If you were standing at the center of curvature in front of a concave mirror, what image would be projected?
The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.
The image would be upright, would look shorter than you, and would be closer to the mirror than you are.
The image would be upside down, would look shorter than you, and would be closer to the mirror than you are.
The image would be upright, would look as tall you, and would be at the same distance from the mirror as you are.

"The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are" is the type of image among the choices given in the question that would be projected. The correct option among all the options that are given in the question is the first option. I hope it helps you.

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aristocles aristocles · Answer 2 · 2019-08-18T05:37:45+02:00

Answer:

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

Explanation:

As we know that the object position is at center of curvature of the mirror

So here the distance is given as

[tex]d = R = 2f[/tex]

now by mirror formula of relation between distance of image and distance of object

[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]

[tex]\frac{1}{d_i} + \frac{1}{2f} = \frac{1}{f}[/tex]

[tex]d_i = 2f[/tex]

so image will form at the position of object itself

Now we will say that the magnification of the object will be

[tex]M = -\frac{d_i}{d_o} = -1[/tex]

so image will be upside down and same as that the height of object

so correct answer would be

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.