Ican't remember how to calculate this.
match the object positions with the image positions for a convex lens. (4 points)

You need to draw two lines from the head of the arrow: one parallel to the axis (which will be bent) and the other one passing through the center of the converging lens (which will continue straight) and see where they intersect.

If the arrow is at an infinite distance, the image will be point-like in F.
If the arrow is at a distance greater than 2F, the image will be upside down between F and 2F.
If the arrow is at a distance of 2F, the image will be upside down in 2F.
If the arrow is at a distance between F and 2F, the image will be upside down beyond 2F.
If the arrow is at a distance F, all rays proceed parallel to each other and no image is formed.
If the arrow is at a distance smaller than F, the image will be virtual.

Therefore, the correct matches are 1B, 2C, 3D, and 4A.

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By dividing you can find the answer