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!
Both equations are in fact structurally similar with Abbe limit given by
$d= \dfrac{\lambda}{2\mathrm{NA}}$
And Rayleigh limit given by
$d =1.22 \dfrac{\lambda} {2\mathrm{NA}}= 0.61\dfrac{\lambda} {\mathrm{NA}}$
where lambda is the wavelength and $\mathrm{NA}$ the numerical aperture of the light collecting lens.
The factor 1.22 comes from the definition of Bessel function of 1st kind,the fact that 1st minima of the diffraction pattern appears at 1.22 units from the central zero.
Rayleigh criterion is thus a modification of the Abbe Resolution limit. The Rayleigh criterion states that in order for 2 closely placed PSF to be resolved, the central maxima of one should lie exactly at the first minima of the second one. Since the Airy pattern is defined by the Bessel function, the minimum separation between the 2 patterns should be $1.22 \lambda/ 2\mathrm{NA}$ instead of just $\lambda/2\mathrm{NA}$ considering that the first minima will be at 1.22 times the unit from the central maxima.
Best Answer
An Airy disk having a having a central spot in the focal plane the radius of the first zero is $$d = 1.22f\frac{\lambda}{D}$$
In the object space the angular resolution is then $$\theta = 1.22\frac{\lambda}{D}$$
where $D$ is the diameter of the aperture
You should check your definition once again. https://en.wikipedia.org/wiki/Diffraction#Circular_aperture