Imagine a circle with the center of O. OD is the radius of the circle. There is a line tangent to the circle at point B. AB is parallel to the radius. Point C lies on the radius. If angles ABO and OBC are equal, prove that OC=CD. Can you please help? I need to prove this to show that the focal point in lenses is in the middle of the radius. Thank you
The image I tried to describe:
On
You are using the spherical approximation. The true shape for perfect focus of parallel rays is a paraboloid. As a result, what you want will only hold approximately for small values of $m\angle BOD$. In general you could find $DC$ as a fraction of $OD$, and show that in the limit of small angles, the fraction approaches $\frac{1}{2}$.
$\triangle OBC$ is isosceles by design. So $OC = BC$. As $B$ approaches $D$, $OC + CB$ approaches $OD$, therefore $OC$ approaches $\frac{1}{2} OD.$
If you need the exact expression, a bit of trigonometry shows that $\frac{OC}{OD} = \cos \theta - \frac{\sin \theta}{\tan 2\theta}$.
On
As others have also observed, OC=OD only for paraxial rays incident/ incoming. That is, true only if $\angle ABO \approx 0 $ ie only in approximation.
It is in a parabola that such a property holds.
In spheres there is spherical aberration in geometric optics when reflected rays do not pass through a point focus causing blurring around a caustic curve.
Here is a proof that $OC$ is not equal to half of the radius:
Because $AB \parallel OD$ we have $\widehat{ABO} = \widehat{BOC}$. Also, the problem gives us $\widehat{ABO} = \widehat{OBC}$. Therefore $\widehat{BOC} = \widehat{OBC}$ and the triangle $OBC$ is isosceles, hence $OC = BC$.
Now, note that the shortest distance between two points is a straight line. Therefore in a triangle, the sum of lengths of two sides is always greater than the third side. Applying this to triangle $OBC$ we have $$OC + BC > OB$$ $$2OC > R$$ $$OC > \frac R2$$ As mentioned in other answers, the equality that you are searching for happens in parabolic mirrors and not in circular/spherical mirrors.