# 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.

# 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. 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.