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## Mathematics of depth of field

Current time:0:00Total duration:2:44

# Mathematics of pinhole cameras

## Video transcript

- In the last lesson, we explored the geometric behavior of cameras. We saw how light bounces off objects, passes through a small hole or aperture, and hits an image plane. One of the important concepts we covered was depth of field,
which is a region where objects appear in focus. Outside that region,
objects appear blurry. When a point at an image is out of focus, it grows into a blurry circle known as the circle of confusion. In this lesson, we'll develop
the algebraic equations which tell us exactly where
things come into focus, and how big the circle
of confusion will be if something isn't in focus. And these equations
will allow us to create virtual cameras to create stunning images, such as this shot. To begin, let's return
to our pinhole camera. As usual, when bridging the gap between geometry and algebra, we'll need to introduce
a coordinate system. It's convenient to choose
our coordinate system so that the pinhole is at
the origin, right here. And let's imagine that our
scene is off to the right. Suppose the image plane inside our camera is at some distance to
the left of the pinhole. Let's call this distance i. Now consider a point on
some object in our scene. Call this point x zero, y zero. When light in the
environment hits this point, some of it will bounce toward the camera, through the pinhole and
hit the image plane. Let's call the place it hits
the image plane x one, y one. So our first question is
what is x one and y one? One way to solve this is to use the slope intercept form of a line. The slope of the ray
is y zero over x zero, and the y-intercept is zero because we said this ray
passes through the origin. That is, the equation of the ray is just y equals y zero over x
zero, times x, plus zero. Or, simply, y equals y
zero over x zero, times x. Now the point we're
looking for, x one, y one, is also on this ray, so it must
satisfy that line equation. Meaning y one equals y zero
over x zero, times x one. And notice the diagram tells
us that x one is negative and a distance i away from the origin. That is, we know x one equals minus i. Finally, to get y one, we
just substitute to give us y one equals minus y zero
over x zero, times i. Notice that the point
was originally positive but the corresponding point
on the plane is negative. And that's the image flip. Okay, let's pause here to give
you some experience with this before we add a lens to our camera.