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I took an image of Jupiter through my 8" Dobsonian Telescope, attaching a DSLR and a 1.25" Barlow Lens where the eyepiece goes, as shown in this video: https://www.youtube.com/watch?v=reFxoF3XoaU

Through numerous online sources, I learnt that to find the angle of view of my image, the magnification of my image needed to be calculated, and that this value along with the given field view for my eyepiece.

http://www.rocketmime.com/astronomy/Telescope/Magnification.html:

The above website answers the following question, the calculations of which I attempted to mimic.

My first telescope was a Meade 6600 -- they don't make it any more -- it's a 6-inch f/5 Newtonian scope. It came with a 25mm eyepiece. So... what was the magnification I was getting with this scope?

Here are the calculations I did in hopes of getting the above result:

$$Diameter = 8" = 203.2\ mm$$

$$f_{ratio} = \frac{focal\ length\ of\ objective}{203.2\ mm}$$

$$\therefore focal\ length\ of\ objective = 203.2 \cdot 5.9 = 1200\ mm$$

$$Magnification = \frac{focal\ length\ of\ objective}{focal\ length\ of\ eyepiece} = \frac{1200}{x}$$

As shown in the video (the first link above), I didn't use an eyepiece to take my picture - I used a barlow lens, a couple of adapters, and a DSLR. So at this point, I am not sure what value to use for the "focal length of the eyepiece." How can I proceed to calculate the magnification?

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    $\begingroup$ That’s a nice photo of the moons! Unfortunately, it’s overexposed for Jupiter itself, which has caused the image to bleed & grow. See how large it appears relative to the radii of the moons? You’re going to get an overly large radius of Jupiter because of that. $\endgroup$ – Bob Jacobsen Jul 15 '18 at 18:35
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For a telescope with an objective lens and an eyepiece the magnification $M$ is defines as

$$M = \dfrac{\text{angle subtended by object when viewed through telescope}}{\text{angle subtended by object by naked eye}}$$

If you do not have an eyepiece then you form an image of the object in the focal plane of the objective lens and the angle subtended by the object (visual angle) when viewed through the objective lens is exactly the same as the visual angle subtended by the naked eye so using the magnification formula given above is inappropriate as $M=1$.

Assuming a small angle approximation for a given visual angle the linear size of the image in the focal plane of the objective is proportional to the focal length of the objective.
So the longer the focal length the larger will be the image formed in the focal plane.

The objective of your telescope is acting like the lens of your camera.
To make a comparison and so find a magnification of the arrangement it is often assume that a "standard" 35 mm SLR camera has a 50 mm lens and a magnification of one.

The magnification of your set up with the telescope objective acting as the lens of your camera is given by:

$$M = \dfrac{\text{focal length of telescope objective in millimetres}}{50}$$

When you use a Barlow lens you effective increase the focal length of your objective and the most usual increase is $\times 2$.
So the magnification would be increased by a factor of two.

For the angle of view you need to know the size of the sensor in your camera. The 50 mm focal length lens quoted above as a reference assumed a 35 mm film camera but for other than the most expensive SLR cameras the sensor is less than 35 mm by a factor of at least 1.6 and so you have to make a comparison with a camera of focal length $1.6 \times 50 = 80$ mm to find the magnification.

With your set up you can work out the angle of view if you know the effective focal length of your telescope with the Barlow lens $f_{+B}$ and the size of your sensor $d$.
Then measurements using your image of Jupiter and its moon will give you the angualar size of Jupiter.

Perhaps it might be better to work "backwards" and take a photograph of two objects whose angular separation is known and so calibrate your setup?

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