Image distance / focal length relation

In summary, when projecting an image with an object (O), lens (L), and screen (S), the focal length of the lens (f) must be less than one-fourth of the distance between the object and screen (OS) for a clear image to be formed on the screen. This can be proven by setting the determinant of the quadratic equation for the image distance (d_i) to be positive, resulting in the equation s=\frac{{d^2}_o}{d_o-f}.
  • #1
jmcgraw
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Let O be the object, L be the lense, I be the image, and S be a screen for projecting an image. The lens can be placed anywhere between O and S. Let f be the focal length of the lense. Prove that for an image to be clearly formed on screen S, OS > 4f must be true.

A figure of a "too short" OS distance (since the image would be past the screen) would look like this:

O-----L-----------S----I


I am having a very hard time proving that OS>4f must be true! Any hints?
 
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  • #2
For the image distance [itex]d_i[/itex] subs [itex]s-d_o[/itex] where s is the distance between object and screen. After a bit of work you should get to
[tex]s=\frac{{d^2}_o}{d_o-f}[/tex]
which gives you a quadratic equation in [itex]d_o[/itex]
By requiring that the determinant should be positve the result follows.
 
  • #3


I understand the importance of providing evidence and proof for any claims or statements. In this case, we are looking at the relationship between image distance and focal length. Let us first define some terms for clarity:

- O: Object
- L: Lens
- I: Image
- S: Screen
- f: Focal length

We are given that the lens can be placed anywhere between O and S. This means that the distance between the object and the lens, OL, can vary. Similarly, the distance between the lens and the screen, LS, can also vary.

In order for an image to be clearly formed on the screen, we need to ensure that the image is not too small or too large. This means that the image should be the right size to be projected onto the screen.

Now, let us consider the case where the image distance is equal to the focal length of the lens (I = f). In this case, the image would form at the focal point of the lens, which is not ideal for projecting onto the screen. The image would either be too small or too large for the screen, depending on the size of the object.

To have a clearly formed image on the screen, we need to adjust the distance between the lens and the screen, LS. For an image to be projected onto the screen, the image distance (I) must be equal to the distance between the lens and the screen (LS). This can be represented as I = LS.

Using the thin lens equation, we can also write this as:

1/f = 1/OL + 1/LS

Now, let us consider the case where the image distance is greater than the focal length (I > f). In this case, the image will form beyond the focal point of the lens, which is what we want for projecting onto the screen. However, if the distance between the lens and the screen is too short, the image will be projected past the screen, resulting in an unclear image.

In order to avoid this, we need to ensure that the distance between the lens and the screen is greater than the focal length of the lens. This can be represented as LS > f.

Combining this with the earlier equation (I = LS), we get:

I > f
I = LS
LS > f

Therefore, for an image to be clearly formed on the screen, it is necessary
 

FAQ: Image distance / focal length relation

1. How is the image distance related to the focal length of a lens?

The image distance and focal length are inversely related. This means that as the focal length increases, the image distance decreases, and vice versa. This relationship is described by the formula: 1/f = 1/di + 1/do, where f is the focal length, di is the image distance, and do is the object distance.

2. Can the image distance be greater than the focal length?

No, the image distance cannot be greater than the focal length. The image distance can only be equal to or less than the focal length, depending on the position of the object.

3. How does the focal length affect the size of the image formed?

The focal length of a lens determines the magnification of the image. A longer focal length will result in a larger image, while a shorter focal length will result in a smaller image. This is because a longer focal length allows for a wider range of angles to be captured, resulting in a larger image.

4. What is the difference between a positive and negative focal length?

A positive focal length indicates a converging lens, while a negative focal length indicates a diverging lens. Converging lenses have a thicker center and focus incoming light rays to a point, while diverging lenses have a thinner center and cause incoming light rays to spread out.

5. How does the position of the object affect the image distance in relation to the focal length?

The position of the object relative to the focal length determines whether the resulting image will be real or virtual. If the object is placed beyond the focal length, a real image will be formed on the opposite side of the lens. If the object is placed within the focal length, a virtual image will be formed on the same side of the lens as the object.

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