[Physics] Is it possible to calculate image distance by knowing magnification and focal length

lensesoptics

There is an equation in Physics about photography which is all about lenses.

${\rm magnification} = \frac{\rm image\,distance}{\rm object\,distance}$

Assuming we know the magnification, in order to calculate distance of object we need to know distance of image in the lens.

However, I couldn't find any information on web regarding to what image distance is equal to in a photographic lens.

It should be between $F$ and $2F$ (knowing photographs are taken beyond $2F$), and in between the lens glass and sensor of the camera.

How can we make sure about the distance of image in the camera?

Best Answer

Treating the camera lens as a simple thin lens (this is an approximation), then the image distance, $v$ and the object distance, $u$, are related by$$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$$in which $f$ is a constant for the lens, called its focal length.

Focusing the camera amounts to varying the distance, $v$, between the lens and the light-sensing ccd-array in order to satisfy the equation, for the given distance, $u$ of the object from the lens. So the image distance isn't fixed, as you seem to be suggesting.

Related Solutions

[Physics] Compound lenses and virtual images

Most likely, your problem is the sign of the distance from the image of the object through the first lens, to the second lens. When you use the formula for the ideal thin lens you must be really careful about the convention used.

For example, I usually use $\frac{1}{f} = \frac{1}{o} + \frac{1}{i}$, where $f$ is the lens' focal distance, $o$ is the distance from the object to the lens, and $i$ is the distance from the lens to the image. The sign convention for this formula is

letter   positive           negative
o        left of the lens   right of the lens
i        right of the lens  left of the lens
f        converging         diverging

This means that the reference system is different for $o$ and $i$, which can be confusing.

Other common error for this kind of problems is that $o$ and $i$ is the distance to the lens, so you can't use the number you get for $i$ from the first lens as the $o$ for he second lens. You need to take into account the distance between the lenses.

Optics – How Focal Lengths Impact Magnification

It's like leverage. The longer the distance from the objective lens to the virtual image, the larger the virtual image.

Imagine there's a piece of frosted glass at the focal point. It will show the virtual image.

Now the eyepiece looks at that virtual image with a magnifying glass. That also makes it look bigger.

Best Answer

Related Solutions

[Physics] Compound lenses and virtual images

Optics – How Focal Lengths Impact Magnification

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