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Geometric Optics

An optical image is formed when all the rays from a point of an object are brought back together at a point. This is shown in the figure below. The figure shows two of the many, many rays that emanate from the tip of the tree diverging until they impinge upon the lens. The lens refracts the rays changing their direction slightly. The refracted rays then come together at a point behind the lens. The same thing happens with the rays coming from every point on the front of the object tree. The collection of points where the divergent rays from all the points of the object tree come back together is the optical image.


The Lens Equations

Consider the rays from one point of an object which pass through a lens. Two such rays can travel two paths to a point in the image, as is shown in the diagram below.

Now consdier the relationship of the triangles on the left of the lens:



BE/DC = BC/DF = CE/FC
where
BE = BC + CE
 

From this it follows that


BC = DF(CE/FC)
 

From the triangles to the right of the lens


CE/GI = BC/CG
and hence
BC = CG(CE/GI)
 

Combining the equations for BC gives


BC = DF(CE/FC) = CG(CE/GI)
and therefore, after cancelling the CE
DF/FC = CG/GI
or, equivalently
DF*GI = FC*CG
 

The points F and G are called focal points. The previous equation says that the product of the distances between the object and image to the focal points is equal to the product of the distances of the focal points to the lens (center). This is known as Newton's form of the lens equation.

There are other relationships that can be derived from the diagram. The ratio IH/AD is known as the magnification m. From the diagram it is seen that


m = IH/AD = CE/BC
but from the similarity of the triangles
m = CE/BC = FC/DF
m = CE/BC = CG/GI
 

Also from the similarity of the triangles


BE/CE = DC/FC
and therefore
BE = CE(DC/FC)
 

From the triangles on the right


BE/BC = CI/CG
and hence
BE = BC(CI/CG)
 

Therefore


CE(DC/FC) = BC(CI/CG)
and therefore it follows
m = CE/BC = (FC/CG)(CI/DC)
 

Then


(BC+CE)/CE = BC/CE + 1 = (1/m) + 1
= (DC/CI)(CG/FC) + 1
and also
(BC+CE)/CE = DC/FC
therefore
DC/FC = (DC/CI)(CG/FC) + 1
 

Dividing this equation by (DC/FC) gives


1 = (CG/CI) + (FC/DC)
 

DC and CI are the distances from the object and the image to the lens, which hereafter are denoted as s and s', respectively. FC and CG are the distances of the focal points to the lens and they are hereafter denoted as f and f', respectively. The previous equation can therefore be expressed as


f/s + f'/s' = 1
 

This is known as Gauss's form of the lens equation.

If the lens is symmetric then f and f' are equal and this length is known as the focal length of the lens. For this case, Gauss' form of the lens equation is the more familiar


1/s + 1/s' = 1/f
 

The Lens Equation for a Concave Lens

The diagram below shows the relationship of the object and image for a symmetric concave lens.

The lens equation applies to this case but only if the proper sign convention is adhered to. Since for the concave lens the image is on the same side of the lens as the object the distance from the image to the lens is taken to be a negative number. The so-called magnification, which may be a reduction rather that a magnification, is also a negative number.

A Compound Lens

(To be continued.)


Geometric optics is the derivation of equations and relationships for optics that presume the angles involved in ray tracing are small enough that the approximation that sin(θ)=θ may be used. Rays that are nearly parallel to the optical axis are called paraxial.



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