A telescope gives a clear view of a distant object when the angular displacement between the edges is at least $9.7 × 10^{-6} rad$. The Moon is approximately $3.8 × 10^5$ km from the Earth. Estimate the minimum diameter of a circular crater on the Moon's surface that can be seen using the telescope.

The first way I did it is by using $tan (Q/2)$ = radius / distance from the Earth to the Moon.

The second way is by using $s = rQ$, where $s$ is the arc length ( which can be approximated to be the diameter since '$Q$' is small ) and $r$ is approximated to be the distance from the Earth to the Moon.

They both give the same answer, but I think the method using trignometry is better since we assume less. The 'correct method', however, is the second one according to the answer. If the first were acceptable, I reckon it would've been mentioned.

Is there something wrong with it? I thought that it was better since it worked for large angles too.

Also, a far more difficult question I have is this :

Suppose that we have 2 bodies $A$ and $B$ joined together into a 'circular crate'. Now, the angular displacement between the edges of the entire body is $10\mu rad$, which we assume to be the minimum angular displacement required for a clear view. Thus, we have a clear view of the entire body, which of course includes body A.

Remove body B, and then note that the angular displacement between the edges of body A is less than $10\mu rad$. Thus we will not have a clear view of body A.

But this is a contradiction, since we were earlier having a clear view of the entire body. How can removing body B affect the visibility of body A?

Why am I getting this absurd result?



Thanks to Sammy Gerbil, some parts of my question have been clarified. I, however, now have another question.

In this diagram: https://i.stack.imgur.com/iI2nL.png

We cannot distinguish between the points $(A,A_0)$ or $(A,A_1)$ or $(A,A_2)$ and basically any 2 points which do not lie on the diameter. Then how is it that we can see the shape clearly?

  • 4
    $\begingroup$ In the 2nd example, don't you mean the angular separation is $10\mu rad$? A separation of $10 rad$ is greater than $360^{\circ}$. $\endgroup$ – sammy gerbil Apr 22 '17 at 18:35
  • $\begingroup$ Yes, sorry. I just used it as an example. $\endgroup$ – Saad Apr 23 '17 at 11:12

You are wrongly interpreting the criterion. Objects do not suddenly become "clear" when the angle they subtend increases above $10\mu rad$. The criterion is only a rough indication of when details are too blurred to distinguish. There is a gradual blurring of the image when looking at smaller scales.

What the criterion means is that if the edges of A and B are greater than $10\mu rad$ apart then we can probably distinguish that they are two separate overlapping circles, whereas if the edges are less than $10\mu rad$ apart we cannot tell if we are seeing only one circle or 2 overlapping circles. In the latter case if you now remove A or B the image does not become clearer, and is only marginally more fuzzy : we still cannot tell if we are looking at 1 circle or 2 overlapping.

enter image description here

In the diagram above, the grid squares are $10\mu rad$ x $10\mu rad$. We cannot distinguish the difference between the large red and blue rectangles in the telescope, because the edges differ by less than the limit of $10\mu rad$. If they overlap we could not tell if we are seeing a single rectangle or two overlapping rectangles. The blurred images would look too similar for us to tell them apart.

Neither would we be able to tell if the red and blue circles were distinct circles or part of the same larger shape. Neither could we decide if the small red square is a different shape from the small red circle : the difference is less than the limit of $10\mu rad$ so we cannot distinguish the details. All that we could say here is that the small red circle and square are 2 separate objects.

enter image description here

In the illustration above, in the image on the left the resolution is about $0.3\mu m$. We can tell that we are not seeing a single circle, but we cannot tell if we are seeing a single object (eg a rhombus) or a composite figure (a group of spheres). If the resolution limit were say $1\mu m$ the image would be so blurred that we would not be able to tell if we were looking at a rhombus, square or circle. The image on the right is the same object viewed at a resolution of less than $0.1\mu m$. Now that we can see finer details we can tell that we are looking at a group of 6 spheres, although we cannot tell if they are joined together in some way - for that we would need a resolution of less than perhaps $0.03\mu m$.

  • $\begingroup$ Thank you for your answer, but I remain unclear. What exactly does it mean for an object to appear clearly? For example, if we have a rectangle with an angular displacement of 10u rad between the top and bottom edges, and 8u rad between the left and right edge, then would the image be clear? If not, then doesn't a similar analogy apply to the circular crate? $\endgroup$ – Saad Apr 23 '17 at 11:17
  • $\begingroup$ Whether an image is "clear" depends on what scale you look at it. If the criterion for a particular telescope and wavelength is that points in the image which are closer than 10urad cannot be distinguished, then a rectangle of 8 x 10 urad cannot be distinguished from a 5 x 12 urad rectangle or a 6 urad diameter circle. All of these shapes are so blurred that we cannot tell the difference between their outlines : the differences in the sizes are less than the 10 urad limit. ... $\endgroup$ – sammy gerbil Apr 23 '17 at 11:40
  • $\begingroup$ ... We could tell the difference between 80 x 100 urad and 50 x 130 urad rectangles, because the differences in size exceed 10 urad. We could also tell the difference between 80 x 100 and 100 x 80. But we not tell the difference between 80 x 100 and 75 x 105 because the differences between corresponding sides are less than 10 urad. $\endgroup$ – sammy gerbil Apr 23 '17 at 11:42
  • $\begingroup$ But the question simply says that if the angular displacement between two edges of an object exceeds 10u rad, it's clearly visible. In my example, will body A be clearly visible when the two bodies are merged to make the circular crate? $\endgroup$ – Saad Apr 23 '17 at 11:59
  • $\begingroup$ I have added an image. Please let me know if you are still confused. $\endgroup$ – sammy gerbil Apr 23 '17 at 12:22

Is there something wrong with it? I thought that it was better since it worked for large angles too.

No, there's nothing wrong with the formula using the tangent function. In fact, the other formula only works for small angles; it is an approximation of the tangent formula. That's because tangent can be expanded as a polynomial (see Maclaurin Series expansion): $$\tan\theta = \theta+\frac{1}{3}\theta^3+\frac{2}{15}\theta^5 + \ldots $$ where $\theta$ is in radians. If $\theta$ is small ($<<1$), then the $\theta^3$ and higher power terms are small enough to ignore for most purposes. That means $$\tan\theta \simeq \theta.$$

Applying that to your tangent formula we get $$\tan\frac{Q}{2}=\frac{s}{2r}\simeq \frac{Q}{2}$$ which is, by the small angle approximation, the same as your second formula. The trig method is correct.

  • $\begingroup$ But s=rQ is exact, not an approximation, right? Do you mean to say that the arc length can be approximated to be the diameter because Q is small? And could you please also answer the second part? Thanks! $\endgroup$ – Saad Apr 23 '17 at 11:15
  • $\begingroup$ If s is the arc length, s=rQ is exact. If s is the diameter, s=rQ is an approximation. And the arc length and diameter are approximately the same when the angle is small, and tan Q is approximately Q when Q is small. One could say both are exact in their own derivations and they are approximately equal to each other when applied to a specific situation. I believe sammy gerbil gave a good answer to the 2nd part: to resolve the true size of an object it must subtend a large enough visual angle. If the visual angle is too small, the distance from one edge to another is too uncertain. $\endgroup$ – Bill N Apr 23 '17 at 12:41
  • $\begingroup$ I see. Do you have any idea why the official solution is the s=rQ one then? $\endgroup$ – Saad Apr 23 '17 at 16:13
  • $\begingroup$ Depends on who defines "official." They are both correct. s=rQ was easier to use before scientific calculators because you don't need trig functions. It can also be useful when taking ratios involving small angle relations such as this formula. What's the official was to double 3? $2\times 3$ or $3\times 2$. They're both correct. $\endgroup$ – Bill N Apr 24 '17 at 20:15

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