Practice a College Board-style free response question on Geometric Optics. Write your response, then reveal the model answer to see exactly what earns each point.
A converging lens has a focal length of 12 cm. An object 4.0 cm tall is placed 20 cm from the lens, along the principal axis.
Quantity
Value
Focal length, f
12 cm
Object distance, d_o
20 cm
Object height, h_o
4.0 cm
A
Calculate the image distance, d_i, formed by this lens.
✓ Model answer (earns the point)
Using the thin lens equation: 1/f = 1/d_o + 1/d_i, so 1/d_i = 1/f − 1/d_o = 1/12 − 1/20. Finding a common denominator (60): 1/d_i = 5/60 − 3/60 = 2/60 = 1/30. d_i = 30 cm.
Why it scores: Correctly states the thin lens equation, correctly rearranges to solve for 1/d_i, correctly finds a common denominator, AND arrives at the correct numerical value with units.
B
Calculate the magnification of the image, and determine the height of the image. State whether the image is upright or inverted.
✓ Model answer (earns the point)
Magnification: m = −d_i/d_o = −30/20 = −1.5. Image height: h_i = m × h_o = (−1.5)(4.0 cm) = −6.0 cm, meaning the image is 6.0 cm tall and inverted (the negative sign on both m and h_i indicates inversion).
Why it scores: Correctly calculates magnification using the formula, correctly applies it to find image height, AND correctly interprets the negative sign as indicating an inverted orientation (not just reporting a negative number without explanation).
C
Is the image formed by this lens real or virtual? Justify your answer using your result from Part A, and explain what this means physically.
✓ Model answer (earns the point)
The image is real. Since d_i = 30 cm is a positive value, the image distance corresponds to light rays that actually converge on the far side of the lens (opposite from the object). Physically, this means a screen placed 30 cm from the lens (on the side opposite the object) would show a sharp, in-focus image of the object — the rays of light genuinely meet at that location, rather than only appearing to come from there.
Why it scores: Correctly identifies real (not virtual) based on the positive sign of d_i, explicitly connects the sign convention to the physical meaning (rays actually converge), AND describes what this means in practice (could be projected on a screen) rather than just stating "real" without justification.
How to score points on AP Physics 2 FRQs
Show the thin lens or mirror equation explicitly before plugging in numbers. Graders need to see the correct relationship, not just a final answer.
Always interpret the sign of your answer. A negative magnification means inverted; a positive image distance means real — state this explicitly rather than leaving it implicit.
Connect algebra to physical meaning when asked to "explain" or "justify." "d_i is positive, so light actually converges there, meaning the image could be projected on a screen" earns more than "the image is real" alone.
Double-check your arithmetic with fractions. Finding common denominators is a frequent source of small errors that cascade into wrong final answers.
Keep consistent units throughout — if focal length and object distance are both in cm, image distance and image height should also come out in cm.