Question

1.3 Newtonion Form of the Thin Lens Equation Newton, being Newton, derived an alternative for of the thin lens equation. With reference to figure 2, derive the Newtonian Thin Lens Equation: Toci You can use what we already know in class about the standard form: ss 1 f-1

Answer 1

The distance $s_o$ is the distance of the object from the lens. And $f$ is the focal length of the lens. And $x_o$ is the distance of the object from the focal point. And so,
   $s_o=x_o+f$
And the distance $s_i$ is the distance of the image from the lens. And $f$ is the focal length of the lens. And $x_i$ is the distance of the image from the focal point. And so,
   $s_i=x_i+f$
And so, from the Gaussian form of the thin lens equation
   $\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}$
And so, we have
   $\frac{1}{x_o+f}+\frac{1}{x_i+f}=\frac{1}{f}$
  $\Rightarrow \frac{x_o+f+x_i+f}{(x_o+f)(x_i+f)}=\frac{1}{f}$
  $\Rightarrow \frac{x_o+x_i+2f}{(x_ox_i+x_of+x_if+f^2)}=\frac{1}{f}$
  $\Rightarrow (x_ox_i+x_of+x_if+f^2)=(x_o+x_i+2f)f$
  $\Rightarrow x_ox_i+x_of+x_if+f^2=x_of+x_if+2f^2$
  $\Rightarrow x_ox_i+f^2=2f^2$
  $\Rightarrow x_ox_i=f^2$
  
  ​​​​​​​