Archived Telescope geometric optics problem

[ryan11]

Ok I'm working on my geometric optics homework and this is the last problem and I can't seem to get it right.

An 6 astronomical telescope has a 32 cm focal-length objective lens. After looking at stars, an astronomer moves the eyepiece 1.0 cm farther away from the objective to focus on nearer objects. What is the distance to the nearer objects?

I'm not really sure how to set this up. I know that magnification=fo/fe but I don't know how to find the distance of the nearer object. Thanks for any help.

 

[Maidien]

I can't imagine, telescope is used to see stars which are so far away that we always take theirs distance to objective len infinite in formulae. How can it change @@.

 

[Drakkith]

I can't imagine, telescope is used to see stars which are so far away that we always take theirs distance to objective len infinite in formulae. How can it change @@.

Easy. Just move the eyepiece back and forth. That will change the focus. This is just like using binoculars to view wildlife, sporting events, or something else where the objects in view are at many different distances. You have to adjust the focus if the change in distance is large, such as viewing a bird on a limb where the background is miles away. You can focus on the bird and perhaps the tree at the same time, but the distant mountains will most likely be blurry, and vice-versa. The details of why this is so would require you to delve into geometrical optics.

 

[Charles Link]

I'm hoping I'm not giving more info. than I should, but hints on this one are hard to give without giving almost the complete solution. For a telescope viewing at infinity, the focal point of the eyepiece is at the focal point of the objective lens which is where the image forms. $1/f=1/s+1/m$. When $s=+\infty$, $\$, $m=f =32$ cm. If the eyepiece needs to be moved 1 cm, it means the object focused (made a real image)from the objective lens at m=33 cm. Use the lensmaker's formula $1/f=1/s +1/m$ . $f$ =32 cm and $m$ =33 cm. Solve for $s$ =object distance.

 

[Drakkith]

I'm hoping I'm not giving more info. than I should, but hints on this one are hard to give without giving almost the complete solution.

You're fine with posting complete solutions here in this forum. Just don't do it in the homework forums. See the post stickied at the top.

 

[George Jones]

The last time the original poster logged on was more than 12 years ago, which was when the homework was due..

 

[Charles Link]

The last time the original poster logged on was more than 12 years ago, which was when the homework was due..

It was fun solving the homework problem. This one was one of the easier ones.