3

There are a number of intricacies here which would take hours to get into. Here are a few quick answers: First, define the % contrast at which you are specifying the MTF. The value "145lp/mm" should be specified with a percentage, such as MTF of 20%@145lp/mm. Most people use 10-30% image contrast as the minimum resolvable value. This is subjective ...


3

Technically, the white lines are your image formed by light; the black lines are a lack of light. And what you need is 290 airy disks/mm; which is airy disks of 3.45um in diameter, which requires a perfect lens at f/2.6 or an extremely good lens at f/2 or less. An airy disk looks something like this where the height equates to the illumunance of the light. ...


3

Since your field ov view is square, you only have to worry about the smaller dimension of the sensor, its height. With H and D: height and distance of object, and h and f: height of sensor and focal length: D/H = f/h so f = h * D/H So f = 3.6 * 10/1 = 36mm.


3

It is not possible to calculate the sensor size with that information because the 1/3" measurement refers to the diameter of a hypothetical tube that would enclose the sensor. That sets a limit on the diagonal of the sensor but it not an exact fit. The the sure way to know the sensor-size is to measure it or find the part number and look at ...


2

I found this at the Digital Camera Database Another list is at the Wikipedia article for Image sensor format Which both say for 1/3": 4.8 x 3.6 mm


2

Did I make a correct assumption regarding finding the sensor dimensions using similar triangles? Yes. But you don't need to worry about them. All that matters is the sensor diagonal (16.1 mm in your case), because the lens projects a cone of light that lands on your sensor as an image circle. As long as the sensor diagonal fits within the lens's image ...


2

This is actually a simple ratio problem: We can trace imaginary lines from the boundaries of the object being imaged to the center of the lens. A triangle is traced. The height of this triangle is the object distance = 10 meters = 10,000mm. The base of this triangle is the object’s width = 1 meter =1000mm. The ratio of height to base is 10000 ÷ 1000 =10.0 ...


2

For the same FOV and f-stop, will total luminous flux increase linearly with sensor area? Yes, total luminous flux will increase. But photography doesn't measure exposure or brightness¹ using total luminous flux. It bases exposure on light per unit area. I.e., for the same FOV and f-stop, would the large-sensor camera only need 1/2 the exposure time to ...


1

I'll try to answer based on your second alternative. ISO values were defined using the chemical film model, defining which film exposure (in lumen-seconds per square meter) gives a decent gray or color level. The benefit of ISO values is that you don't need object distance, focal length, sensor size or pixel size to determine the correct exposure, only the ...


1

The part you are missing is the Inverse Square Law (ISL). The inverse square law states that if the distance of a *point light source is 2x, then the density/luminous flux of the light is 1/2^2 (1/4). The opposite of that is at 1/2 the distance it is 4x. So two stops difference, not one. I.e. not 1/2 the SS (which is incorrect anyway). The inverse square law ...


Only top voted, non community-wiki answers of a minimum length are eligible