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

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

Test your understanding with short quizzes. This quiz has 11 questions to check your progress.

Question 1 of 11

A beam of light rays, traveling parallel to the principal axis, strikes a thin convex lens. After passing through the lens, the rays will:

A)converge toward the focal point on the transmitted side.B)diverge as if from the focal point on the incident side.C)pass through the lens undeviated.D)reflect back parallel to the principal axis.

All Questions (11)

A beam of light rays, traveling parallel to the principal axis, strikes a thin convex lens. After passing through the lens, the rays will:

A) converge toward the focal point on the transmitted side.

B) diverge as if from the focal point on the incident side.

C) pass through the lens undeviated.

D) reflect back parallel to the principal axis.

Correct Answer: A

Based on the provided content, "Incident light rays parallel to the principal axis of a thin convex (converging) lens will be refracted and converge toward a common location on the transmitted side of the lens, called the focal point."

A beam of light rays, traveling parallel to the principal axis, strikes a thin concave lens. After passing through the lens, the rays will:

A) converge toward the focal point on the transmitted side.

B) diverge as if they originated from a focal point on the incident side.

C) converge toward the focal point on the incident side.

D) pass through the lens undeviated.

Correct Answer: B

The content states, "Incident light rays parallel to the principal axis of a thin concave (diverging) lens will be refracted and diverge as if they originated from a focal point on the incident side of the lens."

An object is placed 30 cm from a thin converging lens with a focal length of 10 cm. At what distance from the lens will the image be formed?

A) 7.5 cm

B) 15 cm

C) 20 cm

D) 30 cm

Correct Answer: B

Using the thin-lens equation: 1/s_i + 1/s_o = 1/f. Given s_o = 30 cm and f = 10 cm. 1/s_i + 1/30 = 1/10. So, 1/s_i = 1/10 - 1/30 = 3/30 - 1/30 = 2/30 = 1/15. Therefore, s_i = 15 cm.

An image is formed by a thin lens at a distance of 40 cm from the lens. The object was placed 20 cm from the lens. What is the magnification of the image?

A) -2.0

B) -0.5

C) +0.5

D) +2.0

Correct Answer: A

Using the magnification equation: M = -s_i / s_o. Given s_i = 40 cm and s_o = 20 cm. M = -(40 cm) / (20 cm) = -2.0.

A thin lens produces an image with a magnification of -0.5. Which of the following correctly describes the image?

A) Upright and larger than the object.

B) Inverted and larger than the object.

C) Upright and smaller than the object.

D) Inverted and smaller than the object.

Correct Answer: D

The magnification M gives two pieces of information. The sign indicates orientation: a negative sign means the image is inverted. The magnitude indicates size: a magnitude less than 1 (in this case, 0.5) means the image is smaller than the object. Therefore, the image is inverted and smaller.

An object with a height of 4 cm is placed 15 cm in front of a converging lens with a focal length of 10 cm. What is the height of the image?

A) -2 cm

B) -4 cm

C) -8 cm

D) -12 cm

Correct Answer: C

First, find the image distance s_i using the thin-lens equation: 1/s_i + 1/15 = 1/10. This gives 1/s_i = 1/10 - 1/15 = (3-2)/30 = 1/30, so s_i = 30 cm. Now, use the magnification equation M = h_i/h_o = -s_i/s_o. So, h_i / (4 cm) = -(30 cm) / (15 cm) = -2. Therefore, h_i = -2 * 4 cm = -8 cm.

An object is placed at a distance equal to twice the focal length (2f) from a thin converging lens. According to the thin-lens equation, where will the image be formed?

A) At the focal point, f.

B) Between f and 2f.

C) At a distance of 2f on the other side of the lens.

D) Farther than 2f from the lens.

Correct Answer: C

Let s_o = 2f. Using the thin-lens equation: 1/s_i + 1/s_o = 1/f. Substituting s_o = 2f gives 1/s_i + 1/(2f) = 1/f. Rearranging, 1/s_i = 1/f - 1/(2f) = 2/(2f) - 1/(2f) = 1/(2f). Therefore, s_i = 2f.

For a thin converging lens, an object is moved from a position very far away from the lens toward the focal point. How do the location and size of the real image change?

A) The image moves closer to the lens and gets smaller.

B) The image moves farther from the lens and gets larger.

C) The image moves closer to the lens and gets larger.

D) The image moves farther from the lens and gets smaller.

Correct Answer: B

According to the thin-lens equation 1/s_i + 1/s_o = 1/f, as the object distance s_o decreases (moving toward the lens), the term 1/s_o increases. Since 1/f is constant, 1/s_i must decrease to maintain the equality. A smaller 1/s_i means a larger image distance s_i. According to the magnification equation M = -s_i/s_o, since s_i is increasing and s_o is decreasing, the magnitude of the magnification |M| increases, meaning the image gets larger.

The focal point of a lens is described as the location on the incident side from which parallel incident rays appear to diverge after passing through the lens. This description applies to which type of lens?

A) A thin convex lens.

B) A thin concave lens.

C) Any spherical lens.

D) A flat plane of glass.

Correct Answer: B

The provided content explicitly states that for a thin concave (diverging) lens, parallel rays "diverge as if they originated from a focal point on the incident side of the lens." This matches the description in the question.

An object is placed at the focal point of a thin converging lens (s_o = f). According to the thin-lens equation, what can be concluded about the image distance, s_i?

A) The image is formed at the focal point (s_i = f).

B) The image is formed at the location of the object (s_i = -s_o).

C) The image is formed at an infinite distance.

D) No image is formed because the equation is undefined.

Correct Answer: C

Using the thin-lens equation: 1/s_i + 1/s_o = 1/f. If s_o = f, the equation becomes 1/s_i + 1/f = 1/f. This simplifies to 1/s_i = 0. For this to be true, the denominator s_i must approach infinity. This means the refracted rays emerge parallel and never converge to form an image at a finite distance.

By combining the thin-lens equation and the magnification equation, which of the following expressions correctly represents the magnification (M) in terms of only the object distance (s_o) and the focal length (f)?

A) M = f / (s_o + f)

B) M = (s_o - f) / f

C) M = f / (f - s_o)

D) M = s_o / (f - s_o)

Correct Answer: C

Start with the thin-lens equation: 1/f = 1/s_o + 1/s_i. Solve for 1/s_i: 1/s_i = 1/f - 1/s_o = (s_o - f) / (f * s_o). Invert to find s_i: s_i = (f * s_o) / (s_o - f). Now substitute this into the magnification equation, M = -s_i / s_o: M = - [(f * s_o) / (s_o - f)] / s_o = -f / (s_o - f). This can be rewritten by multiplying the numerator and denominator by -1: M = f / -(s_o - f) = f / (f - s_o).