# Magnification of an astronomical telescope not in normal adjustment?

I am stuck on this question:

A telescope consists of two thin converging lenses of focal lengths 100cm and 10cm respectively. It is used to view an object 2000cm from the objective. What is the separation of the lenses if the final image is 25cm from the eye-lens (in front of I.e. A virtual image)? Determine the magnifying power for an observer whose eye is close to the eye lens.

I am fine with the first part of which the answer is $\frac{14950}{133}$ but am really stuck on the second (finding the magnification).

Here are my thoughts: The magnification is given by the angle subtended at the eye by the image (b) / the angle subtended at the unaided eye by the object (a). This using small angle approximations gives me $a=\frac{h}{2000/19}$ where 2000/19 is the distance of the intermediate image from the objective lens and h is the hight of the intermediate image and $b=\frac{h}{50/7}$ where 50/7 is the distance from the intermediate image to the eyepiece lens. This gives me a magnification of 14.73. Whilst my textbook gives the answer as 15.6. I have also tried finding the linear magnification of both lenes and timesing them together, but this to gives me the wrong answer.

Please can someone explain where I am going wrong or direct me to any relevant formulas. thanks.

• This seems a bit odd: if the telescope is forming a real image, there isn't an afocal "magnification" as there is when a telescope does not form an image. Perhaps what is meant is "At this setting, for an object at infinity, what is the afocal magnification seen by the observer?" – Carl Witthoft Aug 27 '14 at 11:40

A telescope with two convex (converging) lenses is a Keplerian telescope. The lens with the longer focal length is the objective, and the shorter focal length lens is the eyepiece. Since it is explicitly stated that the lenses are thin, you can use the thin lens equations:

$$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$$

where $d_i$ is the distance to the image, $d_o$ is the distance to the object, and $f$ is the focal length. You will also need

$$M = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$$

where $M$ is the magnification, $h_i$ is the image height, and $h_o$ is the object height. Note the minus sign: you need to follow sign conventions on the image and object distances -- see any introductory physics textbook for coverage on these. A negative magnification represents an inverted image.

Whenever you encounter a problem like this, it is always best to draw a ray diagram. Consider the diagrams below: The first diagram shows the typical situation, where the intermediate image is outside of the focal length of the second lens. The second diagram shows the situation you are interested in, where the image of the first lens falls inside the focal length of the second lens. This yields a virtual image between the two lenses.

Using the first equation, we calculate $d_{i1}$: $$d_{i1} = \frac{d_o f_1}{d_o - f_1} = 105.263 \ \text{cm}$$ Now, this yeilds a magnification for the first image of $$M_1 = -\frac{d_{i1}}{d_o} = -0.0526$$

As you have already found, the separation between the lenses is $L=112.406 \ \text{cm}$ and you are given that $d_{i2} = -25 \ \text{cm}$. The minus sign tells you the image is formed on the "wrong side" (left in the image) of the lens, i.e., it is a virtual image. This implies that $$d_{o2} = L - d_{i1} = 7.143 \ \text{cm}$$

Next, we compute the second magnification: $$M_2 = -\frac{d_{i2}}{d_{o2}} = 3.500$$

Now, the total magnification is simply the product of the two:

$$M_{total} = M_1 M_2 = -0.184$$

However, what you really want is the angular magnification.

Assume, for simplicity, that the original object height is $1 \ \text{cm}$. That means that the final virtual image will be $0.184 \ \text{cm}$ tall and inverted. However, the original image is $d_o + L = 2112.406 \ \text{cm}$ away, which implies an angular size (using small angle approximation) of: \begin{equation} s = \theta r \implies \theta_1 = \frac{s}{r} = \frac{1 \ \text{cm}}{2112.406 \ \text{cm}} = 4.734 \times 10^{-4} \ \text{radians} \end{equation}

The virtual image, on the other hand, will have an angular size of \begin{equation} \theta_2 = \frac{s}{r} = \frac{0.184 \ \text{cm}}{25 \ \text{cm}} = 7.36 \times 10^{-3} \ \text{radians} \end{equation}

Thus the angular magnification will be \begin{equation} M_{angular} = \frac{\theta_2}{\theta_1} = \frac{7.36 \times 10^{-3} \ \text{radians}}{4.734 \times 10^{-4} \ \text{radians}} = 15.547 \end{equation} I have been sloppy with rounding off my numbers, but this should give you the idea. Cheers!