Alpha Centauri A which has an about 50% higher luminosity than the Sun is obviously undangerous to look at through a handheld telescope, at about 4.4 ly distance. How close would a yellow dwarf have to be in order to make it dangerous to look at it through an average spyglass? Or to feel some heat on the eye through the spyglass? This question can also be seen reversed as "how far from the Sun would you have to be in order to look at the Sun through a spyglass without danger".

  • $\begingroup$ @RobJeffries This is also about the temperature from the star's light shining through the spyglass. And at what relative brightness would it be undangerous to look at the Sun or a Sunlike star through an average spyglass? $\endgroup$
    – Giovanni
    Aug 15, 2020 at 12:15
  • $\begingroup$ astronomy.stackexchange.com/questions/22231/… is close to being a duplicate $\endgroup$
    – ProfRob
    Aug 15, 2020 at 14:23
  • $\begingroup$ @RobJeffries Unfortunately the question is answered under the premise that it is safe to look at the Sun when it appears point like because of saccades. Further on, it is assumed that without saccades it would be safe to look at the Sun in the mid-Kuiper belt. But the following answer by usernumber states that it's even dangerous to look at the Sun from Eris' aphelion which is at 100 AU distance and beyond the Kuiper belt: astronomy.stackexchange.com/questions/34929/… $\endgroup$
    – Giovanni
    Aug 15, 2020 at 14:35

2 Answers 2


This is more a question of "what is the brightest light I can look at", and subsequently this answer is more physiology, geometry than astronomy which only gives us the distances.

According to the wikipedia page on laser safety it is around 1mW/cm². A binocular - or any optics - again change the equation as so far that they accumulate all light falling onto its entry pupil into your eye.

Stars are sufficiently far away so that they are point sources (that's why they wobble around in the viewfinder due to atmospheric disturbance, unlike solar-system objects).

Energy emitted by a star is obtained by a simple black-body assumption $E=4\pi R^2_\star \sigma T^4$ and energy $W$ per on an area in distance $r$ thus is $W_\star = \frac{4\pi R^2_\star}{4\pi r^2}\sigma T^4 = \frac{L_\star}{4\pi r^2}$ (using the stellar luminosity $L_\star$ further on). Let's assume that your pupil has 0.5cm opening, thus the area there is $A_e = \pi 0.25^2 = 0.2cm^2 = 2\cdot 10^{-5}m^2$. Let's assume a binocular with 5cm aperture, thus the area there is $A_b=\pi\cdot 0.05^2m^2 = 8\cdot 10^{-3} m^2$, thus the ratio $A_e / A_b = 2.5\cdot 10^{-3}$. Thus the light at the entry of the binocular needs to be $W_{max} = 1mW/cm^2 \cdot 2.5\cdot 10^{-3} = 2.5\mu W/cm^2 = 2.5\cdot 10^{-2} W/m^2$ at most to not become dangerous. So, what distance does that imply? Let's equate:

$W_{max} = W_\star $ and solve for $r$: $r = \sqrt{\frac{L_\star}{4\pi W_{max}}}$. So for a sun-like star ($L_\star = 4\cdot 10^{26} W$) we get a minimum distance to safely look at its disk of roughly r=240 AU or 0.004 light years.

You can modify that equation for other stars with higher or lower energy output. Alpha centauri is not much brighter than our sun. But for a star like Vega ($L=40L_{sun}$), the minimum safe distance would already be more than 6 times as distant - yet still very near for any distances when considering distances to other stars. Thus any star becoming dangerous to look at with the naked eye or a binocular would pose a threat to our solar system.

And even if you consider not binoculars, but telescopes, say 1m diameter, thus 400x the light gathering power compared to the considered binocular, we would start to get distances of the order of a light year... even then all stars are further away.

(this still sounds crazy... is there an error in the estimates? But likely this errs on the side of caution. While you should not look at the sun even with the unaided eye, you're not immediately blind - and it's intensity is $1400W/m^2$ compared to the assumed safe power of $0.025W/m^2$, thus there's a difference of $10^5$)

  • $\begingroup$ Note that laser safety guidelines also depend on the wavelength of the laser. The damage from the Sun comes from the UV light and probably not the "heating" effect. I calculate that the laser safety threshold is an order of magnitude higher than the flux at the back of the eye from the Sun viewed from Earth. $\endgroup$
    – ProfRob
    Aug 16, 2020 at 13:46

Laser safety guidance does suggest that beams of power greater than 1 mw/cm$^2$ ($= 10$ W/m$^2$) should not be viewed continuously and perhaps a few times higher than this if you were only going to glance at the Sun for a second. I think these levels would be uncomfortably bright, but perhaps not dangerous.

The assumptions behind these maximum exposure levels are that the laser beam is focused into a spot on the retina and that the eye has an open pupil of 0.39 cm$^2$.

The angular resolution of someone with good eyesight is about 1 arcminute. Since the Sun subtends an angular diameter of 32 arcminutes, the Sun will become an unresolved, star-like object once you get beyond about 32 au from the Sun.

It is simple to work out the power received at the eye as a function of distance: $$ P_{\rm eye} \simeq 1400 \times \left(\frac{d}{{\rm au}}\right)^{-2},\ {\rm W}/{\rm m}^2 \tag*{(1)}$$ where $d$ is measured in astronomical units. Thus in order to reduce the beam power below 10 W/m$^2$ you would have to go beyond 12 au.

However, at this distance, the Sun is not an unresolved disc. When an object is resolved like this, it means the amount of flux per unit area hitting the back of the eye is actually unchanged by distance - the flux goes down, but the area of the image also goes down by the same amount. That means the Sun will do just as much damage to a small portion of the back of the eye as if you viewed it from Earth.

What we need to do is translate the safety limit to a power per unit area in a focused spot at the back of the eye. If we use a 1 arcminute circular diameter spot at the back of the eye, then the laser safety requirement is that we receive $<5\times 10^{-4}$ W/arcmin$^2$ at the back of the eye. If we compare that with the Sun, then when viewed from the Earth we get 1400 W/m$^2$ at the eye, projecting to a 32 arcminute diameter image, which is (assuming the same unrealistic 0.39 cm$^2$ pupil area) $7\times 10^{-5}$ W/arcmin$^2$ at the back of the eye.

In this sense it seems that the laser safety thresholds are contradictory to advice against viewing the Sun. The actual amount of power per unit area received at the back of the eye is about an order of magnitude smaller when viewing the Sun from Earth than for a collimated laser with a beam of 10 W/m$^2$. However, this does not factor in the spectrum of the Sun compared with say the (monochromatic) spectrum of a laser pointer. The damage by sunlight is done by the UV part of the spectrum, which isn't present in a a red or green laser.

So, I am figuratively crossing out the above calculation and doing another one which says that to be safe, you need to reduce the solar flux per unit area at the back of the eye by at least a factor of 10 from that we view from Earth. i.e. $f_{\rm eye} < 7 \times 10^{-6}$ W/arcmin$^2$.

As explained above, there is no change in the flux per unit area at the back of the eye, until the Sun becomes and unresolved point at about 32 au. In practice what would happen is that it would remain almost constant out to 32 au and then as the image of the Sun, which is the convolution of the angular resolution of the eye with the angular diameter of the Sun, became point-like it would smoothly merge towards an inverse square relationship with distance. An appropriate model for this might be $$f_{\rm eye} \simeq 7\times 10^{-5} \left[ 1 + \left(\frac{d}{32\ {\rm au}}\right)^2\right]^{-1}\ {\rm W}/{\rm arcmin}^2. \tag*{(2)}$$ When $d \ll 32$ au then $f_{\rm eye} = 7 \times 10^{-5}$ W/arcmin$^2$, as calculated earlier, but then when $d > 32$ au, $f_{\rm eye}$ decreases as $1/d^2$.

Using equation (2), to get $f_{\rm eye} <7\times 10^{-6}$ W/arcmin$^2$, requires $d>96$ au. So 96 au is my estimate for the naked eye.

With a "spy glass" you have the added complication of collecting the flux over a bigger area. This increases $f_{\rm eye}$ by a factor equal to the ratio of the area of the spy-glass objective lens to the area of the pupil of the eye. For example, with a $D=5$ cm diameter objective lens, you would have about 50 times the collecting area of the pupil of the eye (though individuals vary and it varies under illumination conditions).

However, at the same time, the spy-glass provides a linear magnification (for a resolved object) by a factor equal to the ratio of the objective to the eyepiece focal lengths. A typical value for a spyglass magnification would be $M=10-20$. If the object is fully resolved, this decreases $f_{\rm eye}$ by a factor $M^2$ because the light is spread over a larger area at the back of the eye.

This modifies equation (2) as: $$f_{\rm eye} \simeq 7\times 10^{-5}\left( \frac {\pi D^2}{4\times 0.39M^2}\right) \left[ 1 + \left(\frac{d}{32M\ {\rm au}}\right)^2\right]^{-1}\ {\rm W}/{\rm arcmin}^2, \tag*{(3)}$$ where $D$ is in cm. Note that the exact magnification becomes unimportant at large distances (i.e. once $d > 32M$ au).

Using equation (3) with $D=5$ cm and $M=10$, I find that $f_{\rm eye}<7\times 10^{-6}$ W/arcmin$^2$ when $d>640$ au. So 640 au is my estimate for a spy-glass.

The heating effect you ask about is negligible at this distance. You are focusing $7\times 10^{-4}$ W/m$^2$ (from equation (1)), using an objective with a diameter of 5 cm. So the total heating power is just $7\mu$W. This can be compared with a number that would be about 400,000 times larger if you used the same spy-glass to look at the Sun from Earth (don't ever do that).


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