AP Physics 2 Flashcards: Images Formed By Lenses

What this deck covers

This deck focuses on Images Formed By Lenses, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What happens to the image if the object is moved closer to a lens than the focal point?

Answer: Image becomes virtual and upright. Object inside focal length produces virtual, upright image.

Flashcard 2: If $f = 20$ cm and $d_o = 30$ cm, calculate $d_i$ for a convex lens.

Answer: $d_i = 60$ cm. Using lens formula: $\frac{1}{20} = \frac{1}{30} + \frac{1}{d_i}$

Flashcard 3: Find the image distance if $f = 10$ cm and $d_o = 15$ cm for a convex lens.

Answer: $d_i = 30$ cm. Using lens formula: $\frac{1}{10} = \frac{1}{15} + \frac{1}{d_i}$

Flashcard 4: What does a negative magnification value indicate?

Answer: Image is inverted. Negative magnification means image is flipped vertically.

Flashcard 5: Find the image distance if $f = 12$ cm and $d_o = 24$ cm for a convex lens.

Answer: $d_i = 24$ cm. Using lens formula: $\frac{1}{12} = \frac{1}{24} + \frac{1}{d_i}$

Flashcard 6: Determine the magnification for $d_o = 25$ cm and $d_i = 50$ cm.

Answer: $m = -2$. Magnification equals negative distance ratio: $m = -\frac{50}{25}$

Flashcard 7: Find the focal length if power of lens is $+5$ D.

Answer: $f = 0.2$ m. Power equals reciprocal of focal length: $f = \frac{1}{5}$

Flashcard 8: Calculate magnification if $d_o = 10$ cm and $d_i = -5$ cm.

Answer: $m = 0.5$. Magnification equals negative ratio: $m = -\frac{(-5)}{10} = 0.5$

Flashcard 9: Determine the power of a lens with focal length $-0.5$ m.

Answer: $P = -2$ D. Negative power indicates diverging lens: $P = \frac{1}{-0.5}$

Flashcard 10: Which type of lens corrects myopia?

Answer: Concave lens. Diverging lens spreads light to correct nearsightedness.

Flashcard 11: Calculate magnification if $d_o = 10$ cm and $d_i = -5$ cm.

Answer: $m = 0.5$. Magnification equals negative ratio: $m = -\frac{(-5)}{10} = 0.5$

Flashcard 12: What is the unit of power for lenses?

Answer: Diopter (D). Standard unit for measuring lens refractive power.

Flashcard 13: Calculate focal length if the power of the lens is $-4$ D.

Answer: $f = -0.25$ m. Focal length equals reciprocal of power: $f = \frac{1}{-4}$

Flashcard 14: How does image distance relate to object distance for a real image in a lens?

Answer: Image distance $d_i$ is positive. Real images form on opposite side of lens from object.

Flashcard 15: What is the outcome if an object is placed at the center of curvature of a convex lens?

Answer: Image is real, inverted, and same size. Object at center of curvature equals twice focal length.

Flashcard 16: Find the image height if $h_o = 2$ cm and $m = 3$.

Answer: $h_i = 6$ cm. Image height equals object height times magnification.

Flashcard 17: Which lens type converges light rays?

Answer: Convex lens. Brings parallel rays together at focal point.

Flashcard 18: Identify the type of lens used in a magnifying glass.

Answer: Convex lens. Converging lens creates enlarged virtual image for close objects.

Flashcard 19: When is an image formed by a lens virtual?

Answer: When image distance $d_i$ is negative. Negative $d_i$ means image forms on same side as object.

Flashcard 20: When is an image formed by a lens virtual?

Answer: When image distance $d_i$ is negative. Negative $d_i$ means image forms on same side as object.

Flashcard 21: Identify the type of image formed by a convex lens when object is beyond 2F.

Answer: Real, inverted, and reduced. Object farther than twice the focal length produces these characteristics.

Flashcard 22: What type of image is formed by a lens if the object is placed between the lens and the focal point?

Answer: Virtual and magnified. Object inside focal length produces enlarged virtual image.

Flashcard 23: What is the lens formula?

Answer: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$. Relates focal length to object and image distances.

Flashcard 24: Identify the image when an object is placed at infinity for a convex lens.

Answer: Real, inverted, and at F. Parallel rays from infinity converge at focal point.

Flashcard 25: Determine the power of a lens with focal length $-0.5$ m.

Answer: $P = -2$ D. Negative power indicates diverging lens: $P = \frac{1}{-0.5}$

Flashcard 26: What type of image is formed at 2F by a convex lens?

Answer: Real, inverted, and same size. Object at twice focal length creates unit magnification.

Flashcard 27: Find the image distance if $f = 10$ cm and $d_o = 15$ cm for a convex lens.

Answer: $d_i = 30$ cm. Using lens formula: $\frac{1}{10} = \frac{1}{15} + \frac{1}{d_i}$

Flashcard 28: Identify the nature of the image formed by a concave lens.

Answer: Virtual, upright, and reduced. Concave lenses always diverge light, creating virtual images.

Flashcard 29: Which lens type converges light rays?

Answer: Convex lens. Brings parallel rays together at focal point.

Flashcard 30: Find the image distance if $f = 12$ cm and $d_o = 24$ cm for a convex lens.

Answer: $d_i = 24$ cm. Using lens formula: $\frac{1}{12} = \frac{1}{24} + \frac{1}{d_i}$

Flashcard 31: What is the effect of increasing object distance on image size in a convex lens?

Answer: Image size decreases. Magnification decreases as object distance increases.

Flashcard 32: What is the relationship between curvature and power of a lens?

Answer: More curvature, higher power. More curved surfaces bend light more, increasing power.

Flashcard 33: Identify the image when an object is placed at infinity for a convex lens.

Answer: Real, inverted, and at F. Parallel rays from infinity converge at focal point.

Flashcard 34: Identify the type of image formed by a convex lens when object is beyond 2F.

Answer: Real, inverted, and reduced. Object farther than twice the focal length produces these characteristics.

Flashcard 35: How does image distance relate to object distance for a real image in a lens?

Answer: Image distance $d_i$ is positive. Real images form on opposite side of lens from object.

Flashcard 36: What does a negative magnification value indicate?

Answer: Image is inverted. Negative magnification means image is flipped vertically.

Flashcard 37: State the effect of lens curvature on focal length.

Answer: More curvature, shorter focal length. Greater curvature increases light bending, reducing focal length.

Flashcard 38: What is the unit of power for lenses?

Answer: Diopter (D). Standard unit for measuring lens refractive power.

Flashcard 39: Which type of lens corrects hyperopia?

Answer: Convex lens. Converging lens focuses light to correct farsightedness.

Flashcard 40: Identify the image characteristics when object is between F and 2F of a convex lens.

Answer: Real, inverted, and magnified. Object between focal points creates enlarged real image.

Flashcard 41: What is the nature of the image formed when an object is placed at 2F of a concave lens?

Answer: Virtual, upright, and reduced. Concave lenses always produce diminished virtual images.

Flashcard 42: Which type of lens corrects myopia?

Answer: Concave lens. Diverging lens spreads light to correct nearsightedness.

Flashcard 43: Find the focal length if power of lens is $+5$ D.

Answer: $f = 0.2$ m. Power equals reciprocal of focal length: $f = \frac{1}{5}$

Flashcard 44: State the sign convention for focal length of a concave lens.

Answer: Focal length is negative. Concave lenses diverge light, making $f$ negative by convention.

Flashcard 45: What is the power of a lens formula?

Answer: Power $P = \frac{1}{f}$ (in meters). Reciprocal of focal length measured in meters.

Flashcard 46: What is the magnification formula for lenses?

Answer: Magnification $m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$. Ratio of image to object height equals negative distance ratio.

Flashcard 47: Which lens type always forms a virtual image?

Answer: Concave lens. Diverging lens cannot form real images regardless of object position.

Flashcard 48: Explain the sign of $d_i$ for virtual images.

Answer: $d_i$ is negative. Virtual images form on same side as object.

Flashcard 49: Describe the image when an object is at the focal point of a convex lens.

Answer: No image is formed. Parallel rays emerge, creating no convergence point.

Flashcard 50: Determine the magnification for $d_o = 25$ cm and $d_i = 50$ cm.

Answer: $m = -2$. Magnification equals negative distance ratio: $m = -\frac{50}{25}$

Flashcard 51: What is the nature of the image formed when an object is placed at 2F of a concave lens?

Answer: Virtual, upright, and reduced. Concave lenses always produce diminished virtual images.

Flashcard 52: State the effect of lens curvature on focal length.

Answer: More curvature, shorter focal length. Greater curvature increases light bending, reducing focal length.

Flashcard 53: Describe the image when an object is at the focal point of a convex lens.

Answer: No image is formed. Parallel rays emerge, creating no convergence point.

Flashcard 54: Identify the image characteristics when object is between F and 2F of a convex lens.

Answer: Real, inverted, and magnified. Object between focal points creates enlarged real image.

Flashcard 55: What is the magnification formula for lenses?

Answer: Magnification $m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$. Ratio of image to object height equals negative distance ratio.

Flashcard 56: State the relation between object distance, image distance, and magnification.

Answer: $m = -\frac{d_i}{d_o}$. Negative sign accounts for image orientation relative to object.

Flashcard 57: Define focal length for a lens.

Answer: Distance from lens center to focal point. Where parallel rays converge or appear to diverge.

Flashcard 58: Identify the nature of the image formed by a concave lens.

Answer: Virtual, upright, and reduced. Concave lenses always diverge light, creating virtual images.

Flashcard 59: What type of image is formed at 2F by a convex lens?

Answer: Real, inverted, and same size. Object at twice focal length creates unit magnification.