We can write the conclusion:
■ When the object is placed between F and 2F:
(i) Position of the image: Beyond 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Enlarged

Case 4: Object at F
This is shown in fig.7.23 below:

Fig.7.23

Fig.7.24

As before, we will draw the ray diagram on a graph paper.
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 2.5 units to the left of P
■ So we can write: u = -2.5 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays diverge from each other. They will not meet on the right side of the lens
• But if they are drawn backwards, they will meet at a point on the same side
♦ This is shown by the dashed lines
• This meeting point is the image of the 'top of the object'
• From there we can draw the image downwards
8. The position of the image:
The image is formed at 6.66 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = -6.66 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 6 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the image in our present case is 6.0 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 6.0 cm
• So the height of the image has increased.
• Not only the height, the thickness also will increase. So a thicker yellow line and a bigger square are shown as the image
11. We can write the conclusion:
■ When the object is placed between F and lens:
(i) Position of the image: Same side of the object
(ii) Nature of the image: Virtual and erect
(iii) Size of the image: Enlarged

Case 5: Image formation by a concave lens
This is shown in fig.7.25 below:

Fig.7.25

As before, we will draw the ray diagram on a graph paper.
• The steps are as follows:
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. We will not be needing 2F for this case
5. The object is placed at 6.25 units to the left of P
■ So we can write: u = -6.25 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 3 that we saw above.
♦ After refraction, it will diverge away
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays diverge from each other. They will not meet on the right side of the lens
• But if they are drawn backwards, they will meet at a point on the same side
♦ This is shown by the dashed line for ray (i)
♦ For ray (ii), the dashed line will fall on that ray itself. So it is not visible
• This meeting point is the image of the 'top of the object'
• From there we can draw the image downwards
8. The position of the image:
The image is formed at 2.21 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = -2.21 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 0.79 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the image in our present case is 0.79 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 0.79 cm
• So the height of the image has decreased.
• Not only the height, the thickness also will decrease. So a thinner yellow line and a smaller square are shown as the image
11. We can write the conclusion:
■ When the object is placed anywhere in front of a concave lens:
(i) Position of the image: Same side of the object between F and the lens
(ii) Nature of the image: Virtual and erect
(iii) Size of the image: Diminished

In the next section, we will see an experiment based on the above results.

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In the previous section we saw the details about refractive index. In this section we will learn about lens.

Lens is a transparent medium having spherical surfaces
From the above definition, we can say that, a lens should satisfy the following two conditions:
(i) It must be transparent
• That is., a lens can be made using glass, plastic or any similar materials. But it must be transparent
(ii) It should have spherical surfaces
• That is., any surface of a lens should be a part of a sphere.
• It is just like saying: An arc is a part of a circle

Let us see the details about those spherical surfaces. We will write them in steps:
1. Consider the two spheres in fig.7.10 (a) below:
• They are identical spheres. That is., both have the same radii

Fig.7.10

• The centre of the left sphere is C₁
• The centre of the right sphere is C₂
• The red line joins C₁ and C₂
2. In fig.7.10(b), the two spheres are brought closer to each other.
• So close that, they intersect each other
• The distance C₁C₂ is now lesser than the C₁C₂in fig.a.
3. The 'portion common to both the spheres' is retained as such
• The rest of the two spheres are erased off
• The result is shown in fig.c
• The retained portion in fig.c forms a convex lens
• Note that, the distance C₁C₂ is fib.c is the exact same as that in fig.b
4. The surfaces of the lens are thus parts of the surfaces of spheres

Now we will see the concave lens:
1. Consider the two spheres in fig.7.11 (a) below:
• They are identical spheres. That is., both have the same radii

Fig.7.11

• The centre of the left sphere is C₁
• The centre of the right sphere is C₂
• The red line joins C₁ and C₂ .
2. In the fig.a, the spheres are close to each other. But they are not touching each other
• In fig.b, the space between them is filled up with a transparent material
3. In fig.c, the two spheres are removed.
• The newly added transparent material is left behind. This forms a concave lens.
• Note that, the distance C₁C₂ is fib.c is the exact same as that in fig.b
4. The surfaces of the lens are thus parts of the surfaces of spheres

Let us see the properties of convex and concave lenses:
1. Optic centre
• Optic centre is the midpoint of a lens.
• It is denoted by the letter 'P'. It is shown in fig.7.12 below:

Fig.7.12

2. Centre of curvature
• We have seen that each lens is associated with two spheres with centres C₁ and C₂
• Both C₁ and C₂ are eligible to be called as the 'centre of curvature' of the lens
• So the definition is:
Centre of curvature is the centre of the imaginary spheres of which the surfaces of the lens are parts. It is denoted by the letter 'C'
3. Principal axis
• The definition is:
Principal axis is the imaginary line that passes through the optic centre joining the two centres of curvatures
• So the principal axis should satisfy both the conditions given below:
(i) The principal axis must pass through both C₁ and C₂
(ii) The principal axis must pass through P
4. Principal focus
(a) Convex lens
• The definition is:
Light rays parallel and close to the principal axis converge at a point on the principal axis of a convex lens. This point is it's principal focus. It is denoted by the letter 'F'
• This will be clear from fig.7.13(a) below

Fig.7.13

• The red line is the principal axis.
• Note that light rays will be travelling in all directions. In the fig.a above, we are considering only those which satisfies the following two conditions:
(i) Light rays must be parallel to the principal axis
(ii) Light rays must be close to the principal axis
• A ray which travel exactly along the principal axis will also satisfy the two conditions. So it is also included in the diagram
(b) Concave lens
• The definition is:
Light rays parallel and close to the principal axis diverge from one another after refraction. These rays appear to originate from a point on the same side. This point is the principal focus of a concave lens. It is denoted by the letter 'F'
• This will be clear from fig.7.13(b) above.
5. Focal length

It is the distance between the principal focus F and the optic centre P. This distance is denoted by the letter 'f'

■ Why do we say that the principal focus of a concave lens is virtual?
Let us analyse:
• Consider the dashed green lines diverging from F in the fig.7.13(b).
• The lines are dashed from 'F' up to the outer boundary of the lens.
• Why are they 'dashed'?
• The answer can be written in steps:
1. The actual light rays pass through the lens and after refraction, leaves the outer boundary of the lens
• They diverge from each other
2. These rays cannot travel back to give us the position of 'F'
• That is., there are no actual rays along the dashed lines behind an actual concave lens in 'real life'
♦ So they are drawn with dashed lines
• Consequently, there is no actual 'F' behind an actual concave lens in 'real life'
3. We can only draw them on paper.
• So the principal focus of a concave lens is said to be virtual
■ In short, we can write:
It is impossible to converge light at a point using a concave lens. Therefore the principal focus of a concave lens is virtual

Now we will learn about image formation. Consider fig.7.14 below:

Fig.7.14

1. A convex lens is placed in front of a candle. A screen is placed in front of the lens.
• In this arrangement we have two distances:
(i) Distance between the object (candle) and the lens. This is denoted by the letter 'u'
(ii) Distance between the image (screen) and the lens. This is denoted by the letter 'v'
2. By adjusting the above two distances, we can form a clear image of the candle on the screen

■ How is the image formed on the screen?
We will write the answer in steps:
1. Consider the fig.7.15 below:

Fig.7.15

• An object (consisting of an yellow pole and a cube at the top) is placed on the left side of a lens
2. We know that rays of light reflect from the object in all directions. When those rays fall on our eyes, we see the object.
• Two such rays are considered here. Those two rays start from the cube and travel in two different directions.
• That is., after starting from the cube, they diverge from each other.
• But after passing through the lens, they converge at a point on the other side of the lens.
3. At that point of convergence, we get an image of the cube.
■ So we can note a point:
To define a single point on the image, we will need at least two rays.
4. In the fig.7.15 above, we defined the cube in the image. In the same way, we can define all other points on the image.

• The rays emerging from the object, passes through the lens and then converge to give points on the image.
• To draw the exact paths taken by the rays, we must first familiarise ourselves with some rules. We will see those rules in the next section.

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High school Physics Lessons

Pages

Tuesday, April 10, 2018

Chapter 7.3 - Image formation and Ray diagrams

Monday, April 9, 2018

Chapter 7.2 - Convex and Concave Lens