Telescopes

Telescopes are designed to collect light from distant objects and present an angularly magnified image. Two families: refracting (lens-based) and reflecting (mirror-based).

Concept

Astronomical (refracting) telescope

Two converging lenses:

• Objective of large focal length $f_o$ and large aperture (to collect light).
• Eyepiece of small focal length $f_e$.

Parallel rays from a distant star form an image at the focal plane of the objective. The eyepiece magnifies this image.

Magnification at relaxed eye (image at infinity):

$M = -\frac{f_o}{f_e}$

(negative sign means inverted; for astronomy this doesn't matter.)

Length of telescope tube:

$L = f_o + f_e$

For image at near point:

$M = -\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)$

Reflecting telescope

Uses a large concave mirror as the objective. Light reflected by the primary mirror is intercepted by a small secondary mirror and brought to the eyepiece.

Advantages over refracting telescope:

• No chromatic aberration (mirrors do not disperse).
• Easier to support a large mirror (held from behind) than a large lens (held only at the rim).
• Spherical aberration can be reduced with a parabolic mirror.
• Much higher light-gathering power for the same cost.

The Cassegrain configuration places the eyepiece behind the primary mirror (small hole in it).

Derivation

For the astronomical telescope at relaxed eye:

A distant object subtends angle $\alpha$ at the unaided eye. With the telescope, the objective forms an image of size $h = f_o \alpha$ at its focal plane. The eyepiece, positioned so that this image lies at its focal plane (output at infinity), creates an angle at the eye of $\beta = h/f_e = (f_o/f_e)\alpha$.

$M = \frac{\beta}{\alpha} = \frac{f_o}{f_e}$

with a sign indicating inversion.

Worked Example

An astronomical telescope has $f_o = 200$ cm and $f_e = 5$ cm. Find magnification and length.

$M = -\frac{200}{5} = -40\text{ (inverted, 40x)}$

$L = 200 + 5 = 205\text{ cm}$

Common Confusions

• Telescopes need large $f_o$ (long objective focal length), microscopes need small $f_o$.
• The length of an astronomical telescope is $f_o + f_e$; for a microscope it is much shorter and the formula is different.
• Magnification is angular (ratio of angles), not linear. The image at the eyepiece is virtual and seemingly "far away" but appears angularly larger.

Key Takeaways

• Astronomical telescope, relaxed eye: $M = -f_o/f_e$; length $L = f_o + f_e$.
• Reflecting telescopes (Newtonian, Cassegrain) avoid chromatic aberration and scale better.
• Large $f_o$ for telescopes, small $f_o$ for microscopes.