Essay,
Pages 6 (1264 words)

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Aim:

To find the power of a given diverging lens by using a pre-selected converging lens.

Outline:

The power of two lenses placed together in line to an object will allow both of their powers to add together.

P1 + P2 = Ptotal

Therefore, by using a combination of 2 lenses, one of which the power is known, I can find the focal length of the combination of lenses and then calculate the total power. The power of the converging lens can be subtracted from the total power to find the power of the diverging lens.

Variables:

Independent: Object distance

Dependent: Image distance

Control: Object size and shape, power of both lenses, material of both lenses

In my pre-tests, I have found that I obtain a reasonable range of results using a converging lens of 10 Dioptres that is larger than the power of the diverging lens therefore giving reasonable results.

Method:

1. Set up apparatus as shown in diagram.

2. Adhere the unknown diverging lens to the known, pre-selected converging lens with plasticine ensuring that the plasticine does not interfere with the light i.

e. the plasticine does not cover the main central portion of the lenses.

3. Fix the lenses on to the stand with plasticine, making sure that they are perpendicular to the ground therefore the light that falls upon the light will be in line with the normal at the centre of the lenses.

4. Use a piece of card with a triangular hole in it with mesh stuck over the top to be used as the object, one, to block out most of the light to prevent shadows forming, two, it discern whether the image was formed inverted or not, three, the mesh helps me judge when the image is at its clearest.

5. The light, object, lens stand, lens and screen will be arranged in a straight line and kept in a straight line, so that there won’t be any undue error caused in the measuring of object and image distances from the centre of the lens, parallel to the metre ruler used to measure the distances.

6. The object distance can then be varied to give difference image distances from which the power of the combined lenses can be calculated.

7. The screen will be held perpendicular to the table to ensure that the measuring is accurate and the image will be formed properly.

8. There will be a range of distances from the lens, which the image will be judged as to being clear and this range will be recorded.

9. Repeat experiment with at least 6 difference object distances

10. Carefully repeat all measurements

11. Use the results to plot a graph to find the power of the combined lenses.

Safety:

Having considered the safety aspects of the experiment, I am confident that normal safety procedures are adequate and there is no particular danger in this experiment apart from the fact that the light could heat up to rather high temperatures. The electrical equipment should also be handled with more care because of the possible electrical hazard.

Uncertainties and Errors

Extensive pre-testing proves that the following are likely sources of uncertainty and error.

1. Uncertainties in measurements

Length. The metre ruler is accurate to ï¿½ 0.001 metres, however, the range over which the image is clear vary around ï¿½ 0.020 metres – this value may change as clarity of the image may change with distance.

A range can be measured, but those upper and lower boundaries will also be affected by the error in measurement by the ruler therefore if the range is ï¿½ 0.020 metres, the range including measurement errors is ï¿½ 0.021 metres.

2. Errors in the Experiment

Distance between lenses and the thin lens formula. Errors will occur because the light will converge inside the space between the two lenses, therefore the focal distance will be incorrect. If the light converges between the lenses, then the light that hits the second lens will have converged and therefore the focal length will change, as shown in the diagram,

Large Gap

Small Gap

There must always be a space between them, because they are not one lens – the only possibility is that if both their curvatures were equal, then their powers would cancel out (the converging lens has a positive power, diverging has same power, but negative) and then the light emitted would be parallel, therefore no image will be formed at any point and the power cannot be calculated.

Also, the thin lens formula is only an approximate model to finding the power of thin lenses, this is because the formula applies 100% accurate only to a infinitely thin lens, but because all lenses will have a certain thickness, the formula can only give a good approximation. The thinner the lens, the better the approximation given by the formula.

Errors in judgement of which position creates the clearest image is the probably the most significant error in the entire investigation because there may be a range of positions where the image appears to be clear. Therefore a range of values will be taken as the position for the maximum clarity of an image thereby reducing the error that could be involved and also means, graphically, that a best fit line can be drawn within the error bars that fit the hypothesis.

Analysis

After collecting all the results, I can plot the range of values of each different object distance with their corresponding image distance. To draw this graph and obtain a linear relationship, I can rearrange the thin lens formula to the linear graph form y = mx + c where y and x are variables, m is the gradient of the line and c is the y-intercept.

Where and but, because the formula can also be arranged thus,

That means that will be both the y-intercept and the x-intercept, therefore giving a graph like this, where both intercepts are equal to the power (Power = reciprocal of focal length, P = f-1)

I will also include error analysis, which is shown here with a set of pre-test data using a converging lens of 10 dioptres,

TABLE 1

Object Distance (m)

Minimum Length for Clarity (m)

Maximum Length for Clarity (m)

Mean Length for Clarity (m)

Subsequent Power of Diverging Lens (D)

Error in measurement (m)

0.550

0.660

0.700

0.680

-6.71

ï¿½ 0.001

0.500

0.750

0.790

0.770

-6.70

ï¿½ 0.001

0.450

0.830

0.870

0.850

-6.60

ï¿½ 0.001

0.400

0.930

0.970

0.950

-6.45

ï¿½ 0.001

0.350

2.230

2.270

2.250

-6.69

ï¿½ 0.001

Then the error in the object, image minimum, image maximum and image mean length will be ï¿½0.001 but because the lengths are made into reciprocals, their errors become neither absolute nor percentage, therefore to calculate it, the maximum and minimum values must be used, so there will be 2 more columns where the error is added to the values,

TABLE 2

Absolute Minimum Length for Clarity (m)

Absolute Maximum Length for Clarity (m)

0.659

0.701

0.749

0.791

0.829

0.871

0.929

0.971

2.229

2.271

If the values themselves are used to calculate the power of the diverging lens mathematically, it would be as such, including Table 1

TABLE 3

Lower Boundary Power (D)

Upper Boundary Power (D)

Mean Power (D)

Error of Mean Power (D)

-6.670

-6.750

-6.710

ï¿½0.040

-6.670

-6.700

-6.685

ï¿½0.015

-6.570

-6.620

-6.595

ï¿½0.025

-6.420

-6.460

-6.440

ï¿½0.020

-6.690

-6.700

-6.695

ï¿½0.005

Average

-6.625

ï¿½0.021

Therefore the power of the diverging lens is -6.625 dioptres with an error of ï¿½ 0.021.

The graph can be drawn including these errors, and it makes it possible to draw a graph with a linear relationship where the x and y intercepts are equal – within the range of the error of the results – to find the correct power of the diverging lens.

See graph of pre-test results.

From the graph, the intercepts are at 3.3 D, therefore the resultant power of the combined lenses is 3.3 D. The pre-selected lens is 10 D, therefore,

10 + d = 3.3

? d = 3.3 – 10

? d = -6.7 D

So, the power of the diverging lens is -6.7 dioptres.

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