core25 min studyLesson 208 in Physics

Microscopes, telescopes, and angular resolution

Relate angular magnification to objective and eyepiece geometry, then separate image enlargement from resolved detail.

On this page

What you’ll understand

  • Calculate the signed angular magnification of an afocal telescope.
  • Combine objective magnification with a relaxed-eye eyepiece estimate.
  • Estimate circular-aperture Rayleigh resolution with its assumptions.

A telescope makes distant directions appear farther apart, while a microscope turns a small nearby structure into a larger angular pattern for the eye. In both instruments an objective forms an intermediate image and an eyepiece changes how the eye views it. Enlargement alone cannot restore detail already lost at the objective. We will trace the geometry behind angular magnification and compare it with a diffraction resolution scale, keeping the observation convention and the role of numerical aperture explicit.

An afocal telescope changes angles

Consider two positive thin lenses in air: an objective of focal length fof_o and an eyepiece of focal length fef_e. Put them a distance fo+fef_o+f_e apart so the objective’s back focal plane coincides with the eyepiece’s front focal plane. Parallel input from a distant point becomes a focused intermediate bundle and then parallel output. The system is afocal: a relaxed eye can receive the output without accommodating to a finite virtual-image distance.

A small input direction beta places the intermediate image at height approximately foβf_o\beta. The eyepiece sends that height into an output direction approximately −foβ/fe-f_o\beta/f_e. The minus sign denotes angular inversion in this Keplerian telescope. The objective also forms an image of itself through the eyepiece, giving an exit pupil where the eye should be placed. Magnification without an accessible pupil can leave much of the intended ray bundle outside the observer’s eye.

ℓ=fo+fe,Mθ=βoutβin=−fofe,Dexit≃Do∣Mθ∣\ell=f_o+f_e,\qquad M_{\theta}=\frac{\beta_{\mathrm{out}}}{\beta_{\mathrm{in}}}=-\frac{f_o}{f_e},\qquad D_{\mathrm{exit}}\simeq\frac{D_o}{|M_{\theta}|}
These relations concern an ideal two-positive-lens afocal telescope in air with the objective as aperture stop. All focal lengths and pupil diameters use common length units; angles are paraxial.↔ Scroll wide equations horizontally.

A microscope multiplies two distinct gains

A finite-conjugate microscope objective forms a real enlarged intermediate image. Its transverse magnification is mo=−si/som_o=-s_i/s_o using positive real object and image distances. Place that intermediate image in an eyepiece’s front focal plane for relaxed-eye viewing. Compared with direct viewing of the original object at a conventional near distance DnearD_{\mathrm{near}}, the eyepiece contributes angular magnification approximately Dnear/feD_{\mathrm{near}}/f_e. This reference distance is a viewing convention, not a universal anatomical constant.

The total angular gain is the product of objective transverse magnification and eyepiece angular gain. Infinity-corrected objectives introduce a tube lens and require the corresponding system specification rather than direct substitution of an arbitrary tube length. In either architecture, the objective’s numerical aperture largely determines the accepted fine spatial detail. A stronger eyepiece can enlarge an already blurred intermediate image without increasing the number of distinguishable object features, an effect often described as empty magnification.

Mmic≃moDnearfe,θR≃1.22λ0Do (air),dR≃0.61λ0NAM_{\mathrm{mic}}\simeq m_o\frac{D_{\mathrm{near}}}{f_e},\qquad \theta_R\simeq1.22\frac{\lambda_0}{D_o}\ \text{(air)},\qquad d_R\simeq0.61\frac{\lambda_0}{\mathrm{NA}}
Theta R is the circular-pupil angular Rayleigh separation of two incoherent point sources. d R is the corresponding object-space microscope estimate, using vacuum wavelength and object-side NA. These are criteria, not universal information bounds.↔ Scroll wide equations horizontally.

Resolution asks what detail survives

An ideal unobstructed circular aperture produces an Airy intensity pattern rather than a geometric point. Rayleigh’s criterion places the central maximum from one equal-brightness incoherent point at the first dark ring of the other. It supplies a useful conventional separation scale. Coherent sources add fields rather than intensities, and unequal brightness, noise, detector sampling, and reconstruction methods change what can be distinguished. State the source and measurement assumptions before calling a number a resolution limit.

A larger objective diameter improves distant angular resolution when diffraction dominates, while a larger object-side NA improves microscope spatial resolution. Aberrations, turbulence, defocus, and detector pixels can prevent the instrument from reaching either ideal estimate. Increasing angular magnification can make available detail easier for an eye or sensor to sample, but it does not shrink the objective’s diffraction pattern in object coordinates. A useful instrument matches collected detail, output pupil, magnification, and receiver sampling instead of maximizing only one specification.

A modest astronomical telescope

An ideal Keplerian telescope has fo=500 mmf_o=500\,\mathrm{mm}, fe=25 mmf_e=25\,\mathrm{mm}, and objective diameter 100 mm. At vacuum wavelength 500 nm in air, find angular magnification, exit-pupil diameter, and the Rayleigh angular scale.

  1. The lenses are separated by 500+25=525 mm500+25=525\,\mathrm{mm} for afocal operation. Signed angular magnification is −500/25=−20-500/25=-20, indicating inversion and twentyfold angular enlargement.

  2. With the objective as aperture stop, the exit pupil diameter is 100/20=5 mm100/20=5\,\mathrm{mm}. The observer must intercept that pupil to receive the full intended bundle.

  3. Convert wavelength and diameter to the same unit: θR=1.22(500×10−9)/0.100=6.1×10−6\theta_R=1.22(500\times10^{-9})/0.100=6.1\times10^{-6} radians. This is the ideal objective’s diffraction criterion, independent of the chosen eyepiece magnification.

The result

The instrument gives −20 angular magnification, a 5 mm exit pupil, and an ideal Rayleigh scale of 6.1 microradians. Real aberrations or atmospheric seeing can worsen the useful separation.

Explore the model

Interactive lab

Interactive lab · waves

Coherence changes how signals combine

Compare two equal-frequency waves rather than two telescope point images. Reset, set Phase difference = 0° and Amplitude ratio = 1, and predict coherent addition. For a fresh run Reset, set Phase difference = 90° and Amplitude ratio = 1, and predict a summed sinusoid with amplitude square root of 2.

Coherence changes how signals combineTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forPhase matters for coherent field addition. The instrument does not display Airy patterns, incoherent source intensities, telescope magnification, or a Rayleigh resolution test.

Check your understanding

An ideal telescope’s eyepiece magnification doubles while objective diameter and wavelength stay fixed. What happens to its objective Rayleigh angular scale?

Try it yourself

A finite-conjugate microscope objective has so=5.5 mms_o=5.5\,\mathrm{mm} and si=55 mms_i=55\,\mathrm{mm}. Its eyepiece has fe=25 mmf_e=25\,\mathrm{mm}, and the conventional near distance is 250 mm. Find total angular magnification and Rayleigh object separation for NA 0.50 at λ0=0.50 μm\lambda_0=0.50\,\mathrm{\mu m}.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Calculate objective transverse magnification first, then the eyepiece gain. Resolution uses NA rather than total magnification.

Compare with the solution

The objective has mo=−55/5.5=−10m_o=-55/5.5=-10. The relaxed-eye eyepiece gain is 250/25=10250/25=10, so total angular magnification is approximately −100. The Rayleigh object separation is 0.61(0.50)/0.50=0.61 μm0.61(0.50)/0.50=0.61\,\mathrm{\mu m}. A different eyepiece would change apparent angular size while leaving this ideal objective-side resolution estimate unchanged.

Take it with you

  • A Keplerian telescope is afocal when its focal planes coincide.
  • Microscope angular gain combines objective imaging with an eyepiece viewing convention.
  • Magnification and resolution answer different questions and depend on different limiting factors.

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