Kepler's telescope (refractor)

The astronomical telescope is made of two converging lenses arranged in series: the objective, with large focal length $f_\text{ob}$, facing the celestial object; and the eyepiece, with small focal length $f_\text{oc}$, close to the eye. The two lenses are separated by a distance $d \approx f_\text{ob} + f_\text{oc}$, so that the back focus of the objective coincides with the front focus of the eyepiece.

The celestial object is very far away, so $p_\text{ob} \to \infty$: the objective forms a real, inverted image at its focus. The eyepiece receives this image at its own front focal point and sends it to the eye as a virtual image at infinity, further magnified.

Ray path in Kepler’s telescope: the objective forms the image at the common focus, the eyepiece sends it to the eye as a virtual image at infinity.

Key formula

The angular magnification of Kepler’s telescope (Walker 2013) is $\ev{G_\text{cann} = \frac{f_\text{ob}}{f_\text{oc}}}$

Note

To magnify $50\times$, with an eyepiece of $f_\text{oc}=2$ cm an objective of $f_\text{ob}=100$ cm is needed. This is why astronomical telescopes are so long.

Example — A telescope for the Moon

We want $G = 80\times$ with an eyepiece of $f_\text{oc} = 25$ mm. What objective focal length is needed? And how long will the tube be?

$f_\text{ob} = G\cdot f_\text{oc} = 80\cdot 25 = 2000$ mm $= 2{,}0$ m. The tube is $d \approx f_\text{ob} + f_\text{oc} = 2{,}025$ m long, just over two metres.

Collegamenti

Argomenti: Ottica Concetti: Lenti e specchi

Esercizi collegati: Problema — Raggi per oggetto oltre 2f (lente) · Problema — Raggi per specchio concavo con p minore di f · Esercizio svolto — Specchio concavo