This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2019-07-12 15:40 +0800.

The center of the circular luminous object is CC, and its radius is rr. The center of the thin lens is OO, and its focal length is ff. The object is in the same plane as CC, and OCOC is perpendicular to the thin lens, and ∥OC∥=2f\left\|OC\right\|=2f.

Set up Cartesian coordinate with OO being the origin and COCO being the xx-axis. Then, C(−2f.0)C\left(-2f.0\right). The luminous object is described by the parametric equations

{x=−2f+rcos⁡t,y=rsin⁡t.\begin{cases} x=-2f+r\cos t,\\y=r\sin t. \end{cases}

Pick point P(−2f+rcos⁡t,rsin⁡t)P\left(-2f+r\cos t,r\sin t\right) on the object. According to the formula for imaging of thin lenses

{−1x+1x′=1f,xx′=yy′,\begin{cases} -\frac1x+\frac1{x'}=\frac1f,\\ \frac x{x'}=\frac y{y'}, \end{cases}

the point PP is transformed to P′(f(1+ff−rcos⁡t),−frsin⁡tf−rcos⁡t)P'\left(f\left(1+\frac f{f-r\cos t}\right),-f\frac{r\sin t}{f-r\cos t}\right). Therefore, we can have the parametric equations of the image:

{x=f(1+ff−rcos⁡t),y=−frsin⁡tf−rcos⁡t.\begin{cases} x=f\left(1+\frac f{f-r\cos t}\right),\\y=-f\frac{r\sin t}{f-r\cos t}. \end{cases}

Cancel tt, and we have

y2=(rf)2(x−f)2−(x−2f)2,y^2=\left(\frac rf\right)^2\left(x-f\right)^2-\left(x-2f\right)^2,

which means that the image is a conic section with the focus being (2f,0)\left(2f,0\right), the directrix being line x=fx=f, and the eccentricity being rf\frac rf.

Alternatively, let ρ≔(x−2f)2+y2\rho\coloneqq\sqrt{\left(x-2f\right)^2+y^2}, and we have

ρ=r1−rfcos⁡t,\rho=\frac r{1-\frac rf\cos t},

and we may have the same conclusion through this equation.