If you were standing at the center of curvature in front of a concave mirror, what image would be projected?
A. 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.

B.The image would be upright, would look shorter than you, and would be closer to the mirror than you are.

C.The image would be upside down, would look shorter than you, and would be closer to the mirror than you are.

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

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akshayrao091 akshayrao091 · Answer 1 · 2016-06-22T20:37:09+02:00

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.

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aristocles aristocles · Answer 2 · 2019-06-13T03:14:59+02:00

Answer:

A. 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 by the mirror formula

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

now we know that object is placed at distance equal to the radius of mirror

[tex]d_o = -R[/tex]

also we know that focal length of mirror is half of the radius of mirror

[tex]f = \frac{R}{2}[/tex]

now we have

[tex]\frac{1}{d_i} - \frac{1}{R} = -\frac{2}{R}[/tex]

so we have

[tex]d_i = -R[/tex]

so image will be at same position as that the position of object and it is inverted in position so correct answer will be

A. 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.