Rayleigh's Criterion + Moon (What am I doing wrong?)

[Erickly]

Homework Statement

Suppose you wanted to be able to see astronauts on the moon. What is the smallest diameter of the objective lens required to resolve a 0.60 m object on the moon? Assme the wavelength of the light is near the middle of the visible spectrum: 550 nm yellow light.
(in m)

A: 4.25×10^2 B: 4.97×10^2 C: 5.82×10^2 D: 6.81×10^2 E: 7.96×10^2 F: 9.32×10^2 G: 1.09×10^3 H: 1.28×10^3

Homework Equations

Rayleigh's Criterion:
=> θ1 = 1.22λ/D
=> w = (2.44*λ *L)/D
w is width; L is length from screen to object.

The Attempt at a Solution

So I know we are calculating the diameter of the lens needed so I solved the equation for D and got:

w = (2.33*λ *L)/D

after this step I converted all my values to meters and got: these values:

(2.44*5.5E-7m*3.97E8m) / 0.6m = 888.6m which is in the range I need but It has to be an exact answer.

Can anyone help me understand what I did wrong?

 

[Charles Link]

Where did the 2.33 come from? I think they often use 2.44, but this is somewhat subjective. Also, the distance to the moon can vary, so they should really supply you with the number they want you to use.

 

[Erickly]

Where did the 2.33 come from? I think they often use 2.44, but this is somewhat subjective. Also, the distance to the moon can vary, so they should really supply you with the number they want you to use.

Opps I meant 2.44 but wrote 2.33 on the post. Sadly I got no such info for the distance. I played around with the values though and think that might be the reason why...

 

[haruspex]

When I look up the distance to the moon, www hits tell me the average is 384,400 km.
Not sure why you used 2.44 instead of 1.22. Can you explain your reasoning there?

 

[Erickly]

When I look up the distance to the moon, www hits tell me the average is 384,400 km.
Not sure why you used 2.44 instead of 1.22. Can you explain your reasoning there?

My lecture notes say to use 2.44 in the equation for width. It says use 1.22 to find theta and 2.44 for width.

 

[haruspex]

My lecture notes say to use 2.44 in the equation for width. It says use 1.22 to find theta and 2.44 for width.

Ok, but that would be in the context of a specific problem. What is that problem and what do "theta" and "width" represent in it?

 

[Erickly]

Ok, but that would be in the context of a specific problem. What is that problem and what do "theta" and "width" represent in it?

It is to measure the height of a circular aperture (one circle only, not 2 merging ) . Theta is the angle between the top part of the circle and bottom if you are looking at it as a flat circular surface. Width is top of the circle to bottom so diameter of the circle.

 

[haruspex]

It is to measure the height of a circular aperture (one circle only, not 2 merging ) . Theta is the angle between the top part of the circle and bottom if you are looking at it as a flat circular surface. Width is top of the circle to bottom so diameter of the circle.

If I have understood that correctly, this is viewing a circle, full on, at some distance s from the observer. width is the diameter of the circle and theta the angle that width subtends at the observer.
$w=2s\sin(\frac{\theta}2)≈s\theta$.
If so, I cannot understand why you would not use 1.22 for both.

If you use 1.22 and the shorter distance to the moon that I posted, do you get close to one of the offered answers? What about if you use 2.44?

 

[Erickly]

If I have understood that correctly, this is viewing a circle, full on, at some distance s from the observer. width is the diameter of the circle and theta the angle that width subtends at the observer.
$w=2s\sin(\frac{\theta}2)≈s\theta$.
If so, I cannot understand why you would not use 1.22 for both.

If you use 1.22 and the shorter distance to the moon that I posted, do you get close to one of the offered answers? What about if you use 2.44?

Right so when I use 1.22 and 3.84E8 for distance form Earth to moon, I get 429.44

When I use the same distance but use 2.44 I get 858.88.

Neither are possible answer. I'm pretty lost :)

 

[Charles Link]

It is for this extra detail that (the 1.22 vs. 2.44), that the Rayleigh criteria (and I think I have seen the resolution results calculated using both numbers) very subjective. They want you to compute something accurate to decimal places that does not have two-decimal place accuracy.

 

[Charles Link]

Additional item: They would actually do better to teach and emphasize the fundamentals than to just present the formulas. The Rayleigh criteria comes from the result that a faraway point source will create a diffraction pattern upon being incident and focused by the lens in the focal plane so that it focuses to a circular spot/blob instead of a point that covers a $\Delta \theta \approx 1.22 \frac{\lambda}{D}$. (I believe this is the $\theta$ measured from straight on to the first zero of the diffraction intensity pattern (the Airy disc), so that the blob has radius $r=f (1.22) \frac{\lambda}{D}$, and the diameter of the blob is $d=f (2.44) \frac{\lambda}{D}$). A second faraway (small=point size type) object that is located an angle $\theta_o$ from the first object will come to a focus, also with a similar circular blob, in the focal plane of the lens, at an angle of $\theta_o$ away from the center of the first blob, which is a distance $d_1=f \theta_o$ from center of the first blob. The question is, how big does $\theta_o$ need to be to tell that we have two separate blobs as opposed to a single blob? (Oftentimes, the answer is given as $\theta_o=2 \, \Delta \theta = 2.44 \frac{\lambda}{D}$). Or alternatively, given $\theta_o$, how big does the lens diameter $D$ need to be to make each blob (which is a diffraction pattern also known as an Airy disc) small enough, that we can basically see that there are two distinct blobs. $\\$ Given the formula, anyone can do a little arithmetic, but it has been my experience that only a limited number of students are able to pick up the details of this concept, because they often are not presented with enough of the details. $\\$ One additional detail: The circular aperture of the lens basically creates a circular diffraction pattern from a single faraway point source that could be viewed in the far-field if it were just an aperture and not a lens. The lens makes it so that the far-field diffraction pattern is brought to a focus and can be viewed in the focal plane of the lens, instead of needing to view it in the far-field. This is because parallel rays that are incident on the lens at angle $\theta$ come to a focus at a position $x=f \, \theta$ in the focal plane. (You can think of it as the rays\wavefront from the source first encounter the circular aperture, and then the lens, placed just after the aperture, brings the far field diffraction pattern into view in the focal plane of the lens if a screen is placed there). The far-field pattern from the circular aperture is a diffraction pattern that has a spread (angular distance to off-center to the first zero of the diffraction pattern) $\Delta \theta$. The larger the aperture the narrower the angular spread of this circular (Airy disc) pattern. So, instead of the point source focusing to a point, it instead makes a finite sized blob. A second point source needs to be far enough from the first source that the result isn't two overlapping blobs that looks like one blob. $\\$ Edit: And a google of the topic "Rayleigh criteria" suggests that 1.22 and not 2.44 is the number to use here.

 

[haruspex]

Right so when I use 1.22 and 3.84E8 for distance form Earth to moon, I get 429.44

When I use the same distance but use 2.44 I get 858.88.

Neither are possible answer. I'm pretty lost :)

The offered answers form a geometric progression, the increase being 17% at each step.
429.44 differs from one of them by a little over 1%. Given the uncertainty in the distance to the moon that strikes me as close enough. On the other hand, 859 falls half way between E and F.

 

[Charles Link]

All indications are that I would go with "A" as well. See also my Edit=last sentence in post 11 above.

 

[haruspex]

Additional item: They would actually do better to teach and emphasize the fundamentals than to just present the formulas. The Rayleigh criteria comes from the result that a faraway point source will create a diffraction pattern upon being incident and focused by the lens in the focal plane so that it focuses to a circular spot/blob instead of a point that covers a $\Delta \theta \approx 1.22 \frac{\lambda}{D}$. (I believe this is the $\theta$ measured from straight on to the first zero of the diffraction intensity pattern (the Airy disc), so that the blob has radius $r=f (1.22) \frac{\lambda}{D}$, and the diameter of the blob is $d=f (2.44) \frac{\lambda}{D}$). A second faraway (small=point size type) object that is located an angle $\theta_o$ from the first object will come to a focus, also with a similar circular blob, in the focal plane of the lens, at an angle of $\theta_o$ away from the center of the first blob, which is a distance $d_1=f \theta_o$ from center of the first blob. The question is, how big does $\theta_o$ need to be to tell that we have two separate blobs as opposed to a single blob? (Oftentimes, the answer is given as $\theta_o=2 \, \Delta \theta = 2.44 \frac{\lambda}{D}$). Or alternatively, given $\theta_o$, how big does the lens diameter $D$ need to be to make each blob (which is a diffraction pattern also known as an Airy disc) small enough, that we can basically see that there are two distinct blobs. $\\$ Given the formula, anyone can do a little arithmetic, but it has been my experience that only a limited number of students are able to pick up the details of this concept, because they often are not presented with enough of the details. $\\$ One additional detail: The circular aperture of the lens basically creates a circular diffraction pattern from a single faraway point source that could be viewed in the far-field if it were just an aperture and not a lens. The lens makes it so that the far-field diffraction pattern is brought to a focus and can be viewed in the focal plane of the lens, instead of needing to view it in the far-field. This is because parallel rays that are incident on the lens at angle $\theta$ come to a focus at a position $x=f \, \theta$ in the focal plane. (You can think of it as the rays\wavefront from the source first encounter the circular aperture, and then the lens, placed just after the aperture, brings the far field diffraction pattern into view in the focal plane of the lens if a screen is placed there). The far-field pattern from the circular aperture is a diffraction pattern that has a spread (angular distance to off-center to the first zero of the diffraction pattern) $\Delta \theta$. The larger the aperture the narrower the angular spread of this circular (Airy disc) pattern. So, instead of the point source focusing to a point, it instead makes a finite sized blob. A second point source needs to be far enough from the first source that the result isn't two overlapping blobs that looks like one blob. $\\$ Edit: And a google of the topic "Rayleigh criteria" suggests that 1.22 and not 2.44 is the number to use here.

Thanks for the explanation of how the factor of 2 arises. The question is, does the present problem equate to discriminating two blobs 0.6m apart? It doesn't seem that way to me. We only want to know if we can see the astronauts, not whether we can form an idea of how tall they are. So I would say this problem calls for the 1.22 factor.

Edit: I probably should have said "how wide they are"

 

[Charles Link]

Thanks for the explanation of how the factor of 2 arises. The question is, does the present problem equate to discriminating two blobs 0.6m apart? It doesn't seem that way to me. We only want to know if we can see the astronauts, not whether we can form an idea of how tall they are. So I would say this problem calls for the 1.22 factor.

A google of the "Rayleigh criteria" basically says that 1.22 is the number to use, even if it might be just as logical to use 2.44. Resolving something in a telescope is a somewhat subjective problem, but when they say to use the Rayleigh criteria, that would mean to use 1.22.

 

[Erickly]

Thanks for the explanation of how the factor of 2 arises. The question is, does the present problem equate to discriminating two blobs 0.6m apart? It doesn't seem that way to me. We only want to know if we can see the astronauts, not whether we can form an idea of how tall they are. So I would say this problem calls for the 1.22 factor.

Edit: I probably should have said "how wide they are"

One thing to note, I did a calculation with 2.44 & the distance from Earth to moon being 3.56E8m and got 795.35 which is very close to choice E. Do you guys think E could be a possibility?

[source for distance (surface to surface) : https://socratic.org/questions/what...en-the-moon-s-surface-and-the-earth-s-surface ]

 

[Charles Link]

One thing to note, I did a calculation with 2.44 & the distance from Earth to moon being 3.56E8m and got 795.35 which is very close to choice E. Do you guys think E could be a possibility?

[source for distance (surface to surface) : https://socratic.org/questions/what...en-the-moon-s-surface-and-the-earth-s-surface ]

When using the Rayleigh criteria, the answer is no. The google search of a couple articles seems to be fairly clear that the Rayleigh criteria says that 1.22 is the number to use in determining resolution. $\\$ Edit: I re-read the question. The Rayleigh criteria doesn't look like it was specified in the statement of the problem. (You @Erickly introduced it in part 2 of the homework template). If they wanted two-decimal place accuracy, or even one sig-fig, they really should have stated to use the Rayleigh criteria in the statement of the problem. Alternatively, I do think the Rayleigh criteria is the universally accepted criteria to use in a problem such as this. Otherwise, any of their answers is really fairly reasonable.

 

[haruspex]

One thing to note, I did a calculation with 2.44 & the distance from Earth to moon being 3.56E8m and got 795.35 which is very close to choice E. Do you guys think E could be a possibility?

[source for distance (surface to surface) : https://socratic.org/questions/what...en-the-moon-s-surface-and-the-earth-s-surface ]

As I posted, based on Charles' explanation for how the 2.44 factor is derived, that would tell you the size of lens required to judge the size of each astronaut. The lens needed to detect an astronaut, as no more than a blob, would obviously be a lot smaller, and the 1.22 factor is reasonable.

As to the distance to the moon to be used, certainly the Earth and Moon radii should be excluded. I hadn't expected it would make so much difference - about 2%. And if you interpret the question as being able to see the astronauts some of the time then you can defend using the perigee, but I think that's a bit of a stretch. It is more reasonable to use the average distance.
The question ought to have provided the distance to be used or made the options more widely spaced.

 

[Erickly]

As I posted, based on Charles' explanation for how the 2.44 factor is derived, that would tell you the size of lens required to judge the size of each astronaut. The lens needed to detect an astronaut, as no more than a blob, would obviously be a lot smaller, and the 1.22 factor is reasonable.

As to the distance to the moon to be used, certainly the Earth and Moon radii should be excluded. I hadn't expected it would make so much difference - about 2%. And if you interpret the question as being able to see the astronauts some of the time then you can defend using the perigee, but I think that's a bit of a stretch. It is more reasonable to use the average distance.
The question ought to have provided the distance to be used or made the options more widely spaced.

I see, so 1.22 then. Thanks guys, I really appreciate!