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Investigating the Power of a Diverging Lens

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Introduction

Christopher Yu

Investigating the Power of a Diverging Lens

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.
...read more.

Middle

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.

image00.png

image01.png

Where image02.pngandimage03.png but, because the formula can also be arranged thus,

image04.png

That means that image05.pngwill be both the y-intercept andthe 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

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Conclusion

± 0.001

0.350

2.230

2.270

2.250

-6.69

± 0.001

 Then the error in the object, image minimum, image maximum and image mean length will be ±0.001 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

±0.040

-6.670

-6.700

-6.685

±0.015

-6.570

-6.620

-6.595

±0.025

-6.420

-6.460

-6.440

±0.020

-6.690

-6.700

-6.695

±0.005

Average

-6.625

±0.021

Therefore the power of the diverging lens is –6.625 dioptres with an error of ± 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.

...read more.

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