Analyzing 2-Lens System: Image Position, Magnification & Nature

[anizet]

Homework Statement

A convex lens (f1=300mm) is placed 200 mm from a concave lens (f2=-50mm). An object is placed 6m away form the convex lens. (the order is as follows: object - 6m - convex lens - 200mm - concave lens).
Determine the position, magnification and nature of a final image.

Homework Equations

for a single lens:
1/x+1/y=1/f (where x is distance from the lens to the object, y is the distance from the lens to the image),

The Attempt at a Solution

Should I analyze 2 lenses separately?
At first first lens, which gives the image at y1:
1/6m+1/y1=1/300mm
y1=~31.5cm, so it is placed behind the second lens! Should I now consider the second lens, starting from the point, where is the image? I mean:
1/x2+1/y2=1/f2
where x2=20cm-31.5 cm=-11.5cm ?
I'm a little bit confused..
And finally, how to calculate the magnification?

 

[Doc Al]

Should I analyze 2 lenses separately?
At first first lens, which gives the image at y1:
1/6m+1/y1=1/300mm
y1=~31.5cm, so it is placed behind the second lens! Should I now consider the second lens, starting from the point, where is the image? I mean:
1/x2+1/y2=1/f2
where x2=20cm-31.5 cm=-11.5cm ?

Exactly right. Solve it in steps. The image from the first lens becomes the object for the second lens (virtual object, in this case, since it's behind the second lens).

And finally, how to calculate the magnification?

Find the magnification for each, then multiply them.

 

[baba89]

I would first clarify the terminology used in the problem. It is important to note that the term "nature" is not typically used in optics and may refer to the type of image formed (real or virtual) or the orientation (upright or inverted). I would suggest using more specific terminology to avoid confusion.

To solve this problem, it is important to understand the behavior of light as it passes through a 2-lens system. In this case, the convex lens will form an image of the object at a distance y1 from the lens, and this image will then act as the object for the concave lens to form a final image at a distance y2 from the concave lens.

To determine the position of the final image, we can use the thin lens equation for each lens separately. For the convex lens, we have 1/x + 1/y1 = 1/f1, where x is the distance from the object to the lens, y1 is the distance from the lens to the image, and f1 is the focal length of the convex lens. Plugging in the given values, we get 1/6 + 1/y1 = 1/0.3, which gives y1 = 0.315 m. This means that the image formed by the convex lens will be located 31.5 cm to the right of the lens.

Now, we can use this value of y1 as the object distance for the concave lens. Using the thin lens equation again, we get 1/y1 + 1/y2 = 1/f2, where y2 is the distance from the concave lens to the final image and f2 is the focal length of the concave lens. Plugging in the given values, we get 1/0.315 + 1/y2 = 1/-0.05, which gives y2 = -0.06 m. This means that the final image formed by the concave lens will be located 6 cm to the left of the lens, which confirms our initial assumption that the final image will be formed behind the concave lens.

To calculate the magnification, we can use the formula M = -y2/y1, where M is the magnification, y1 is the distance from the convex lens to the image, and y2 is the distance from the concave lens to the final image. Plugging in the values