# Relationship Between U and V For a Convex Lens

August 7, 2017 September 1st, 2019 Free Essays Online for College Students

I think that when the length of ‘u’ increases (object distance), the length ‘v’ will decrease. (image distance.) Only when the object is greater than the focal length of the lens, is a real image is produced. If the object distance is nearer the focal length than a virtual image is produced. The virtual image has a negative distance from the lens, which means it can’t be focused onto the screen. When plotting my results on graph, I will expect them to produce a curved line for u over v, producing a reciprocal graph. However a straight line when plotting 1/u over 1/v.

Scientific Ideas

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This formulae for the focal length of an object:

1/focal length = 1/object distance + 1/image distance

The following formulas are rearranged from the one above. This will help me throught my experiment, and with my scientific ideas.

u = object distance f = focal length v = image distance

(20.0 cm) (15.0 cm) (60.0 cm)

f = 1 / (1/u + 1/v) 15 = 1 / (1/20 + 1/60) (FOCAL LENGTH)

f = (uv) / (u + v) 15 = (20*60) / (20 + 60)

u = 1 / (1/f – 1/v) 20 = 1 / (0.06 – 0.016) (OBJECT DISTANCE)

u = (vf) / (v-f) 20 = (60*15) / (60-15)

v = 1 / (1/f – 1/u) 60 = 1 / (0.06 – 0.05) (IMAGE DISTANCE)

v = (uf) / (u-f) 60 = (20*15) / (20-15)

FOCAL LENGTH = 15

cm

u (cm)

v (cm)

10.00

– 30.00

20.00

60.00

30.00

30.00

40.00

24.00

50.00

21.43

60.00

20.00

70.00

19.09

80.00

18.46

90.00

18.00

100.00

17.65

110.00

17.37

120.00

17.13

130.00

16.96

140.00

16.80

150.00

16.67

160.00

16.55

170.00

16.45

180.00

16.36

190.00

16.29

Above shows a table that I have produced from the previous equations. The table shows what results I am to expect, along with proving my prediction that:

‘I think that when the length of ‘u’ increases (object distance), the length ‘v’ will decrease. (image distance.) Only when the object is greater than the focal length of the lens, is a real image is produced. If the object distance is nearer the focal length than a virtual image is produced. The virtual image has a negative distance from the lens, which means it can’t be focused onto the screen.’

When I increase the object distance (u), the image distance (v) decreases. This is shown on the table from 20.00cm of u to 200.00cm of u, where each time v decreases. However the image distance at 10.00cm meant v was recorded as negative. This is because the object distance is less than the focal point. So the image produced is virtual, and has a negative distance from the lens.

The graph that I have produced from the table above proves my prediction that;

‘When plotting my results on graph, I will expect them to produce a curved line for u over v, producing a reciprocal graph.’

As you can see from that the graph I have is produced a reciprocal graph, as predicted. Due to the fact the graph produced is a reciprocal graph; it means that there is a tighter distribution of results at higher object distances, resulting in a greater distribution of results at lower object distances. Until the object distance is below the focal point, when the image produced is virtual. The equation of the graph produced which is a prediction of the actual results, when plotting u over v is:

1 + 15

(v – 15)

As u increases, v decreases can also be shown by the ray diagrams above. The first two diagrams show that as u increases, v decrease. This is because the first diagram the object is placed outside 2F, which means it is placed over twice the focal length away from the lens. However the second diagram, the object has been moved nearer the convex lens which means u has decreased. The ray diagram illustrates that the v distance has increased. This proves that as u increases, the v distance decreases.

The third diagram shows that when the object is placed nearer the lens than the focal point (u distance is less than the focal point), a virtual image is produced. Using the formula v distance is negative, because it appears on the same side of the lens.

Preliminary Investigations

1. The first preliminary investigation done was to choose a suitable lens

METHOD

First of we shall to choose a group of lenses. We shall then find out which lens to use by producing images on the screen at different u distances, and testing which lens will produce clearer distances. This will be done by setting up the apparatus, by placing a cross wire window in front of the lamp, then placing the different lenses on the lens holder. To see which images were best produced over different u distances.

RESULTS

LENS

Average v(cm)

RED (small)

29.6

YELLOW (small)

14.6

BLUE (fat)

– (too small)

YELLOW (fat)

– (too small)

We decided to choose the RED lens, because it gave a reasonable image distance. Which would mean it would enable us to get a wider and clearer range of results.

‘RED LENS CHOOSEN’

ANALYSIS

The fat lenses instantly weren’t chosen because they produced images that were to small to measure, and could only be measured at large distances for v.

The red lens was chosen firstly it was a thin lens, and not a fat lens. This is because the fatter the lens the more rays of light refract as they go through, the more the rays of light bend the nearer they will cross over. The crossing over point is called the focal length. So this means the focal point of a fat convex lens closer to the lens than in a thin convex lens.

The focal length of a lens affects the size of the image transmitted to the screen and the depth of field of the image, along with the relative sharpness of objects in the image.

Secondly red was chosen over the other thin lens. This was because the other thin lens produced an image much further away, making it harder to take results of other distances of u below it. The other thin lens (yellow) would have given a much larger distance of results, due to having a much bigger focal length. Red produced clearer and easy to measure distances. Lastly, choosing the other lens would have meant that other v distances are hard to measure.

2. The second preliminary investigation done was to find the focal length of the lens chosen.

METHOD

To find the focal length of the lens chosen, we shall use a equation. This is done firstly by taking some measurements. First of all we shall take the object distance as 30.0cm. Then we shall measure the image distance produced. Lastly we shall use the formula below to work out the focal length.

1/f = 1/u + 1/v

RESULTS

We used the formula to work out the focal length:

1/f = 1/u + 1/v

1/f = 1/30.0 + 1/29.6

1/f = 0.03 + 0.3378

f = 14.9

FOCAL LENGTH = 14.9 cm

ANALYSIS

This is focal length of the lens which I am going to use throught the experiment. It tells me that the minimum distance u can be is 14.9cm, where the object distance is between F and the lens, and a virtual image is produced. The virtual image has a negative distance from the lens, which means the image it is virtual. However at 14.9cm the largest distance of v is produced, after which the distance of v decreases.

Variables

* Independent variable

My independent variable is u (object distance).

* Dependent variable

My dependent variable is the v (image distance).

* Controlled variables

Same lens, same focal length of lens, same light intensity, same position of window in front of the light.

Range

* Independent variable

This will be 2 metres maximum, measuring at intervals of 10cm until 2m.

* Dependent variable

This will determined in the experiment as this will be measured to get the results, measured in cm.

* Controlled variables

Which lens to be used will be determined in the preliminary investigations, due to different lenses able to give different results. The preliminary investigations also determine the focal length of what lens I will have chosen. The light intensity will stay at 12V; it means the image is clearer and easily seen. The window simulates the object, which will be much brighter in front of the lamp,

making the image produced easier to be seen.

Fair Test

I shall take the following precautions were made to the experiment, to ensure it was a fair test:

* I shall measure in one unit (cm), for the distances between u and v.

* I shall make sure that the brightness of the lamp stays the same throughout the experiment (12V).

* I shall use the same convex lens throught the experiment, the one determined in the preliminary investigations.

* I shall us the same sort of equipment throught the experiment, same sort of image screen, same sort of rulers, same les holders and same window images.

Safety

I shall take the following safety precautions were taken, to ensure that the experiment was safe to do. They were:

* I will put my things out the way, so they don’t come in the way of other people (e.g. when the lights are switched off.)

* I will not play with the instruments (e.g. rulers, or swing wires).

* I will take care with my lamp so that it doesn’t shine into other people’s eyes, which can damage their site. I will also take care in touching them (e.g. when I pick them up), because they are hot after being turned on.

* I will take care with the lenses so they don’t break in the dark, which could cut somebody when broken.

* Lastly, I will take care when the lights are switched off and I am in the dark, so that I don’t trip up and hurt myself.

Apparatus

The instruments which I shall use in this experiment are:

* I metre Ruler X2

* Convex lens

* Lamp (12V)

* Image Screen

* D.C supply

* Lens holder

* Object window

Method

The experiment will be done on a table allowing a flat base, so that straight rays of light can travel to the convex lens and produce an image, along with providing an easy measuring surface. We will use a lamp as a bright object, so that an image can be easily produced on to the screen. A lamp will also be used in this experiment to transmit rays of light into the lens, and focus them to form a sharp image on the screen. To supply electricity to the lamp a D.C supply will be used. The D.C supply is kept at 12V (the brightest), so that the brightest image can be produced and easily focused. In front a window will be placed; this is so that the object distance can be easily located. Without the use of a window it will mean that only light rays go through the lens, and not produce a picture, meaning the object distance is harder to find. Then, the lens will be placed in a lens holder; the lens holder with the lens will be first placed 16.0cm away from the object. A convex lens will be used to focus the image on a screen, and not to divert the image like a concave lens. If a real image is formed, it will be focused onto a screen in front of the lens holder so that it can be seen on the screen. A lens holder is used because so that the lens can be held upright without falling. Also because the lamp with the bulb in it is held up by a few inches, with the use of a stand so it doesn’t fall. So for the light rays to pass through the lens, the lens holder where the lens is placed is also a few inches of the ground. This means the window or the image is placed in front of the lamp, is at the correct height so that the light rays from the lamp can go through the image and through the lens. The image produced is focused on to a screen, allowing you to produce a sharp image of the bright object easily. The lights will be turned off, so that the image can be easily seen.

We will measure v, the distance from the screened image to the lens, and record this result. All distances between u and v will be measured using two 1 metre rulers. One for the image distance (v) and object distance (u). Then, firstly we will move the lens holder with the lens, and the object so that the distance from the object to the lens is 16cm. Starting with 16cm because it is the next available size after the focal point. Anything smaller than the focal point means a virtual image is produced. We will then shine the light onto the object and measure the distance from the lens to the image on the screen and record this result in centimeters.

This will be repeated for distance of 17.0cm and then distances of 30.0cm, 40.0cm, 50.0cm, 60.0cm, 70.0cm, 80.0cm, 90.0cm, 100.0cm, 110.0cm, 120.0cm, 130.0cm, 140.0cm, 150.0cm, 160.0cm, 170.0cm, 180.0cm, 190.0cm, 200.0cm, 210.0cm, 220.0cm, 230.0cm, 240.0cm and 250.0cm of u. In each case the distance v from the lens to the image on the screen will be recorded. Also, for each u distance v will be measured twice, and an average v distance for each u distance will be recorded in centimeters.

Results Table

u

(cm)

v (cm)

Attempt 1

Attempt

2

Attempt 3

Average

16.0

215.1

216.7

218.3

216.7

17.0

120.6

120.2

120.8

120.5

20.0

65.5

62.0

63.0

63.5

30.0

29.3

30.7

30.3

30.1

40.0

23.0

24.3

24.0

23.8

50.0

21.7

21.8

22.0

21.8

60.0

19.8

19.5

19.7

19.7

70.0

18.9

18.3

19.5

18.9

80.0

18.1

18.8

18.9

18.6

90.0

17.8

17.6

18.0

17.8

100.0

17.6

17.9

17.0

17.5

110.0

17.2

17.6

16.8

17.2

120.0

16.7

17.0

17.1

16.9

130.0

16.8

16.4

17.1

16.8

140.0

16.3

17.0

16.5

16.6

150.0

15.9

16.8

16.1

16.3

160.0

15.8

16.5

15.8

16.0

170.0

15.7

15.8

15.8

15.8

180.0

15.6

15.9

15.6

15.7

190.0

15.5

15.7

15.6

15.6

200.0

15.2

15.5

15.7

15.5

(cm)

Expected

Results

1/u

1/v

216.7

0.063

0.0046

120.6

0.06

0.008

58.4

0.05

0.015

29.6

0.03

0.03

23.7

0.025

0.04

21.2

0.02

0.045

19.8

0.016

0.05

18.9

0.014

0.052

18.3

0.013

0.053

17.9

0.011

0.056

17.5

0.01

0.057

17.3

0.009

0.058

17.0

0.0083

0.059

16.8

0.0076

0.059

16.6

0.007

0.060

16.5

0.006

0.061

16.4

0.006

0.062

16.3

0.0058

0.063

16.2

0.0055

0.063

16.1

0.0052

0.0641

16.1

0.005

0.0645

Firstly, my results table proves my prediction that:

‘I think that when the length of ‘u’ increases (object distance), the length ‘v’ will decrease. (image distance.)’

This happened throught the experiment, which can be showed by the results. From the start starting at 16cm for u where a large image distance is produced, until 200cm for u where a small image distance is produced.

Graphs

Analysis

I have proven that as u increases v decreases. This happens because as v increases (the object distance), it means the middle rays of light from the top of the image changes. It changes so that as u increases, the middle ray goes through the principal axis and convex lens at a much smaller angle. This is because as the line is bigger it becomes less steep from the top of the image to when it crosses. Causing it to cross at a much smaller angle each time, therefore crossing the top rays from the image at a nearer distance. This happens every time u increases, along with the image becoming smaller. This happens because as the two top and middle rays cross, because u becomes bigger they cross at a nearer distance. This is point nearer to the principal axis; therefore a smaller image is produced.

The two diagrams have proven why as u increases v decreases. As you can see from the first diagram, u is much greater than the second diagram. Meaning the middle ray as you can see is much less steeper than the second diagram. Shown by the first diagram having a smaller angle going through the convex lens than the second. Causing the image to be produced a much shorter distance than the second. You can also see that a smaller image is produced by the first diagram, due the rays having met at a smaller distance.

When the object is placed before the focal point, it means that the image becomes on the side of the lens, upright, magnified and virtual. This is because the middle rays passing through the centre of the convex lens, becomes so steep. Which means that it can’t cross with the top rays from the object, to give an image. The object has to be placed after the focal point; this is because the middle ray hits the principal axis at an angle. If the object is placed before the focal point it means that this angle becomes larger, larger than the angle created when the top rays hit the principal axis. This means that the middle rays are moving away from the top ray, so a real image can’t be created.

The diagram proves that when the object is placed before the focal point, the middle rays from the top of the object create a large angle. Larger then the angle created from the top rays hitting the principal once going through the convex lens. This means that the rays of light are diverted and never join, to give a real image. Therefore a virtual image is produced that is upright and magnified, because it is on the same side of the lens.

However when the object is after the focal point it is inverted, and real. This is because when the rays of light cross over, they cross over on the other side of the principal axis, therefore producing an inverted image.

A trend and pattern that I noticed in the graph was that as u increased, the difference in v results also decreased. This like, at u distance of 20.0cm, the v distance is 58.4cm. At u distance of 40.0cm, the v distance is 23.7cm. The difference between 58.4cm and 29.6cm is 28.8cm and the difference between 29.6cm and 23.7cm is 5.9cm.This means that as well as v decreasing when u increases; the difference between each v distance also decreases. The graph never becomes parallel to the y / x axis; this is because the graph is never less than the focal point, when v is positive.

The graph that I have produced proves my prediction that;

‘When plotting my results on graph, I will expect them to produce a curved line for u over v, producing a reciprocal graph.’

As you can see from that the graph I have is produced a reciprocal graph, as predicted. Due to the fact the graph produced is a reciprocal graph; it means that there is a tighter distribution of results at higher object distances, resulting in a greater distribution of results at lower object distances. Until the object distance is below the focal point, when the image produced is virtual. It also means that when v is positive it never is less than the focal point, along with never being parallel to the y/x axis. The equation of the graph produced which is a prediction of the actual results, when plotting u over v is:

1 + 14.9

(v – 14.9)

Using the formula also proves that when u increases v decreases.

Using the formula, rearranged from the original, it shows that when u increases v decreases. However when a distance for u is put in, which is less than the focal point, it means a negative distance is given. This shows the image produced is, virtual.

When the a virtual image is formed, one ray goes straight through the middle of the lens and another passes through the focal point and emerges parallel to the principal axis. To work out the focal length from a virtual image, the equation below is used:

1/v – 1/u = 1/f

However to work out the focal length from a real image, the equation below is used:

1/v + 1/u = 1/f

Using the Cartesian sign convention, the linear magnification is given by:

V/U

During the practical I made some errors, firstly I didn’t use the same equipment throughout the practical. This is because I had to take the results down in four lessons. This meant that different instruments were used for the different days, which meant that the experiment could not to be a fair test. This means that my controlled variables weren’t accurate. Lastly another mistake I made was that my work base (table) wasn’t long enough for the distance v.

The anomalies that I have discovered are, are from my results. These were found by looking at the expected results, worked out by the formula. Which showed that my average reading for v, when u was 20.0cm was 10cm out then expected. It also showed that my last few results were also out, but this time only by 1cm.

However overall, my experiment shows that when the distance u increases, the distance v decreases. This meant my prediction was correct and proved. I found my results were fairly accurate, and close to my expected results using the lens formula.

Evaluation

My plan has worked out as expected this is because my method, readings, range, safety and equipment were used how I expected in my plan. However most of my references expected were used, along with most of my hypotenuse being true. This is because the basic structures of the graphs were turned out as expected. However they were not perfect, when looked at the graph and results in detail. The suggestions predicted in the fair test, were also followed mostly, with only one suggestion that I was not able to follow.

I think that my results followed the basic structure that was predicted. This is because when the results were plotted on to a graph. Here I expected a curved line, which basically was achieved. However if looked at the graph in detail, I would see that only a few results weren’t totally accurate. I think that my results have achieved, what I predicted, however if I was to look at them in great detail I would see that, only a few results weren’t totally accurate.

I think that the evidence obtained is good enough to support the conclusion. That when u increases, v decreases. The graphs produced also allows this evidence to be supported.

Overall, I think that the experiment can be improved so that the results obtained can become more accurate. Along with extension added, to the experiment so that, a wider range of results is available. Overall, I think that the results obtained were sufficient and well done.

The following ways I could improve this experiment are:

* Use the same equipment again by labeling your own equipment, reassuring a fair test.

* I could also improve on this experiment, by correcting my mistake (errors), which is explained above. Which will ensure a more accurate experiment

* Having the experiment done on the floor, so that the focal length could be easily measured.

* A better screen should have been used so that the image produced, can be seen easily.

The following ways I could add on to this experiment are:

* I could have other independent variables, which include having the lamp turned on different brightness each time. Or different strength of the lenses.

* I could also compare results of other people, and comment on how there results are the same or differ to mine.

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