AP Physics 2: Algebra-Based Practice Quiz: Images Formed by Mirrors

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 12 questions to check your progress.

Question 1 of 12

According to the provided text, which of the following best describes the formation of a real image by a mirror?

A)Reflected light rays diverge as if they originated from a point behind the mirror.B)Reflected light rays from a common point on an object intersect at another common point.C)The image is always upright and the same size as the object.D)The image can only be seen by looking into the mirror and cannot be projected.

All Questions (12)

According to the provided text, which of the following best describes the formation of a real image by a mirror?

A) Reflected light rays diverge as if they originated from a point behind the mirror.

B) Reflected light rays from a common point on an object intersect at another common point.

C) The image is always upright and the same size as the object.

D) The image can only be seen by looking into the mirror and cannot be projected.

Correct Answer: B

Content point 2 explicitly states that a real image is formed when light rays emanating from a common point are reflected and then intersect at a common point.

A virtual image is formed by a mirror when:

A) Reflected light rays converge to a focal point in front of the mirror.

B) The magnification of the image is always negative.

C) Reflected light rays diverge in a way that they appear to have come from a single point.

D) The image is formed by the actual intersection of light rays.

Correct Answer: C

Content point 3 defines a virtual image as one formed when reflected light rays diverge such that they appear to have originated from a common point.

Which equation correctly relates the object distance ($s_o$), image distance ($s_i$), and focal length ($f$) for a mirror?

A) s_i + s_o = f

B) M = s_i / s_o

C) 1/s_i + 1/s_o = 1/f

D) f = s_i - s_o

Correct Answer: C

Content point 4 provides the relevant equation for relating image location, object location, and focal length as 1/s_i + 1/s_o = 1/f.

An object is placed 20 cm in front of a mirror with a focal length of 5 cm. At what distance from the mirror will the image be formed?

A) 4.0 cm

B) 6.7 cm

C) 15.0 cm

D) 25.0 cm

Correct Answer: B

Using the mirror equation 1/s_i + 1/s_o = 1/f: 1/s_i + 1/20 = 1/5. Solving for 1/s_i gives 1/5 - 1/20 = 4/20 - 1/20 = 3/20. Therefore, s_i = 20/3 cm, which is approximately 6.7 cm.

The magnification of an image formed by a mirror is calculated to be M = -3. What does the negative sign indicate about the image?

A) The image is virtual.

B) The image is smaller than the object.

C) The image is inverted relative to the object.

D) The image is located behind the mirror.

Correct Answer: C

According to the magnification equation M = -s_i/s_o (Content point 5), a negative value for M indicates that the image orientation is inverted compared to the object's orientation.

An object of height 2 cm is placed in front of a mirror. An image is formed with a height of 4 cm. If the object is 10 cm from the mirror, how far from the mirror is the image located?

A) 5 cm

B) 10 cm

C) 20 cm

D) 40 cm

Correct Answer: C

First, find the magnification: M = h_i/h_o = 4 cm / 2 cm = 2. Then use the magnification equation M = -s_i/s_o. We have 2 = -s_i/10 cm. Solving for s_i gives s_i = -20 cm. The distance is 20 cm.

What is the primary purpose of using a ray diagram when analyzing images formed by mirrors?

A) To calculate the exact focal length of the mirror.

B) To determine the location, type, size, and orientation of the image.

C) To measure the intensity of the light reflected from the mirror.

D) To verify the material composition of the mirror.

Correct Answer: B

Content point 6 states that ray diagrams are used to determine the location, type, size, and orientation of images formed by mirrors.

An object is placed in front of a mirror, and an image is formed with a magnification of M = +0.5. Which of the following statements about the image is correct?

A) The image is real, inverted, and smaller than the object.

B) The image is virtual, upright, and smaller than the object.

C) The image is real, upright, and larger than the object.

D) The image is virtual, inverted, and larger than the object.

Correct Answer: B

A positive magnification (M > 0) indicates an upright image. A magnification with a magnitude less than 1 (|M| < 1) indicates the image is smaller than the object. From M = -s_i/s_o, a positive M requires a negative s_i (since s_o is positive), which defines a virtual image.

A real image is formed 30 cm from a mirror. The image is inverted and is the same size as the object. What is the focal length of the mirror?

A) 10 cm

B) 15 cm

C) 30 cm

D) 60 cm

Correct Answer: B

If the image is the same size as the object, the magnitude of the magnification |M| is 1. Since it is inverted, M = -1. Using M = -s_i/s_o, we get -1 = -30/s_o, which means s_o = 30 cm. Now use the mirror equation: 1/f = 1/s_o + 1/s_i = 1/30 + 1/30 = 2/30 = 1/15. Therefore, f = 15 cm.

An object is placed at a distance s_o from a mirror with focal length f. If the resulting image is virtual, what must be true about the image distance s_i?

A) s_i must be positive.

B) s_i must be negative.

C) s_i must be equal to f.

D) s_i must be equal to s_o.

Correct Answer: B

A virtual image is one where the reflected rays appear to diverge from a point behind the mirror. By convention in the mirror equation, distances behind the mirror are negative. Therefore, a virtual image always has a negative image distance (s_i < 0).

The mirror equation is 1/s_i + 1/s_o = 1/f. Which of the following correctly expresses the focal length, f, in terms of s_i and s_o?

A) f = s_i + s_o

B) f = (s_i * s_o) / (s_i + s_o)

C) f = (s_i + s_o) / (s_i * s_o)

D) f = s_i - s_o

Correct Answer: B

To solve for f, first find a common denominator on the left side: (s_o + s_i) / (s_i * s_o) = 1/f. Then, take the reciprocal of both sides to isolate f, which gives f = (s_i * s_o) / (s_i + s_o).

An object is placed in front of a mirror with focal length f. A real, inverted image is formed that is half the size of the object. At what distance, in terms of f, was the object placed from the mirror?

A) s_o = f

B) s_o = 1.5f

C) s_o = 2f

D) s_o = 3f

Correct Answer: D

The image is inverted and half the size, so M = -0.5. From M = -s_i/s_o, we have -0.5 = -s_i/s_o, which simplifies to s_i = 0.5 * s_o. Substitute this into the mirror equation: 1/(0.5*s_o) + 1/s_o = 1/f. This becomes 2/s_o + 1/s_o = 1/f, which simplifies to 3/s_o = 1/f. Therefore, s_o = 3f.