Newtonian Form of the Thin-Lens Formula

The formula \frac{1}{p}+\frac{1}{i}=\frac{1}{f} is called the Gaussian form of the thin-lens formula. another form of this formula, the Newtonian form, is obtained by considering the distance x from the object to the first focal point and the distance x' from the second focal point to the image. I am going to show that xx'=f^{2} is the Newtonian form of the thin-lens formula.

In this situation, my diagram is going to look as follows below:

 

Starting with the thin lens formula, \frac{1}{p}+\frac{1}{i}=\frac{1}{f} , I note that the object distance is x+f , and the image distance is x'+f , where the focal distances for the two lenses are equivalent. After making these substitutions into the thin lens formula, and solving for f I obtain the equation f=\frac{xx'+fx+fx'+f^{2}}{x'+x+2f} . Now I can multiply each side of the equation by x'+x+2f and solve for f^{2}  to obtain the final solution f^{2}=xx' , which is the Newtonian form of the thin lens equation.

Optics Proof

A luminous point is moving at speed V_0 toward a spherical mirror with radius of curvature r , along a central axis of the mirror. I am gong to prove that the image of this point is moving at speed V_{I}=-(\frac{r}{2p-r})^2 V_{0}

Since it is a spherical mirror I am going to start with a given equation whose derivation I will not provide. This equation relates the focal length, image distance, and object distance of a spherical mirror and that equation is \frac{1}{p}+\frac{1}{i}=\frac{1}{f} . Since I am trying to find the speed of the image produced by the spherical mirror, I will solve for i , which is the image distance. After doing this, I obtain the following equation i=\frac{pf}{p-f} . Since f is related to r by the equation f=\frac{r}{2} , I can plug this into my equation for i , which gives me i=\frac{pr}{2p-r} . Since i is a “position”, if i take the first time derivative of it, I will obtain the image velocity or V_{I} . After taking this derivative by applying the quotient rule, and treating r , the distance to the center of curvature, as a constant, I obtain the following formula: \frac{\frac{dp}{dt}(2p-r)+2\frac{dp}{dt}pr}{(2p-r)^2} . Here, I notice that \frac{dp}{dt} is simply the velocity of the object. Since the two velocities will have different directions, I let the object distance be negative. After making the substitution V_{0}=-\frac{dp}{dt} , I obtain the following equation \frac{-V_{0}(2p-r)+2V_{0}pr}{(2p-r)^2} . Now I can distribute -V_{0} , and cancel certain like terms. After doing so the final equation becomes V_{I}=-(\frac{r}{2p-r})^2 V_{0} which is what I sought to prove.