Same focal length = same angle of view?

<p>Does same focal length = same angle of view, or is there something else involved in determining what a lens shows you? It seems like there must be another factor involved in determining angle of view, because you get a huge angle of view from fisheye lenses even when their focal length is longer than that of normal lenses (like the Nikon 16mm fisheye vs. the Nikon 14-24mm at 14mm).</p>

<p>So does this same factor create differences in what you see among ordinary, non-fisheye lenses, too? For example, will the Nikon 14-24mm, the 24-70mm, and the 24-120mm, all set to 24mm, and the 16-85mm DX lens at 16mm (with an <em>effective</em> FX focal length of 24mm), all give you the same angle of view, or are there differences?</p>

<p>Fish-eye lenses are very different beasts. They are designed to be wierd. The other lenses, on the other hand should all look the same at the same focal length. The only problem is that some of the zooms are not dead accurate on their focal length, sometimes. They may be a few mm off at one end or the other.</p>

 

<blockquote>

<p>The angle of view is determined by the type of lens on the camera, its focal length and the sensor size. </p>

</blockquote>

<p>Among all the lenses you listed, most are rectilinear lenses. They show straight lines as straight lines on the sensor. The 15mm lens you mentioned is not a rectilinear lens. It will show straight lines as curved lines. It is called a "Fisheye" lens for this reason. Fisheye lenses generally have a wider field of view than rectilinear lens of the same focal length. </p>

<p>Focal length is the main facter determining what you see. Short focal lengths provide a wider field of view than long focal lengths. The sensor size is the final determiner of what you see. Smaller sensors have a smaller angle of view than larger sensors. </p>

 

<blockquote>

<p>So does this same factor create differences in what you see among ordinary, non-fisheye lenses, too? For example, will the Nikon 14-24mm, the 24-70mm, and the 24-120mm, all set to 24mm, and the 16-85mm DX lens at 16mm (with an <em>effective</em> FX focal length of 24mm), all give you the same angle of view, or are there differences?</p>

</blockquote>

<p>These lenses will all give the same field of view at 24mm on a FX camera as will the 16mm DX lens on DX camera. </p>

<p >The angle of view of a lens is a straight math problem. However there is a confusing aspect, somewhat like buying a TV set. You admire a TV with a screen that measures 10 inches height by 16 inches width but you are sold a 19 inch set. That’s because the TV industry has settled on quoting the diagonal measure ignoring the height and width as sales points.</p>

<p > </p>

<p >Cameras fall in the same category. We are always told the diagonal angle of view. The Vertical and Horizontal values are withheld. </p>

<p > </p>

<p >For the 35mm full frame the dimensions of the format are 24mm vertical 36mm horizontal 42.3mm diagonal.</p>

<p > </p>

<p >Fit this format with a 50mm lens and the angles of view will be:</p>

<p >27° Vertical -- 40° Horizontal 47º Horizontal</p>

<p > </p>

<p >Fit this format with a 25mm lens and the angles of view will be:</p>

<p >50° Vertical -- 72° Horizontal 82º Horizontal</p>

<p > </p>

<p >Fit this format with a 100mm lens and the angles of view will be:</p>

<p >14° Vertical -- 20° Horizontal 24º Horizontal</p>

<p > </p>

<p >For the smaller D-SLR senor the dimension of the format are 16mm vertical 24mm horizontal 28.8 diagonal.</p>

<p > </p>

<p >Fit this format with a 50mm lens and the angles of view will be:</p>

<p >18° Vertical -- 27° Horizontal 32º Horizontal</p>

<p > </p>

<p >Fit this format with a 25mm lens and the angles of view will be:</p>

<p >35° Vertical -- 51° Horizontal 60º Horizontal</p>

<p > </p>

<p >Fit this format with a 100mm lens and the angles of view will be:</p>

<p >9° Vertical -- 14° Horizontal 16º Horizontal</p>

<p > </p>

<p > </p>

<p >Below is Excel</p>

<table border="0" cellspacing="0" cellpadding="0" width="415">

<tbody>

<tr >

<td width="64" valign="bottom">

<p align="right">1</p>

</td>

<td width="156" valign="bottom">

<p align="center">B</p>

</td>

<td width="64" valign="bottom">

<p align="center">C</p>

</td>

<td width="131" valign="bottom">

<p align="center">D</p>

</td>

</tr>

<tr >

<td width="64" valign="bottom">

<p align="right">2</p>

</td>

<td width="156" valign="bottom">

<p >Focal length mm =</p>

</td>

<td width="64" valign="bottom">

<p align="right">50</p>

</td>

<td width="131" valign="bottom">

<p > </p>

</td>

</tr>

<tr >

<td width="64" valign="bottom">

<p align="right">3</p>

</td>

<td width="156" valign="bottom">

<p >Height format mm =</p>

</td>

<td width="64" valign="bottom">

<p align="right">16</p>

</td>

<td width="131" valign="bottom">

<p align="right">18.2</p>

</td>

</tr>

<tr >

<td width="64" valign="bottom">

<p align="right">4</p>

</td>

<td width="156" valign="bottom">

<p >Length format mm =</p>

</td>

<td width="64" valign="bottom">

<p align="right">24</p>

</td>

<td width="131" valign="bottom">

<p align="right">27.0</p>

</td>

</tr>

<tr >

<td width="64" valign="bottom">

<p align="right">5</p>

</td>

<td width="156" valign="bottom">

<p >Calculate diagonal mm</p>

</td>

<td width="64" valign="bottom">

<p align="right">28.84</p>

</td>

<td width="131" valign="bottom">

<p align="right">32.2</p>

</td>

</tr>

</tbody>

</table>

<p > </p>

<p >Formula cell D5</p>

<p >=((ATAN((C5/2/$C$2)))*180/PI())*2</p>

<p>As I'm attempting to make sense of the math (why - I don't know!) it seems to me something is left out. What about distance from the rear element to the sensor? Wouldn't the AOV change with this distance? Or, would that be the FOV? Are FOV and AOV two different animals?</p>

<p>Distance to the sensor or film plane doesn't change anything by angle. What it does require that the sensor size, film area, etc., be increased or decreased proportionally. If you don't proportionally change the film/sensor, you get a crop (or empty space/lesser performance?). Consider that the "projected image is a quadrilateral pyramid (or cone if you will). As you move up or down the cone, slicing of the base or adding height/distance, as long as you slice straight across the central axis, nothing changes.</p>

<p >The focal length of a simple (single element symmetrical) lens is a measurement made from the center of the lens to the focal plane measured when the lens is focused on infinity. Infinity is an object (disk) viewed from a distance of 3000 times its diameter whereby is seen (resolved) as a point not a disk.</p>

<p > </p>

<p >A modern complex lens is constructed of several elements. Further some elements will be positive (converging) and some will be negative (diverging). The accumulated power of the array will be positive (converging). The geothermic center of the array will likely not be a valid point from which to take measurements and establish the focal length.</p>

<p > </p>

<p >In a true telephoto design this (measuring) point may fall far forward of the lens. Such a design shortens the barrel as apposed to a long lens whereby the measuring point is in the vicinity of the geothermic center. A wide-angle design likely will require the rear element to be quite close to the focal plane. This is unacceptable if the camera is an SLR with moving mirror. Thus short wide-angles are actually inverted or reversed telephoto’s. This is achieved by mounting the diverging components forward. This design allows the back focus to be elongated; now the rear element clears the reflex mirror. </p>

<p > </p>

<p >All of the above generates a point called the rear nodal. Again the rear nodal can be at barrel center or fore of aft. Nevertheless it is the rear nodal from which measurements are made. One can simply draw (ray trace) the angle-of-view using the rear nodal as the meeting point for the lines of the trace. The formula in my post above yields authentic angles of view. It takes into account the focal length which is measured from the rear nodal and the size of the film or digital chip. </p>