Calculating Multiple Images in a Plane Mirror

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In a room with two plane mirrors on opposite walls, a person standing 5 feet from the left mirror and 10 feet from the right mirror can see multiple images. The first image in the left mirror is located 10 feet behind it, while the first image in the right mirror appears 5 feet behind, resulting in a total perceived distance of 20 feet from the left mirror. The second and third images are calculated as being 30 feet and 40 feet away, respectively. Diagrams can aid in visualizing the distances and understanding the reflections. This method effectively demonstrates how multiple images are formed by plane mirrors.
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Plane Mirror
A person walks into a room that has, on opposite walls, two plane mirrors producing multiple images. Find the distances from the person to the first three images seen in the left-hand mirror when the person is 5ft from the mirror on the left wall and 10 ft from the mirror on the right wall.

I know the first image is 10 ft because in the plane mirror, the image is as far behind the mirror as the object is in front. I don’t know how to figure out the second and the third images? I need you help please. Thank you.

Vu95112
 
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Try making a diagram like that found in scrolling down the page at http://dev.physicslab.org/Document.aspx?doctype=3&filename=GeometricOptics_PlaneMirrors.xml":

It may help to keep you from getting confused. The first image in the left mirror is the virtual image of the person and as you say 10' behind the mirror. Apply the same logic to the first image in the right mirror. It would be 5' behind the mirror and since the mirrors are 15' apart, would seem to be 20' away. And so forth.
 
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Hello denverdoc,
Thank you very much for your help. A diagram is very useful.
The first image is 10ft
The second image is 30 ft
The third image is 40 ft

Thank you,

Vu95112
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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