Visual walkthrough — Optical instruments — human eye, simple microscope, compound microscope, telescope

This is the flagship picture-walkthrough for the refracting telescope. The parent note stated $M=f_o/f_e$. Here we build it from nothing — no symbol used before it is drawn.

We will follow one ray-bundle from a star all the way to your eye and watch the angle get fatter. Every step: WHAT we do, WHY we do it, and WHAT IT LOOKS LIKE.

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Step 1 — What "how big something looks" even means

Look at the figure. An object of height $h$ sits a distance $d$ from your eye. Two rays — one from the top, one from the bottom — enter your eye and open a thin wedge. That wedge's opening is the visual angle $\theta$.

WHY tangent and not something else? We have a right triangle (eye at the corner, object standing straight up). The one trig ratio that connects the angle we care about to the two sides we can measure ($h$ and $d$) is the tangent. Sine would need the slanted ray-length (hard to measure); tangent needs the two sides we already have.

Naked-eye limit: a star is at distance $d\to \infty$, so $\theta =h/d\to$ almost zero. Your unaided eye sees a dot. That is the whole problem — and the reason a telescope exists.

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Step 2 — A star sends parallel rays, arriving at a tilt $\alpha$

The figure shows two stars (top and bottom edge of, say, the Moon). Each sends its own bundle of parallel rays. The whole bundle from the top edge arrives tilted by an angle $\alpha$ relative to the bundle from the bottom edge.

• $\alpha$ ::: the true angle the distant object subtends at your eye — the naked-eye visual angle ${\theta }_0$ from Step 1, but now called $\alpha$ because it is our reference.

WHY parallel matters: parallel rays carry no information about distance, only about direction (the tilt $\alpha$). So the instrument cannot "shrink the distance"; all it can ever do is change the tilt the eye finally receives. That single idea drives the whole derivation.

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Step 3 — The objective lens turns a tilt into a spot in its focal plane

The objective is the big front lens with a long focal length $f_o$. In the figure:

• The bundle from the bottom edge (tilt $0$) focuses on the axis.
• The bundle tilted by $\alpha$ focuses off-axis, at a height $h^{\prime }$ below the axis.
• This forms a real, tiny, upside-down image $I$ right in the focal plane.

WHY a long $f_o$? From $h^{\prime }=f_o\alpha$: a longer $f_o$ makes the intermediate image $I$ taller for the same star tilt $\alpha$. A bigger image is easier for the next lens to magnify. This is why telescope objectives are physically long. (Contrast the microscope, where the object is near and we want $f$ short — see Linear vs angular magnification.)

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Step 4 — Place the image exactly at the eyepiece's focus

The figure shows the two lenses lined up on one axis. Their focal points meet at $I$. This shared-focus arrangement is called normal adjustment.

WHY put $I$ at the eyepiece focus? A lens converts rays leaving its focal point into a parallel outgoing bundle. Parallel rays let a relaxed eye (focused at infinity) see the image without straining. So this position = most comfortable viewing. See Lens equation and sign conventions for why "object at focus $\Rightarrow$ image at infinity."

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Step 5 — The eyepiece re-opens the angle to a fat $\beta$

In the figure, rays leave the eyepiece parallel, opening a wide wedge of angle $\beta$ into your eye.

WHY does short $f_e$ give big $\beta$? With $h^{\prime }$ fixed, dividing by a smaller $f_e$ gives a bigger $\beta$. The same little image, viewed from closer up (short focal length), fills more of your view.

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Step 6 — Divide the angles: the magnification appears

The figure overlays the thin naked-eye wedge $\alpha$ (Step 2) and the fat telescope wedge $\beta$ (Step 5) at the same eye. Their ratio is $M$.

WHY the cancellation is the deep point: both angles are built from the same image height $h^{\prime }$. When you divide, $h^{\prime }$ vanishes and you are left with pure geometry — the ratio of the two lenses. That is why the answer is so clean.

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Step 7 — The edge case: relaxed eye vs. straining to the near point

The figure shows both cases side by side: (left) relaxed, image at $\infty$, parallel exit rays; (right) strained, image pulled in to $D$, slightly diverging exit rays and a wider $\beta$.

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The one-picture summary

One diagram, whole story: tilt $\alpha$ in $\to$ objective focuses to image $I$ of height $h^{\prime }=f_o\alpha$ $\to$ eyepiece re-launches it at $\beta =h^{\prime }/f_e$ $\to$ the ratio $\beta /\alpha =f_o/f_e$ pops out because $h^{\prime }$ cancels.

Recall Feynman retelling — the whole walk in plain words

A star is a bright dot because its rays reach us dead parallel, tilted by a tiny angle $\alpha$. My eye only ever notices that tilt — not the star's real size. So I put a big long front lens out front: it takes the parallel bundle and drops it into a tiny real picture $I$ floating in midair, and because the lens is long, that picture is decently tall. Now I look at that midair picture through a little short back lens held so the picture sits right at its focus. That little lens throws the light back out parallel again — so my eye stays relaxed — but tilted much more steeply, angle $\beta$. How much did I win? I just ask "$\beta$ over $\alpha$." Both angles were built from the same little picture height $h^{\prime }$, so $h^{\prime }$ cancels and I'm left with the two lens lengths: $f_o/f_e$. Long front lens, short back lens, and I've made the sky look big.

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Connections

• Optical instruments — human eye, simple microscope, compound microscope, telescope (index 2.5.8) — parent
• Lens equation and sign conventions — why object-at-focus gives image-at-infinity
• Linear vs angular magnification — why the telescope uses angles, the microscope uses heights
• Power of a lens and dioptres — short $f$ = high power (the eyepiece)
• Resolving power and diffraction limit — magnification is not the same as seeing detail
• Reflecting telescope (Cassegrain) vs refracting — same $f_o/f_e$, mirror objective
• Defects of vision — myopia, hypermetropia — why the near point $D$ enters the strained case