How Is the Focal Length of an Ocular Determined in an Astronomical Telescope?

[SwaggerJohn]

A Huygens ocular is constructed of two thin lenses of focal lengths 10cm and 5cm respectively, separated by 7.5cm. The ocular is used in an astronomical telescope whose objective is 30cm to the left of the front lens of the ocular.

I need to find the position of the exit pupil, and then find the focal length of the ocular.

OK so I understand how to find the exit pupil and here is my idea:

Basically the exit pupil will be located where the image of the objective lens is created. So I found where the image of the objective lens will be created by Huygens ocular.

I used the following equation to do this: (x)(x') = f^2

where x is the distance of the object from the 2nd focal point of the thin lens, x' is the distance of the image formed measured from the 1st focal point of the thins lens, and f is the focal length of the thin lens.

So the image from the first lens is the object for the 2nd and eventually this gives that the exit pupil is 3cm to the right of the 2nd lens in the ocular.

Any ideas how to find the focal length of the ocular now?

 

[SwaggerJohn]

Just read the sticky. I know this looks like a homework question but its not, but I guess it should have been posted in the other board. I don't have the ability to move it or I would. Sorry about that.

 

[ogb p]

To find the focal length of the ocular, we can use the thin lens equation:

1/f = 1/do + 1/di

Where f is the focal length, do is the distance of the object from the lens, and di is the distance of the image from the lens. In this case, the distance of the object from the ocular is 7.5cm and the distance of the image from the ocular is 30cm. Plugging these values into the equation, we get:

1/f = 1/7.5 + 1/30

Solving for f, we get a focal length of approximately 6.43cm. This means that the ocular acts as a magnifying glass, with a magnification of approximately 4.67x (30cm/6.43cm).