Here you can see the working principle of a refractive telescope (there are many different kinds but I guess your question is mostly about this type of telescope).
Here what happens is that a far-away object irradiates onto an objective lens, which is set at a specific distance from an eyepiece lens, which in the end should recollimate the light so that it can be observed (if it is not collimated then you also have to account for how light would enter into your eye and that complicates the design by a few degrees of complication).
Here you can see a different interpretation of the same principle, but without the nice imaging into the human eye visual system:
Second thing to understand is the solar radiation irradiance as a function of wavelength:
Here you can see that, roughly, solar irradiation is a function of many things: Wavelength (the color of the light), absorption depending on the height with respect to earth position, and in the wikipedia article where I got this picture from, you can also consider what time of the day it is, where in the world you are, etc.
For the sakes of simplification, I will pick just a value: 1 W/m² / nm. To make things even simpler, I will pic 1 W/m² across all wavelengths (meaning that all wavelengths reach the earth with the same power. This is not really physical as there are a number of processes happening in the world at all times, so bear with me that this is not realistic but within reason, approximate enough).
Next thing to consider is what is called optical power damage threshold (typically this is done for lasers so it can also be found as laser power damage threshold) which is a measure to determine how much power and for how long it has to be irradiating the eye to cause considerable damage.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6484998/ In this paper you can find a control test with a 1319 nm continuous wave laser which was used to determine the laser power damage threshold on rabbits (:().
From the conclusions:
The rabbit retinal damage thresholds at 24-h post exposure for the ocular axial lengths of 15.97 and 17.25 mm were 1.06 and 1.79 J respectively. The obvious difference for the damage thresholds resulted from the dependences of pre-retinal absorption and retinal spot size on the ocular axial length. Detailed analysis indicated that a sufficient margin existed between the damage threshold and MPE for adult humans, but for the newborn eyes the safety factor may be less than 2.3. The obtained results could be used in the refinement of the safety standards for transitional NIR lasers.
So we know that laser damage will happen, roughly, at 1.06-1.7 J for 1319 nm. Considering our previous assumptions here, you can see that we have that the sun is reaching with 1 W/m² and damage happens at roughly 1.5J.
Since we want oranges Vs. oranges for our exercise, you need to convert one until somehow. A watt is a unit of energy per time unit, so 1 W/m² is 1 J/s/m² or J/sm². Since we have J in our damage threshold, we need to convert the solar irradiation to a measureable power.
1 W/m² is 1e-6 W/mm² = 1e-6 J/s mm².
A typical 40x commercially available cheap telescope will give you a 60mm objective. That would make, roughly, an area of $pi r^2$ = 11300 mm². The impinging power onto the objective lens would be 0.011 W (1e-6 * 11300 mm²), or 0.011 J/s.
With a magnification of 40x, we can roughly state that the telescope will roughly give 0.44J/s.
You can clearly see from the quick calculation exercise where the problem comes from: While the solar irradiation is relatively low (or low enough not to cause severe damage in case you look at the sun directly for less than a second), if you put it into an optical system that increases the concentration of light to a smaller area. Here you can see a lot of parallels with laser safety regulations where there are some types of lasers where the natural reaction of the eyelids to close is sufficient to avoid permanent damage, but the longer that you keep your eyesight onto the eyepiece, and the higher the magnification, the higher the power that reaches your eye and can therefore damage it irreversibly.
As a final note, here it is important to note that a) some calculations are just desk calculations and may differ from the actual situation. The calculations are just to give a rough idea of the physical processes but in terms of irradiance, magnification, etc. There may be more accurate calcuations in the literature.
There is for example a review on the effects of UV in eyesight: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3872277/.