Image distance of a Keplerian telescope

[guv]

I found a problem where the question is asking for image distance.

The focal lengths of the objective and eyepiece convex lenses are in the ratio 8:1 for aparticular telescope. The telescope is pointed at a building that is 10 km from thetelescope and is 100 m tall. What is the distance to the image?

Normally this type of question only concerns with the magnification factor. The provided solution has a formula for the image distance,

$d_i = d_o^2 (\frac{f_e}{f_o})^2 = 10^2 (\frac{1}{8})^2 = 156.2 m$

Can someone explain this equation? Thanks,

 

[kuruman]

It seems that the provided solution is dimensionally incorrect. The left hand side has dimensions of length whilst the right hand side has dimensions of length squared.

In any case, you should be able to get the correct expression by considering that the telescope is constructed so that the objective creates a real image of the building at distance $f_o+\epsilon$ which becomes an object for the eyepiece at distance $f_e-\epsilon$. The calculation requires the use of the lens equation in tandem.

 

[sophiecentaur]

The left hand side has dimensions of length

I also think it should be distance squared. But then everything could be square rooted . . . .?

The calculation requires the use of the lens equation in tandem.

But the lens equation, for the eyepiece, needs more information (like the intermediate object distance from the EP). It's up to the user to adjust that 'to taste' and the image distance can be chosen over a range from the user's near point to infinity (and beyond!).

Normally the magnification factor is concerned with angles and not heights. Is it possible that there is some more information somewhere in the printed question? There's one clear error so another one is quite possible.

We need to see the actual question of a reference to it.

 

[guv]

I also think it should be distance squared. But then everything could be square rooted . . . .?But the lens equation, for the eyepiece, needs more information (like the intermediate object distance from the EP). It's up to the user to adjust that 'to taste' and the image distance can be chosen over a range from the user's near point to infinity (and beyond!).

Normally the magnification factor is concerned with angles and not heights. Is it possible that there is some more information somewhere in the printed question? There's one clear error so another one is quite possible.

We need to see the actual question of a reference to it.

That's exactly correct, there will be an adjustable delta between the two foci of the objective and eye piece $\delta$. Therefore exact calculation will yield a final image distance from eye piece
$- \frac{f_e (f_e - \delta )}{\delta }$
Note that this has nothing to do with objective lens focal length and it depends on the adjustable distance which is not given.

I posted the exact problem (Physics Bowl 2023 Question 48) without cutting out anything. I quoted the solution wrong, the provided solution is

$d_i = d_o^2 (\frac{f_e}{f_o})^2 = 10^4 (\frac{1}{8})^2 = 156.2 m$

Here is the direct link to the problem and solution https://aapt.org/programs/physicsbowl/currentexam.cfm

I wonder if aapt got this problem/solution wrong.

 

[hutchphd]

That's exactly correct, there will be an adjustable delta between the two foci of the objective and eye piece $\delta$. Therefore exact calculation will yield a final image distance from eye piece
$- \frac{f_e (f_e - \delta )}{\delta }$
Note that this has nothing to do with objective lens focal length and it depends on the adjustable distance which is not given.

I posted the exact problem (Physics Bowl 2023 Question 48) without cutting out anything. I quoted the solution wrong, the provided solution is

$d_i = d_o^2 (\frac{f_e}{f_o})^2 = 10^4 (\frac{1}{8})^2 = 156.2 m$

Here is the direct link to the problem and solution https://aapt.org/programs/physicsbowl/currentexam.cfm

I wonder if aapt got this problem/solution wrong.

$d_i = d_o (\frac{f_e}{f_o})^2 = 10^4 (\frac{1}{8})^2 = 156.2 m$

 

[kuruman]

I wonder if aapt got this problem/solution wrong.

The solution at the site is $d_i=d_o\left(\frac{f_e}{f_o}\right)^2$, not what you have. The $10^4$ is $d_o = 10$ km, not the square of the height of the building.

Here is how one obtains the answer following the outline in post #2. The real image is formed by the objective at distance $f_o+\epsilon$ beyond the focal point.
$$\frac{1}{d_o}+\frac{1}{f_o+\epsilon}=\frac{1}{f_o}\implies \frac{1}{d_0}=\frac{\epsilon}{f_o(f_o+\epsilon)}\approx\frac{\epsilon}{f_o^2}.$$This real image is just inside the focal length of the eyepiece at distance $f_e-\epsilon$. Then $$\frac{1}{d_i}+\frac{1}{f_e-\epsilon}=\frac{1}{f_e}\implies \frac{1}{d_i}=-\frac{\epsilon}{f_e(f_e-\epsilon)}\approx-\frac{\epsilon}{f_e^2}.$$You can finish the solution to get the magnification and the AAPT answer from here. Note that this relation is only approximate.

 

[guv]

That's the same thing I derived earlier,
$- \frac{f_e (f_e - \delta )}{\delta }$
How do you determine $\delta$ or $\epsilon$ from here?

 

[sophiecentaur]

The solution at the site is di=do(fefo)2, not what you have. The 104 is do=10 km, not the square of the height of the building.

I was over thinking it. The "telescope" is assumed to be set up for infinite object and image distances and that's the missing info. As a bino and scope user I would never have assumed that.
Not a fair question because it triggers panic and not 'I can do this'.