GEOMETRICAL OPTICS: Thin Lens Equation Derivation

Of all the various equation derivations one may encounter as an introductory physics student, the ones regarding ray diagrams are as counterintuitive as any. Whether a system at hand consists of thin lenses or mirrors, radiant energy is diagrammed as rays for the sake of simplicity. Never, ever forget that these diagrams are grossly oversimplified depictions of outwardly diverging packets of energy ( photons ) and momentum. In order to justify usage of such visually oversimplified diagrams, an assumption is made that the rays pictured are those that have traveled a considerable distance from the source that omitted them. Under such circumstances, a spherical wavefront will have expanded so much that any small portion of the front that is encountered far away will essentially consist of a planar ( flat ) wavefront. As symmetry would have it, the rays at the boundaries of such an imaginary wavefront are parallel to one another: 

Fortunately, the laws of physics enable parallel rays to converge again if they move through an appropriate medium; thus, a lens that bends rays inward towards a central axis is either called a convex or converging lens. The degree to which a convex lens bends such rays is both a function of a lens’ shape and its optical characteristics, most notable of which is its index of refraction ( n ). In-depth discussions of a material’s refractive index take place during studies of optical scattering. For now, let’s consider a system in which a convex lens focuses light from a distant object:

There are two distances in the diagram that are of importance to us. First, consider the focal length ( f ) on either side of the lens. As we will see, a parallel ray of light emitted from the top of an object of height ( h_o ) positioned left of the lens will be bent inward by the lens so that it passes through the focal length ( f_i ) on the system’s image side. Maximum symmetry between an object’s height and the image’s inverted height ( – hi ) exist when the object is placed at a distance that is twice that of the focal length:

d_o = 2f_o

d_i = 2f_i

Prior to visually confirming this claim to be true, let’s look at a ray that passes from the top of the object and through the center of the lens at hand:

As can be seen, the ray in question moves through the lens undeflected until it reaches the “ top “ of the image that has been formed. As a consequence, we now have two triangles that are related by an angle ( θ ) that they share with the central axis:

Since the distance that separates the object and image are known, as well as the heights associated with each, we use the tangent ( θ ) function to establish a relationship between the left and right side of the system:

tan ( θ ) = ( opp / adj )

tan ( θ ) = ( h_o / d_o ) = ( – h_i / d_i )

We are now ready to add the ray that is bent inward by the lens to the diagram:

Let’s now take a very close look at the right-hand side of the system ( rightward of the lens ). The ray that dissects the central axis at the image’s focal point ( f_i ) creates a new set of symmetrical triangles. This newer pair of triangles share an angle ( ɸ ) with the central axis:

We will now make use of the tan ( ɸ ) function as it relates the two newer triangles, but we must be somewhat creative in our approach. The adjacent side of the first triangle in question is the focal length of the image; however, the adjacent side of the inverted triangular “ twin “ of the pair will have an adjacent side defined in the following manner:

adj = ( d_i – f )

Taking this new derivation into account, the tangent ( ɸ ) and previous tangent ( θ ) relationship can be used to confirm the symmetry of the right-triangle pairs associated with angles ( θ ) and ( ɸ ):

Note: At this point, whether we keep the subscripts attributed to the focal length in question can be omitted for the sake of simplicity.

Alas, we have finally arrived ( almost ) at our destination:

( h_o / – h_i ) = [ f / ( d_i – f ) ]

And,

( h_o / – h_i ) = ( d_o / d_i )

We can now set these equations equal to one another:

( d_o / d_i ) = [ f / ( d_i – f ) ]

We now multiply both sides of the equation by [ ( 1 / d_o ) / ( 1 / d_i ) ] to get all of our variables onto the right-side of the equation:

[ ( 1 / d_o ) / ( 1 / d_i ) ]( d_o / d_i ) = [ ( 1 / d_o ) / ( 1 / d_i ) ][ f / ( d_i – f ) ]

1 = [ ( f / d_o ) / [ ( d_i – f ) / d_i ] 

And,

1 = [ ( f / d_o ) / [ 1 – ( f / d_i ) ] ]

Multiplying both sides of this expression by the denominator of the fraction on the right yields another intermediate expression, but it contains the focal length variable on both sides:

[ 1 – ( f / d_i ) ] = ( f / d_o )

The next step requires us to multiply both sides of the equation by ( 1 / f ):

( 1 / f )[ 1 – ( f / d_i ) ] = ( 1 / f )( f / d_o )

( 1 / f ) – ( 1 / d_i ) = ( 1 / d_o )

( 1 / f ) = ( 1 / d_o ) + ( 1 / d_i )

The lattermost equation is the Gaussian Lens Equation. We must realize that the diagram above will change if the object’s location within the system is changed. These changes will cause changes to occur in related dimensions of interest, namely the ratios of the object and image heights as well as ratios of the distances that separate the image and object from the lens:

Magnification ( M ) = ( h_i / h_o ) = ( – d_i / d_o )

After indulging in some additional algebraic back flips, we’ll end up with the Thin-Lens Equation:

( 1 / d_o ) + ( 1 / d_i ) = ( n₁ – 1 )[ ( 1 / R₁ ) – ( 1 / R₂ ) ]

And,

( 1 / f ) = ( n₁ – 1 )[ ( 1 / R₁ ) – ( 1 / R₂ ) ]

The ( n ) term references the previously mentioned index of refraction, which shall be expanded upon at a more appropriate time.