@Uwe's question and answer seem to illustrate that the resolution of surface-imaging spacecraft orbiting around the Moon has not dramatically improved over the decades, and that this is pretty-much limited by aperture size which in turn has pretty-much been limited by payload size/weight.

Question: What would be the technical challenges to improving spatial resolution by using the same aperture, but a substantially shorter wavelength instead?

For example, the Sun puts out a significant amount of UV. It's less intense than in the visible part of the spectrum, but it's still there.

Maybe I'm planning a landing a small craft with short legs, and I'd like to find and confirm an area with a low frequency of basketball-sized rocks in which to put my landing ellipse. If the resolution of current data is only 1 meter line pairs, that's not going to be good enough.

  • 1
    $\begingroup$ There are other ways to increase resolution (lower orbit), and Uwe's answer only focused on the diffraction limit of the optics, ignoring the resolution limit of the sensor underneath. $\endgroup$
    – Hobbes
    Commented Sep 30, 2018 at 14:37
  • $\begingroup$ @Hobbes go for it, please! I just wanted to add something to one aspect, but certainly there are other aspects to cover here! $\endgroup$
    – uhoh
    Commented Sep 30, 2018 at 14:40
  • 1
    $\begingroup$ How about SAR imaging on moon ? $\endgroup$
    – zephyr0110
    Commented Oct 3, 2018 at 15:23
  • $\begingroup$ @Prakhar that's a good point! Using a synthetic aperture allows for great resolution without a large spacecraft. e.g. How can ICEYE-X1 capture 2D high resolution SAR images in “tens of seconds”? With low lunar orbit SAR just might be able to do 30cm resolution $\endgroup$
    – uhoh
    Commented Oct 3, 2018 at 15:34

3 Answers 3


At first sight it is a good idea to increase optical resolution by using shorter wavelengths. But sunlight is aproximized as black body radiation of 5900 K with the maximum intensity at about 560 nm. To increase resolution, the used wavelength should be much shorter. But at about 250 nm the solar intensity is much smaller, see this graph. Even without the Earth's atmospheric absorbtion of short wavelengths in lunar orbit.

But you need much light for a short exposure time and low noise images from the electronic image sensor. The camera of the lunar orbiter may be rotated for compensation of orbit velocity to the image. In order to get better images with more resolution, the compensation of camera movement should be better to allow longer exposures for more photons detected by the sensor.

But unfortunately solar intensity is more than 10 times smaller at 250 nm when compared to 560 nm. I think this could not be compensated by using longer exposure times only.

Sorry, I only found a graph with german text. Feel free to replace it with a graph with english description.


To add to @Uwe's answer, I found a source for the solar spectrum above the atmosphere (i.e. in space). It's in old-style .xls binary format, so I found a Python library to open it as I don't like to remember the Excel days.

A very broad band from 250nm to 350nm in the UV has about 1/7 the intensity of the 450 to 650nm band.

However, conventional CCDs and other imagers tend to give one photoelectron per photon, independent of photon energy 1 Watt of 300nm has half the number of photons as 1 Watt of 600nm. So even if the system had flat quantum efficiency (which it won't, even with backside-thinned CCDs) that ratio should really be 1/15.

Further factor in the precipitous drop in the Moon's albedo in the UV (the Moon is red, it is not at all white, see Why doesn't a full/gibbous moon high in the sky ever seem to look orange? Shouldn't it? for more details) so that ratio should really be 1/30.

While the idea could be workable, a factor of 30 loss in brightness would be tough to make up for with a factor of 30 longer exposure. At some point you'll need some pretty fancy camera-slewing or multiple image correlation and stacking with a toasty-hot image processing FPGA in order to keep the S/N comparable to the state-of-the-art in visible light.

Moon's albedo from ESA

above: "Figure 8: Averaged geometrical moon albedos measured by GOME from July 1995, November 1995, and September 1996." From ESA's GOME moon measurements, including instrument characterisation and moon albedo.

Solar Spectrum atm=0 from NREL e490_00a_amo.xls

import numpy as np
import matplotlib.pyplot as plt

# https://www.nrel.gov/grid/solar-resource/spectra-astm-e490.html
# older https://www.nrel.gov/grid/solar-resource/spectra-wehrli.html
# FOUND: [Read Excel File in Python](https://stackoverflow.com/q/22169325/3904031)
# works for older .xls, which are binary and pre-2007 conversion to XML-like
# https://pypi.org/project/xlrd/
# https://libraries.io/pypi/xlrd
# HANDY!  http://www.python-excel.org/
# ALSO HANDY!  https://www.datacamp.com/community/tutorials/python-excel-tutorial

clight  = 2.99792458E+08   # m / s
hplanck = 6.626070040E-34  # J s
Coulomb = 6E+18

if False:   # do this once to read the Excel file, then save as .npy

    from xlrd import open_workbook

    wb     = open_workbook('e490_00a_amo.xls')
    sheets = wb.sheets()
    s0     = sheets[0]   # first sheet

    nrows, ncols = s0.nrows, s0.ncols
    print (nrows, ncols)

    name_C0 = s0.cell(0,0).value   # u'Wavelength, microns'
    name_C1 = s0.cell(0,1).value   # u'E-490 W/m2/micron'

    lam = np.array([s0.cell(n,0).value for n in range(1, nrows)])
    I   = np.array([s0.cell(n,1).value for n in range(1, nrows)])

    np.save('lam', lam)
    np.save('I',   I  )

else:  # then you can just read from disk next time

    lam = np.load('lam.npy')
    I   = np.load('I.npy')

n0  = np.argmax(lam>0.15)
n1  = np.argmax(lam>0.75)

lam = lam[n0:n1+1]
I   = I[n0:n1+1]

dlam = lam[1:] - lam[:-1]
lam  = lam[:-1]
I    = I[:-1]

Ephot = hplanck * clight / (1E-06 * lam)

print lam.min(), lam.max()
print dlam.min(), dlam.max()

uv1  = (lam >= 0.25) * (lam <= 0.35)
uv2  = (lam >= 0.35) * (lam <= 0.45)
vis1 = (lam >= 0.45) * (lam <= 0.55)
vis2 = (lam >= 0.55) * (lam <= 0.65)

Iuv1  = I * uv1
Iuv2  = I * uv2
Ivis1 = I * vis1
Ivis2 = I * vis2

Iuv1_tot  = (Iuv1  * dlam).sum()
Iuv2_tot  = (Iuv2  * dlam).sum()
Ivis1_tot = (Ivis1 * dlam).sum()
Ivis2_tot = (Ivis2 * dlam).sum()

print "Iuv1_tot:  {} W/m^2".format(Iuv1_tot)
print "Iuv2_tot:  {} W/m^2".format(Iuv2_tot)
print "Ivis1_tot: {} W/m^2".format(Ivis1_tot)
print "Ivis2_tot: {} W/m^2".format(Ivis2_tot)

if True:
    plt.subplot(2, 1, 1)
    plt.plot(lam, I,    '-k')
    plt.plot(lam, Iuv1, '-m')
    plt.plot(lam, Iuv2, '-b')
    plt.plot(lam, Ivis1, '-g')
    plt.plot(lam, Ivis2, '-y')
    plt.text(0.32, 350,  '52', fontsize=14)
    plt.text(0.405, 750, '139', fontsize=14)
    plt.text(0.48, 1300, '193', fontsize=14)
    plt.text(0.59, 1200, '176', fontsize=14)
    plt.text(0.18, 1900, 'approx W/m^2', fontsize=14)

    plt.subplot(2, 1, 2)
    plt.plot(lam, I,    '-k')
    plt.plot(lam, Iuv1, '-m')
    plt.plot(lam, Iuv2, '-b')
    plt.plot(lam, Ivis1, '-g')
    plt.plot(lam, Ivis2, '-y')
  • $\begingroup$ For amateur photography on Earth, a sunny day gives you a massive excess of photons leading to shutter times of 1/2000 s, so an exposure 30 times longer doesn't seem so bad. (this leads to the question what sort of exposure times you need when photographing from orbit) $\endgroup$
    – Hobbes
    Commented Sep 30, 2018 at 14:33
  • $\begingroup$ @Hobbes For an improved resolution of 30cm at say 70 km, one would need to use a fairly long focal length. "10.5 µradians per pixel)" was mentioned in the question, this would have to be half that! Surface brightness of extended, resolved objects decreases as the square of the focal length, so better do the math first. $\endgroup$
    – uhoh
    Commented Sep 30, 2018 at 14:37
  • $\begingroup$ @Hobbes: Amateur photography on Earth uses typically a medium sensitivity film with medium resolution. But aerial photography uses a low speed high resolution film. Typical exposure (above Earth) about 1/250 at f/8. The exposure times for for Lunar Orbiter was 1/100, 1/50 or 1/25 second. An exposure 30 times longer seems so bad, 1/3 s to over one second. $\endgroup$
    – Uwe
    Commented Oct 1, 2018 at 10:00

I try to estimate the usage of UV light to get more resolution of the NAC by using information about the wide angle camera WAC. The WAC has also two UV ranges.

There is some more information about the WAC of the Lunar Reconnaissance Orbiter here: (1), (2) and (3) (this is behind a paywall, I did not read it).

The wide angle camera image sensor was sensitive for 200 to 1000 nm. Using filters seven ranges could be detected, UV1 320 nm, width 32 nm, UV2 at 360nm with 15 nm width, 3 visible ranges and 2 ranges for near infrared.

The narrow angle camera NAC had no filters for detection of different wavelengths. But the method used for UV detection of the WAC shows what is necessary to get the increased sensitivity to compensate the low intensity of UV radiation within the solar black body radiation.

The resolution of the wide angle camera WAC was better in the visual range with 74.9 m/pixel or 1.498 mrad at a height of 50 km. For UV it was only 383.5 m/pixel = 7.67 mrad. The resolution was more than 4 times better in the visible range than for UV.

For UV 512 x 16 pixels of the sensor were used and binned to only 128 x 4. The mean value for a sub array of 4 by 4 = 16 pixels was computed by on-chip analog summing to get a better signal to noise ratio. SNR was 157 for 320 nm wavelength as well as 565 nm, but the area of 16 pixels had to be used to detect enough UV photons. It was possible to get the same noise level with the low intensity UV radiation by increasing the effective area of a pixel by factor of 16.

If we want to increase the resolution of the NAC by decreasing the wavelength by a factor of two, we need an image sensor with two times the pixel count in each direction. The area of a pixel is thus 4 times smaller and gets 4 times less photons. Due to the low intensity of the UV radiation we need about 16 times (information from the WAC) more light for the same SNR. So we need 4*16 = 64 times more light and 8 times the diameter of the telescope to get the necessary light. But this means an aperture of f/0.64 instead of f/5.1 for the 700 mm telescope of the NAC.

So it is much cheaper to stay at the same wavelength (also the same focal length) and increase the diameter of the telescope by a factor of 2. The image sensor should have 4 times more pixel to make use of the better resolution, but the total size of the sensor should be the same. The area of a single pixel is thus 4 times smaller. To get the same SNR, we need the same count of photons for the smaller pixel. But we already increased the diameter of the telescope by a factor of 2 and get thus 4 times the photons on the total area of the image sensor. Therefore pixel with 1/4 area get the same count of photons. The aperture of the NAC should be only f/2.75 instead of f/5.5.

  • $\begingroup$ WAC uses a lens, Narrow angle NAC used a mirror. WAC wasn't used for high resolution imaging so I don't understand yet how this is helpful, can you explain a bit more? Thanks! $\endgroup$
    – uhoh
    Commented Oct 1, 2018 at 16:39

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