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United StatesPhysics 2Syllabus dot point

How do curved mirrors form images, and what does the mirror equation predict?

Topic 13.2 Images Formed by Mirrors: apply the mirror equation and magnification to images from concave and convex mirrors.

A focused answer to AP Physics 2 Topic 13.2, covering concave and convex mirrors, the focal length and its relation to the radius, the mirror equation, the magnification equation, the sign conventions, and the characteristics of real and virtual images, with full worked examples.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. Concave and convex mirrors
  3. The mirror and magnification equations
  4. Reading the image characteristics
  5. Try this

What this topic is asking

The College Board (Topic 13.2) wants you to apply the mirror equation and the magnification equation to concave and convex mirrors, using the sign conventions to determine whether images are real or virtual, upright or inverted, enlarged or reduced.

Concave and convex mirrors

The two mirror types behave oppositely. A concave mirror gathers parallel rays to a point (like a makeup or telescope mirror), giving it a real focus and a positive focal length. A convex mirror spreads rays apart, so its focus is virtual and its focal length negative; it gives a wide, shrunken view (car and shop security mirrors). The focal length is half the radius of curvature, f=R/2f = R/2.

The mirror and magnification equations

These two equations, with the sign conventions, predict everything about the image. Solve the mirror equation for did_i, then read its sign (real or virtual) and feed it into the magnification to get the orientation and size. The whole skill is disciplined sign-keeping: a positive image distance is a real image you could catch on a screen, a negative one is a virtual image behind the mirror; a negative magnification flips the image upside down.

Reading the image characteristics

The image's nature depends on the mirror and the object position. A convex mirror always produces a virtual, upright, reduced image, whatever the object distance, which is why it gives a safe wide-angle view. A concave mirror is versatile: an object beyond the focal point gives a real, inverted image (enlarged or reduced depending on distance), while an object inside the focal length gives a virtual, upright, enlarged image (the magnifying-mirror case). The strategic role of this topic is that it turns the reflection law of Topic 13.1 into quantitative image prediction, and it shares its entire mathematical structure, the same equation form and sign logic, with the thin-lens equation of Topic 13.4. Learn the sign conventions once and they transfer directly to lenses, making the second half of geometric optics far easier.

Try this

Q1. State the type of image always formed by a convex mirror. [1 point]

  • Cue. Virtual, upright and reduced.

Q2. An image has a magnification of −2.0-2.0. State its orientation and size relative to the object. [2 points]

  • Cue. Inverted (negative sign) and twice the object's size (magnitude 2.02.0).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). An object is placed 0.300.30 m in front of a concave mirror of focal length 0.100.10 m. (a) Use the mirror equation to find the image distance. (b) Calculate the magnification and state whether the image is upright or inverted. (c) State whether the image is real or virtual and justify it from the sign of the image distance.
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A 7-point FRQ on the mirror equation.

(a) Image distance (3 points): 1f=1do+1di\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}, so 1di=10.10−10.30=10−3.33=6.67\dfrac{1}{d_i} = \dfrac{1}{0.10} - \dfrac{1}{0.30} = 10 - 3.33 = 6.67, giving di=0.15d_i = 0.15 m.
(b) Magnification (2 points): m=−dido=−0.150.30=−0.50m = -\dfrac{d_i}{d_o} = -\dfrac{0.15}{0.30} = -0.50. The negative sign means the image is inverted, and it is half the object's size.
(c) Real or virtual (2 points): the image distance is positive, so the image is real (it forms in front of the mirror where light actually converges).

Markers reward the mirror equation, the magnification with the inverted interpretation, and the positive image distance giving a real image.

AP 2023 (style)1 marksSection I (multiple choice). A convex (diverging) mirror always forms an image that is (A) real and inverted (B) virtual, upright and reduced (C) real and enlarged (D) virtual and enlarged. Justify your reasoning.
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A 1-point MCQ on convex mirrors. The answer is (B).

A convex mirror has a negative focal length and always forms a virtual, upright, reduced image behind the mirror, for any object position. This wide field of view is why convex mirrors are used as security and car side mirrors. The trap is (A): a convex mirror never forms a real image.

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