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Given a Tube Containing a Lens, Calculate The Focal Length of The Lens and Where The Lens Lies Within The Tube.

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Introduction

Given a Tube Containing a Lens, Calculate The Focal Length of The Lens and Where The Lens Lies Within The Tube.

Plan

I plan to set up the apparatus as shown below and change ‘b’. To then make the image focus, I will have to change ‘c’. When the image is focused, I will measure ‘c’ and also the magnification of the image. Using this information I will then be able to calculate ‘a’.

image00.png

Hypothesis

        I predict that using the equations stated in the justification, after measuring values for ‘c’ and ‘m’. I will be able to plot them on the graph of ‘c’ against ‘m’

Using the equation c = fm + [f - a]. I believe that if all the equations work, a straight line will be obtained.

Safety

        After considering all aspects of this experiment, I have come to the conclusion that there are no significant safety issues. If normal levels of care are exercised, then there should be no risks involved.

Variables that alter

  • b and c, (from diagram)
  • Magnification

Variables to control

  • a (from diagram)
  • Ambient light
  • Part of object for which magnification is measured each time.
  • Place b is measured from and to

Apparatus list

  • 0.15m tube
  • Lens
  • Object                                                                                                                                        
  • 12V Power pack
  • Light source (lamp) white light average wavelength 550Nm
  • White screen
  • 2 x 1m rulers (0.0001m increments)
  • 0.15m ruler (0.0001m increments)
  • Stand and clamp
  • Spirit level
...read more.

Middle

2.50

0.291

0.380

0.336

±0.045

3.15

0.340

0.437

0.389

±0.049

3.65

0.397

0.519

0.458

±0.061

From this and the graphs,         

f = gradient = dy / dx = 0.315 / 0.0275 =  0.115m

        a = gradient – y-intercept = 0.115 – 0.045 = 0.07m

        Error for c = Σ errors / 7.00 = ± 0.0442

        Error for m = ± 0.000500m

Experiment 2                        

‘c’ in m

Magnification

Minimum

Maximum

Average

Error

0.50

0.980

0.141

0.120

±0.015

1.15

0.165

0.206

0.186

±0.021

1.50

0.211

0.268

0.240

±0.029

2.15

0.252

0.311

0.282

±0.030

2.50

0.296

0.394

0.345

±0.049

3.00

0.342

0.442

0.392

±0.050

3.60

0.397

0.504

0.451

±0.054

From this information and from the graph, this gives:

f = gradient = dy / dx = 0.290 / 0.0260= 0.112 m

        a = gradient – y-intercept = 0.112 – 0.0550 = 0.0570m

        Error for c = Σ errors / 7.00 = ± 0.0361

        Error for m = ± 0.000500m

Experiment 3

‘c’ in m

Magnification

Minimum

Maximum

Average

Error

0.50

0.083

0.127

0.105

±0.022

0.95

0.127

0.179

0.153

±0.026

1.50

*0.187

0.239

0.213

±0.026

2.00

0.231

0.304

0.268

±0.033

2.45

0.286

0.344

       0.315

±0.029

3.15

0.357

0.434

0.396

±0.039

3.50

0.391

0.477

0.436

±0.042


From this information and from the graph, this gives:

f = gradient = dy / dx = 0.275 / 0.250 = 0.110m

        a = gradient – y-intercept = 0.110 – 0.0450 = 0.0650 m

Error for c = Σ errors / 7.00 = ± 0.0310m

Error for m = ± 0.000500m

* The original reading for this measurement was 0.112. I saw

...read more.

Conclusion

        The experiment should be conducted all in one session, as this will reduce the likelihood of the ambient light changing and affecting the results.

        An obvious two anomalies in my work are in experiment 2, with results number 3 and 4. Number three has either a too small magnification or ‘c’ was too large. Number 4 has the opposite problem. This could have been because of inaccurate readings of the measurements, changes in ambient light or a problem with the equipment used. There were no obvious discrepancies between my results and the results of others with the same lenses and so I have no reason to believe that my results are inaccurate

Improvements and further work

        If the experiment were to be done again, it would have to be conducted under complete blackout conditions. Also, the light source would need to be further away and more intense so that the light rays are closer to parallel. The tube containing the lens should be moved instead of the crosshair to avoid changing the angle of the light rays when they reach the crosshair. I could also use an object, which was easier to focus to narrow down the range of values it is focused for and use a lens free from imperfections. A monochromatic light source would also make the results more accurate.         

I could also do the experiment with a larger range and do more repeats.

...read more.

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