With a concave mirror, an object at a distance less than the focal length from the lens will produce an image that is

Question

BlackTom AI · Accepted Answer

For a concave mirror with object distance do less than the focal length f, the mirror equation and ray behavior directly determine the image type.

Relevant concepts: the mirror equation 1/f = 1/do + 1/di (sign convention: f positive for concave mirror; a negative di indicates a virtual image located behind the mirror). Magnification m = -di/do; a positive m means the image is upright, negative m means inverted; |m| gives size relative to the object.

Step-by-step reasoning with intermediate steps (example values):
1. Choose f = +10 cm (concave), set object distance do = 5 cm (< f).
2. Apply mirror equation: 1/di = 1/f - 1/do = 1/10 - 1/5 = 0.1 - 0.2 = -0.1.
3. Solve for di: di = -10 cm. The negative sign shows the image is virtual and located "behind" the mirror (i.e., on the side from which the object appears to be seen).
4. Compute magnification: m = -di/do = -(-10)/5 = 10/5 = 2. Positive m = upright; magnitude 2 means the image is enlarged (twice the object height).

Ray-geometry explanation: when the object is inside the focal length, reflected rays diverge; their backward extensions intersect behind the mirror. Those intersection points form a virtual image that is upright and larger than the object.

Why the correct choice is c and why the others are wrong:
- a. "Virtual and inverted" is incorrect because virtual images produced by a concave mirror (when object is inside f) are upright, not inverted.
- b. "Real and upright" is incorrect because real images formed by mirrors must be inverted.
- d. "Real and inverted" describes the case when the object is beyond the focal length (do > f), not when do < f.

Therefore the image is virtual and upright, confirming answer c.

With a concave mirror, an object at a distance less than the focal length from the lens will produce an image that is单项选择题

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