Will a Small Telescope Actually Resolve Two Distant Lamps?
A student assembles an astronomical telescope with an objective of focal length 100 cm and diameter 5.0 cm and an eyepiece of focal length 5.0 cm. The telescope is focused for a relaxed eye, so the final image is at infinity. The student points it at two sodium lamps, wavelength 589 nm, separated by 10 cm on a wall 2.0 km away. The separation between the lenses is set to 105 cm. The eye’s angular resolution is about 1.0 arcminute, or 2.9×10⁻⁴ rad. A classmate says that because the magnification is 20, the lamps will appear 2 m apart at 2 km and must obviously be resolved. The student must decide whether the actual limit is diffraction by the 5.0 cm aperture, the eye’s resolution, or the magnification. Ignore atmospheric turbulence and lens aberrations.
1. Calculate the angular separation of the lamps as seen by the unaided eye and the Rayleigh diffraction limit of the objective. Is the objective aperture large enough to resolve them?
Read the worked answer
For small angles, θ = separation/distance = 0.10 m/2000 m = 5.0×10⁻⁵ rad.
The Rayleigh limit for a circular aperture is θ_R = 1.22λ/D:
θ_R = 1.22(589×10⁻⁹)/(0.050) = 1.44×10⁻⁵ rad.
Since 5.0×10⁻⁵ rad is about 3.5 times larger than 1.44×10⁻⁵ rad, the 5.0 cm objective aperture is large enough to resolve the lamps. Diffraction by the objective is not the limiting factor in this situation.
2. Calculate the angular magnification for a relaxed eye and the angular separation of the final image. Compare with the eye’s resolution. Would the lamps be seen as separate? What if the eyepiece were replaced by a 10 cm focal length lens?
Read the worked answer
For a telescope focused for a relaxed eye, M = f_o/f_e = 100/5.0 = 20. The final image has angular separation θ_image = Mθ_object = 20(5.0×10⁻⁵) = 1.0×10⁻³ rad. In arcminutes, this is (1.0×10⁻³)(180/π)(60) ≈ 3.4 arcmin, which is larger than the eye’s 1.0 arcmin resolution. The lamps should be seen as separate.
With a 10 cm eyepiece, M = 100/10 = 10, so θ_image = 10(5.0×10⁻⁵) = 5.0×10⁻⁴ rad ≈ 1.7 arcmin. This is still larger than 1.0 arcmin, so the lamps would still be resolved, though with less margin. If the magnification were only about 5, the image separation would be 2.5×10⁻⁴ rad ≈ 0.86 arcmin, below the eye’s resolution, and the eye would then be the limiting factor.
3. Explain why the lens separation is 105 cm for a relaxed eye. What would be wrong with setting it to 100 cm? Does the 5.0 cm aperture determine the magnification?
Read the worked answer
The objective forms a real image of a distant object at its focal plane, 100 cm from the objective. For the final image to be at infinity (relaxed eye), the eyepiece must have that real image at its own focal point, 5.0 cm in front of the eyepiece. Therefore the lens separation must be f_o + f_e = 100 + 5.0 = 105 cm.
If the separation is set to 100 cm, the objective image falls essentially at the eyepiece itself. The eyepiece cannot then produce a final image at infinity; the system is badly focused for a relaxed eye, and the final image would be at an unintended distance with poor definition.
The 5.0 cm aperture does not determine the magnification. It determines the light-gathering power and the diffraction limit, θ_R = 1.22λ/D. The angular magnification is set by the focal lengths, M = f_o/f_e, and is 20 in this telescope.