# What would be the optimal distance of the object from the lense given the camera spec

I am not a photography expert hence i would like for some advice.

The camera im using is a small camera that will be taking pictures of an object at a fixed point, my problem is identifying where this fixed point should be. What would be the optimal distance my object should be from the lense? Here are the camera specs:

• Optical Size: 1/2.43"
• Focus: Motorized
• Depth of Field: approx 10cm to infinity
• focal length: 4.74mm
• Horizontal FOV: 66 degrees
• Vertical FOV: 41 degrees
• Focal Ratio: F1.8
• Max exposure: 112 seconds

I want to give my camera the best chance of taking the clearest picture it can get with the object. I would assume that the best place is at 10cm? assuming that my object fits in the frame. If it doesnt i just have to move it back a little until it will fit inside the frame

Is this correct?

• I rolled back your edit for a couple reasons: first, you should only have one main question in posts here; second, you already have answers, so the additional question makes them incomplete; finally, that question has been asked and answered here many times before.
– scottbb
Jan 11 at 19:26
• Regarding finding image size from angle of view, this question is the inverse math of your question: Can I calculate sensor size from pixel dimensions and number (to find angle of view)?
– scottbb
Jan 11 at 19:27

A depth of field that extends 10cm to infinity would suggest a lens focused at no less than 20cm.

Since the lens has variable focus the optimal distance would be the minimum focus distance, which must be 20cm (or farther). Or the minimum distance where the subject fits w/in the field of view... just move it back as you stated.

It is odd though that a DoF is defined for a variable focus lens as the DoF will vary depending at what distance the lens is focused. However, the DoF extending to infinity means that the minimum focus distance must be no less than the hyperfocal distance (2x minimum DoF limit).

The caveat to all of this is that we don't know what standard was applied to determine the camera's depth of field...

And there is the possibility that the defined DoF is misidentified and should be focus range as noted in the comments by @scottbb... in that case the optimal distance is still the minimum focus distance, only that would then be 10cm instead of 20cm(+).

• As I replied to your comment under my answer, it's very possible that the "DOF" stated in the specs is really misnamed, and should be something like "Focus range", where it can focus from 10 cm to infinity. That makes a lot more sense to me than some specified DOF. Especially knowing that the camera's focus is motorized.
– scottbb
Jan 11 at 19:35
• I somehow missed that the focus was motorized... I'll have to edit my answer to clarify. Jan 11 at 21:16

Assuming you want to maximize the size of the object in your image, then you should place the object close enough to fill the frame of the camera, but no closer than the minimal focus distance of approximately 10 cm.

Your camera's sensor is a 4:3 aspect ratio, meaning there are 4 horizontal pixels for every 3 vertical pixels (when in the landscape orientation). If your object more than 1.33 times wider than it is tall (i.e., > 4/3 width/height), you should frame it for maximal width. Conversely if the object is more than 75% taller than it is wide (i.e., > 3/4 height/width), you should frame it for maximal height.

A 1/2.43" sensor format is about 6.17 mm wide by 4.55 mm tall. It's confusing, but '1/x"' format is not an actual measurement; it's just a name/number, that doesn't specifically indicate a real world dimension. See: Why is a 1" sensor actually 13.2 × 8.8mm? for an explanation.

The magnification $$\M\$$ of a subject's size $$\h_\mathrm{o}\$$, compared to its size in the image sensor $$\h_\mathrm{i}\$$ is:

$$M = {h_\mathrm{i}\over h_\mathrm{o}}\qquad[1]$$

Now, you probably want to leave some head/bottom room around the object, so give it say, 10%, by multiplying $$\M\$$ by 0.9.

Another formula for magnification relates the lens's focal length $$\f\$$ and the subject distance $$\d_\mathrm{o}\$$ (roughly speaking, from the center of the lens to the subject):

$$M = {f\over f-d_\mathrm{o}}\qquad[2]$$

Setting eqs. [1] and [2] equal to each other (and multiplying the right hand side of eq. [1] by the 0.9 "headroom factor") and solving for $$\d_\mathrm{o}\$$:

\begin{align} {f\over d_\mathrm{o}-f} &= {0.9\times h_\mathrm{i}\over h_\mathrm{o}} \\ {d_\mathrm{o}-f\over f} &= {h_\mathrm{o}\over 0.9\times h_\mathrm{i}} \\ d_\mathrm{o} &= f\left({h_\mathrm{o}\over 0.9\times h_\mathrm{i}} -1\right) \qquad [3] \end{align}

For example, with your camera's lens of 4.74 mm, and assuming we're maximing the object's height of say, 10 cm (100 mm), in the sensor, then your object should be placed about

\begin{align} d_\mathrm{o} &= 4.74\,[\text{mm}]\left(\frac{100\,[\text{mm}]}{0.9\times 4.55\,[\text{mm}]} - 1\right) \\ &= 4.74\,[\text{mm}] \times 23.4 \\ &= 111\,[\text{mm}] \end{align}

111 millimeters (11.1 cm), or approximately 4.4 inches. Note that this is pretty close to your minimal focus distance of approx. 10 cm (Note: I'm assuming that the stated "depth of field" is misnamed and not really depth of field; rather, I assume it's probably the focus distance range of the lens). Therefore, objects smaller than this example's 10 cm high can't be magnified any larger. However, objects larger than 10 cm can be moved farther away from the lens and maximally magnified according to eq. [3].

• This is great ! thank you so much. A following question in mind immediately poped up. How would i calculate the frame size at a given distance. So assuming the 11.1cm optimal distance, what is the size of the object there? In your example the so at 11.1cm i can fit a 10cm height by 13.33cm width (since it is a 4:3 ratio) is this correct? Jan 11 at 10:37
• The only caveat is that the minimum DOF spec (10cm) isn't the minimum focus distance. That is the minimum "acceptable focus" distance; the true minimum focus distance (actually in focus) will be at least 2x that (20cm). Jan 11 at 13:11
• @DrakeJest the frame size is what it is. That is, a 1/2.43" sensor has a fixed size. So just look up the specs.
– scottbb
Jan 11 at 19:23
• @DrakeJest Also see, Is the formula for object image size given focal length, etc. independent of sensor size? for the simple math regarding your frame size question
– scottbb
Jan 11 at 19:32