The problems are on the thin lens equation worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Geometric optics treats light as rays. Reflection obeys θᵣ = θᵢ, refraction obeys Snell's law n₁ sin θ₁ = n₂ sin θ₂, and mirrors and thin lenses obey 1/dₒ + 1/dᵢ = 1/f with magnification m = −dᵢ/dₒ. These worked answers correct the 20 problems of the AP Physics 2 worksheet on thin lens equation, from law of reflection to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.
Here are the thin lens equation worksheet answers for AP Physics 2, explained step by step. First try the worksheet on your own, then compare, and reread the lesson if a step is unclear.
Each answer gives the equation, the substitution and the result in bold. Check that your image distance has the right sign and that your description of the image (real or virtual, upright or inverted) matches your sketch.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 13
Thin lens equation worksheet answers: the answers explained
Problem 1 Law of reflection
1. θᵣ = θᵢ = 35°.
2. 90° − 35° = 55°.
3. 35° + 35° = 70°.
Problem 2 Refractive index
1. n = c/v = 3.00 × 10⁸ / 2.00 × 10⁸ = 1.50.
2. v = c/n = 3.00 × 10⁸ / 1.33 = 2.26 × 10⁸ m/s.
Problem 3 Air to glass
1. sin θ₂ = sin 30° / 1.50 = 0.500 / 1.50 = 0.333, so θ₂ = 19.5°.
2. Towards the normal, since the glass has a larger index.
Problem 4 Image in a plane mirror
1. The image is as far behind the mirror as the object is in front: 1.2 m.
2. 1.2 + 1.2 = 2.4 m.
3. The image also moves 0.50 m/s towards the mirror, so the gap closes at 0.50 + 0.50 = 1.0 m/s.
Problem 5 Focal length of a mirror
1. f = R/2 = 40 / 2 = 20 cm.
2. A concave mirror has positive f.
Problem 6 Lens facts
1. f is positive.
2. A real, inverted, reduced image.
3. A virtual, upright, reduced image.
Problem 7 Magnification
1. |m| = 3.0 / 1.5 = 2.0.
2. For an inverted image, m = −2.0.
Problem 8 Real image from a lens
1. 1/dᵢ = 1/15 − 1/45 = 2/45, so dᵢ = 22.5 cm.
2. m = −22.5 / 45 = −0.50.
3. Real, inverted and half the size of the object.
Problem 9 Magnifying glass
1. 1/dᵢ = 1/10 − 1/6.0 = −2/30, so dᵢ = −15 cm.
2. m = −(−15)/6.0 = +2.5.
3. Virtual (negative dᵢ), upright and 2.5 times larger.
Problem 10 Concave mirror
1. 1/dᵢ = 1/12 − 1/36 = 2/36, so dᵢ = 18 cm.
2. m = −18/36 = −0.50.
3. dᵢ is positive, so the image is real and can be projected on a screen.
Problem 11 Convex mirror
1. 1/dᵢ = −1/20 − 1/60 = −4/60, so dᵢ = −15 cm.
2. m = 15/60 = +0.25.
3. A convex mirror always gives a virtual, upright, reduced image, so a wide area fits into the mirror.
Problem 12 Critical angles
1. sin θc = 1/2.42 = 0.413, so θc = 24.4°.
2. sin θc = 1/1.33 = 0.752, so θc = 48.8°.
3. Diamond traps light more easily because its critical angle is smaller.
Problem 13 Refraction data
1. sin 50° / sin 30.7° = 0.7660 / 0.5105 = 1.50.
2. n = sin θ₁ / sin θ₂, so a graph of sin θ₁ (y) against sin θ₂ (x) is a straight line through the origin whose gradient is n.
3. The ratio is about 1.5 in every row, so n = 1.5.
Problem 14 Water to air
1. 1.33 × sin 30° = 1.00 × sin θ₂, so sin θ₂ = 0.665 and θ₂ = 41.7°.
2. Away from the normal, since air has a smaller index.
Problem 15 Slide projector
1. 1/dₒ = 1/8.0 − 1/240 = 29/240, so dₒ = 8.28 cm.
2. m = −240 / 8.28 = −29.
3. hᵢ = 29 × 24 mm = 0.70 m, inverted.
Problem 16 Camera focusing
1. For a distant object, dᵢ = f = 50 mm.
2. 1/dᵢ = 1/50 − 1/2000 = 39/2000, so dᵢ = 51.3 mm.
3. The lens moves 51.3 − 50 = 1.3 mm away from the sensor.
Problem 17 Optical fibre
1. sin θc = n₂/n₁ = 1.46 / 1.48 = 0.9865, so θc = 80.6°.
2. The ray angle to the boundary normal must be at least 80.6°, so the angle with the fibre axis is at most 90° − 80.6° = 9.4°.
Problem 18 Mirror image changes
1. At 20 cm: 1/dᵢ = 1/15 − 1/20 = 1/60, so dᵢ = 60 cm, m = −3.0, hᵢ = −9.0 cm. At 10 cm: 1/dᵢ = 1/15 − 1/10 = −1/30, so dᵢ = −30 cm, m = +3.0, hᵢ = +9.0 cm.
2. The image changes from real and inverted to virtual and upright as the object moves inside the focal length.
Problem 19 Coin in a pool
1. tan θ₁ = 1.0 / 1.0, so θ₁ = 45°.
2. 1.33 × sin 45° = sin θ₂, so sin θ₂ = 0.940 and θ₂ = 70.2°.
3. The eye traces the bent ray back in a straight line, which meets the vertical through the coin above the real position, so the coin seems shallower (about 0.36 m deep for this ray).
Problem 20 Spot the errors
1. Light slows down in glass and bends towards the normal.
2. A diverging lens gives only virtual images of real objects. 1/dᵢ = −1/10 − 1/30 = −4/30, so dᵢ = −7.5 cm, virtual, and m = 7.5/30 = +0.25, upright and reduced.
