Calculating Angular Magnification and Image Position in Compound Lens Systems

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An object placed 1.5m in front of a convex lens with a focal length of 500mm produces a real, diminished, and inverted image located 750mm behind the lens. A second convex lens with a focal length of 25mm, positioned 770mm behind the first lens, results in a virtual image located at -100mm. The discussion includes a request for guidance on calculating angular magnification for the final image when it is formed at infinity. Angular magnification is defined in terms of the angles subtended by the images at the respective lenses. The condition for the final image to be at infinity is also queried.
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Homework Statement


1. An object is placed 1.5m in front of a convex lens of focal length 500mm. Find the position of the image formed and state its nature.

2. A second convex lens of focal length 25mm is placed 770mm behind the first convex lens. Find the position of the final image formed and state whether the image is real or virtual.

3. Find the angular magnification for the final image if it is formed at infinity.

The Attempt at a Solution


1.\frac{1}{u} + \frac{1}{v} = \frac{1}{f}
\frac{1}{1500} + \frac{1}{v} = \frac{1}{500}
v = 750mm

m = \frac{-v}{u} = \frac{-750}{1500} = -0.5

Image is Real as v > 0
Image is Diminished as |m|<1
Image is Inverted as m < 0

2. \frac{1}{u} + \frac{1}{v} = \frac{1}{f}
\frac{1}{20} + \frac{1}{v} = \frac{1}{25}
v = -100

Image is virtual as v < 0

3. This is the part I don't know how to do. Any advice would be appreciated.
 
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You can define angular magnification as
M_A=\frac{tanθ_1}{tanθ_2}

Where θ1 is the angle subtended by the first image at the first lens and θ2 is the angle subtended by the second image at the second lens.(Check out the attachment)

Under what condition is the final image at infinity?
 

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