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I read in a AQA GCSE book that:

A real image is formed by the converging lens if the object is further away from the principal focus/focal point.

I did this experiment in class: Diagram

Here is my table of results: Results

However, from the quotation above, u needs to be larger than v.

  • So why in my experiment did I get that u is 15cm and v is 30cm?

Here is what I did in my experiment:

A crossed wire object was placed in front of the ray box. After that, the converging lens was placed 15 cm in front of the ray box. The screen was moved until a clear, real image was visible. The distance from the lens to the screen was measured and recorded. This process was repeated with the following measurements (from the ray box to the lens): 20 cm, 25 cm, 30 cm & 50 cm.

Thanks in advance

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  • $\begingroup$ Farther away from what? $\endgroup$ – DJohnM Oct 21 '13 at 0:16
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The equation you need to get the distance of the image from the object and focal length is:

$$ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} $$

where $f$ is what your question describes as the principal focus/focal point.

A bit of playing around with this should convince you that if $u = 2f$ then $v = 2f$ i.e. $u$ and $v$ are identical. If $u > 2f$ then $v < 2f$ and $u$ is greater than $v$. Conversely if $u < 2f$ then $v > 2f$ and $u$ is less than $v$. So, as your data shows, $v$ can be greater or less than $u$.

However for the image to be real $u$ must be greater than $f$. If you put in a value of $u$ less than $f$ you'll find $v$ comes out negative, which with the sign convention I've used means the image is to the left of the lens i.e. a virtual image.

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Well if you use Newton's form for the thin lens formula, you have (x) is the distance from the object to the front focal point, (+ve left to right) and (y) is the distance from the second focal point to the image, then x.y = f^2 which gives a minimum object to image distance of 4f for x = y = f when both x and y are positive you have a real image, and when both are negative, you have a virtual image. In general the magnification is y/x.

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  • $\begingroup$ Interesting though it is too difficult for me at the moment! $\endgroup$ – Turbo Nov 1 '13 at 22:38
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The phrase you quote sounds ambiguous. I suspect they actually mean the following: you get a real image with a converging lens when the distance between the object and the lens is greater than the lens' focal length.

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  • $\begingroup$ Here you are saying that u>v. But for one of mine results was 15<30. Why is this so? The teacher said we were making real images. $\endgroup$ – Turbo Oct 20 '13 at 18:12
  • $\begingroup$ I am saying nothing of the kind. I am saying u>f. $\endgroup$ – akhmeteli Oct 20 '13 at 18:30
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To get a real image you need to have the object distance $u$ greater than the focal length of the lens $f$.
If that condition is satisfied then if $u<v$ where $v$ is the image distance, the image is real, inverted and magnified.

For every combination of object distance $u$ and image distance $v$ there is another combination of these distances which form a real, inverted but diminished image.
This occurs when the object and image distances are interchanged.
So the old image distance becomes the new object distance and vice versa.
You can think of this as an example of the reversibility of light.
So you can have $u>v$ as well as $v<u$.

Note that your $u=15,\;v=30$ and $u=30,\;v=15$ data values are, within experimental error, an example of the reversibility.

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