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.

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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.

Magnetic Fields Proof

In the figure below a beam of electrons whose kinetic energy K emerges from a thin-foil “window” at the end of an accelerator tube. A metal plate at distance d from this window is perpendicular to the direction of the emerging beam. I am going to show that the beam can be prevented from hitting the plate if a uniform magnetic field B>\sqrt{\frac{2mK}{e^{2}d^{2}}} is applied.

 

magnetic field

 

 

 

I start by stating that the Kinetic Energy is given by the formula K=\frac{1}{2}mv^2 . Because the beam of particles exist within a magnetic field, they experience a F=ma , directed radially inward.. Because the applied magnetic field causes them to follow a circular path, a=\frac{v^2}{r} and that magnetic force is F=qvB . If we plug all of these values into the radius of a circle that is created when the beam of electrons move perpendicular to the applied uniform magnetic field I obtain the formula r=\frac{mv}{|q|B} . If I solve the Kinetic Energy formula for v and then plug it into the circle equation I obtain r=\frac{m\sqrt{\frac{2K}{m}}}{qB} . Next, if I solve for (r=d)  after squaring both sides and state that the distance of the circle must be less than d , I obtain the forumla d < \sqrt{\frac{m2K}{e^{2}B^{2}}} . If I now take \frac{1}{B} out of the fraction and multiply each side of the equation by its reciprocal, and then bring \frac{1}{d} under the radical, I obtain the following solution which is the end that I seek, namely B>\sqrt{\frac{2mK}{e^{2}d^{2}}} . After using the Right-Hand Rule, I find that the Magnetic Field is pointing out of the page.