Physics 2 Quiz: Ray Tracing
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Ray TracingQuestion 1 of 8

A thin converging lens of focal length f=12 cmf = 12 \text{ cm} is used to project an image of a small object. The object is placed at a distance of do=8 cmd_o = 8 \text{ cm} from the lens.

When ray tracing is performed for this configuration, which of the following correctly describes BOTH the nature and the location of the image formed by the lens?

The image is virtual, upright, and located 24 cm24 \text{ cm} on the same side as the object, because the thin-lens equation yields a negative image distance, indicating the refracted rays diverge as if emanating from a point behind the lens.
The image is real, inverted, and located 24 cm24 \text{ cm} on the opposite side of the lens from the object, because the object is closer than the focal point and the converging rays cross on the far side.
The image is virtual, inverted, and located 24 cm24 \text{ cm} on the same side as the object, because the ray parallel to the axis refracts through the focal point while the ray through the center continues straight, and their extensions cross below the axis.
The image is real, upright, and located 4.8 cm4.8 \text{ cm} on the opposite side of the lens from the object, because the converging lens always produces a real image when the object distance is finite and nonzero.
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Physics 2 Quiz

Physics 2 Quiz: Ray Tracing

Practice Ray Tracing in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Ray Tracing, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A thin converging lens of focal length f=12 cmf = 12 \text{ cm} is used to project an image of a small object. The object is placed at a distance of do=8 cmd_o = 8 \text{ cm} from the lens.

When ray tracing is performed for this configuration, which of the following correctly describes BOTH the nature and the location of the image formed by the lens?

  1. The image is virtual, upright, and located 24 cm24 \text{ cm} on the same side as the object, because the thin-lens equation yields a negative image distance, indicating the refracted rays diverge as if emanating from a point behind the lens. (correct answer)
  2. The image is real, inverted, and located 24 cm24 \text{ cm} on the opposite side of the lens from the object, because the object is closer than the focal point and the converging rays cross on the far side.
  3. The image is virtual, inverted, and located 24 cm24 \text{ cm} on the same side as the object, because the ray parallel to the axis refracts through the focal point while the ray through the center continues straight, and their extensions cross below the axis.
  4. The image is real, upright, and located 4.8 cm4.8 \text{ cm} on the opposite side of the lens from the object, because the converging lens always produces a real image when the object distance is finite and nonzero.
Explanation: When a converging lens has an object placed inside its focal length, the thin-lens equation tells the whole story. Apply it here: 1di=1f−1do=112−18=2−324=−124\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{12} - \frac{1}{8} = \frac{2-3}{24} = \frac{-1}{24}, giving di=−24 cmd_i = -24 \text{ cm}. The negative image distance is the critical result — in the sign convention for thin lenses, a negative did_i means the image forms on the same side as the object, not the far side. Refracted rays diverge after passing through the lens; they never actually converge, so no real image exists. Instead, tracing those diverging rays backward, your eye perceives them as coming from a point 24 cm behind the lens (same side as the object). This makes the image virtual and upright, which is exactly choice A — the correct answer. Choice B is tempting but wrong: it claims the image is real and inverted on the far side, which would only be true if the object were beyond the focal point (do>fd_o > f). Here do=8 cm<f=12 cmd_o = 8 \text{ cm} < f = 12 \text{ cm}, so rays diverge rather than converge after the lens. Choice C incorrectly labels the image as inverted — virtual images formed by a single converging lens with the object inside the focal length are always upright, not inverted. Choice D states a real, upright image at 4.8 cm, which contradicts both the math and the principle: converging lenses do not always produce real images. Your strategy: always compute did_i first. If it's negative, the image is virtual and upright, on the same side as the object — no exceptions for a single thin lens.

Question 2

A convex (diverging) mirror has a focal length of magnitude ∣f∣=25 cm|f| = 25 \text{ cm}. An object is moved continuously from very far away (do→∞d_o \to \infty) toward the mirror surface (do→0d_o \to 0). Which of the following statements correctly describes how the image location and magnification change throughout this process, as determined by consistent application of the mirror equation with the standard sign convention?

  1. The image moves from behind the mirror to in front of the mirror as the object crosses the focal point, transitioning from virtual to real, because a convex mirror behaves like a concave mirror for objects placed very close to its surface.
  2. The image moves from the focal point (25 cm25 \text{ cm} behind the mirror) toward the mirror surface as the object approaches, but the lateral magnification decreases from +1+1 toward zero, because a closer object always produces a smaller image in a convex mirror.
  3. The image starts at the center of curvature (50 cm50 \text{ cm} behind the mirror) when the object is at infinity and moves toward the focal point as the object approaches, with the magnification remaining constant at +0.5+0.5 throughout.
  4. The image moves from the focal point (25 cm25 \text{ cm} behind the mirror) toward the mirror surface as the object approaches, and the lateral magnification increases from near zero to +1+1, remaining positive (upright) throughout the entire range. (correct answer)
Explanation: When working with mirror problems, always anchor yourself to the sign convention: for a convex (diverging) mirror, the focal length is negative, so f=−25 cmf = -25 \text{ cm}. The mirror equation is 1di=1f−1do\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}, and magnification is m=−didom = -\frac{d_i}{d_o}. As the object moves from do→∞d_o \to \infty toward the surface, plug in the extremes. At do=∞d_o = \infty: 1di=1−25−0\frac{1}{d_i} = \frac{1}{-25} - 0, giving di=−25 cmd_i = -25 \text{ cm} — the image forms 25 cm behind the mirror (virtual), right at the focal point, with m≈0m \approx 0. As do→0d_o \to 0: di→0d_i \to 0 as well, and m=−dido→+1m = -\frac{d_i}{d_o} \to +1. Throughout the entire range, did_i stays negative (behind the mirror) and mm stays positive (upright), increasing from near zero to +1. This perfectly matches answer D. A is wrong because a convex mirror never produces a real image for a real object — no crossover from virtual to real occurs. This confuses convex mirror behavior with concave mirror behavior near the focal point. B is wrong in saying magnification decreases toward zero. The magnification actually increases toward +1 as the object gets closer — smaller dod_o means the object and image distances converge, raising mm. C is wrong because di=−25 cmd_i = -25 \text{ cm} (the focal point) is the starting position at infinity, not −50 cm-50 \text{ cm}, and magnification is not constant. A reliable tip: for a convex mirror, the image is always virtual, upright, and smaller than the object — except as do→0d_o \to 0, where the image approaches the object size (m→+1m \to +1). Memorize this behavior and the sign of ff for each mirror type.

Question 3

A small candle flame (object) is placed 60 cm60 \text{ cm} in front of a concave mirror of focal length f=20 cmf = 20 \text{ cm}. A screen is placed to catch the image. The entire setup — mirror, object, and screen — is then submerged in water (index of refraction nwater≈1.33n_{water} \approx 1.33).

How does the submersion in water affect the position and nature of the image formed by the concave mirror, as predicted by ray tracing principles?

  1. The image shifts farther from the mirror after submersion, because the higher refractive index of water relative to air causes the reflected rays to bend more strongly toward the normal at the mirror surface, increasing the mirror's converging power.
  2. The image shifts closer to the mirror after submersion, because water slows light, increasing the effective optical path length, which effectively increases the focal length of the concave mirror and moves the image toward the focal point.
  3. The image position and nature are completely unchanged by submersion in water, because reflection from a mirror obeys the law of reflection (θi=θr\theta_i = \theta_r), which is independent of the medium's index of refraction, so the focal length and all ray-trace results remain the same. (correct answer)
  4. The image position is unchanged in distance from the mirror, but the image becomes virtual and upright instead of real and inverted, because total internal reflection at the mirror surface is disrupted by the water, reversing the curvature effect of the concave mirror.
Explanation: Whenever you see a question involving mirrors and a change in surrounding medium, your first instinct should be to ask: which law governs mirror behavior? The answer is the law of reflection — θi=θr\theta_i = \theta_r — and this is the key to unlocking the entire question. Unlike lenses, which refract light and depend heavily on the index of refraction of the surrounding medium, mirrors work purely through reflection. The law of reflection holds that the angle of incidence equals the angle of reflection at the mirror surface, and this relationship is completely independent of what medium the light is traveling through. Whether the setup is in air, water, or any other transparent medium, the geometry of reflected rays doesn't change. This means the focal length of the concave mirror stays at f=20 cmf = 20 \text{ cm}, and applying the mirror equation 1do+1di=1f\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} with do=60 cmd_o = 60 \text{ cm} still gives di=30 cmd_i = 30 \text{ cm} — a real, inverted image. Choice C is correct. Choice A is wrong because it confuses mirrors with lenses — refraction at a surface does increase converging power, but mirrors don't refract light. Choice B is wrong for the same reason; slowing light through a denser medium affects refraction, not reflection. The focal length of a mirror is purely geometric. Choice D is wrong on two counts: the image position doesn't change, and total internal reflection is irrelevant here — it applies to light passing between media, not reflecting off a silvered surface. Study tip: On any optics question, always identify whether you're dealing with a mirror (reflection, medium-independent) or a lens (refraction, medium-dependent). That single distinction resolves most medium-change scenarios instantly.

Question 4

An object is placed in front of a concave mirror. A ray tracing produces an image that is real, inverted, and the same size as the object. Without changing the mirror, the object is now moved to a new position such that the image becomes virtual, upright, and larger than the object. Which of the following correctly describes the direction the object was moved and the range of object positions that produce the new image type?

  1. The object was moved closer to the mirror to a position exactly at the focal point (do=fd_o = f), because placing the object at the focal point causes the reflected rays to diverge maximally, creating an infinitely magnified virtual image.
  2. The object was moved farther from the mirror to a position beyond the center of curvature (do>2fd_o > 2f), because at very large object distances the concave mirror acts like a flat mirror and produces virtual images of equal or larger size.
  3. The object was moved closer to the mirror to a position between the focal point and the center of curvature (f<do<2ff < d_o < 2f), because this range produces virtual images that are larger than the object in a concave mirror.
  4. The object was moved closer to the mirror, specifically to a position between the focal point and the mirror surface (0<do<f0 < d_o < f), because only objects within this range produce virtual, upright, magnified images in a concave mirror. (correct answer)
Explanation: When analyzing concave mirror problems, your key tool is the mirror equation 1do+1di=1f\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} combined with your knowledge of how image type depends on object position relative to the focal point ff and center of curvature 2f2f. Start with the initial condition: a real, inverted, same-size image. For a concave mirror, this only occurs when do=2fd_o = 2f — the object sits exactly at the center of curvature. You can verify this: if do=2fd_o = 2f, the mirror equation gives di=2fd_i = 2f, and magnification m=−di/do=−1m = -d_i/d_o = -1, confirming unit size and inversion. Now the object moves to produce a virtual, upright, magnified image. In a concave mirror, this only happens when the object is placed between the focal point and the mirror surface (0<do<f0 < d_o < f). When do<fd_o < f, the mirror equation yields a negative did_i, meaning the image forms behind the mirror — that's a virtual image. The magnification becomes positive and greater than 1, giving you upright and enlarged. Since the object started at 2f2f and moved inside ff, it was moved closer. This confirms D is correct. Choice A is wrong because placing the object exactly at ff produces no image at all — reflected rays are parallel and never converge or appear to diverge from a point. Choice B is wrong because moving farther from the mirror (do>2fd_o > 2f) produces real, inverted, diminished images — the opposite of what's described. Choice C is wrong because the range f<do<2ff < d_o < 2f still produces real, inverted images (just magnified ones), not virtual images. Your memory anchor: inside the focal point = virtual, upright, magnified for concave mirrors. This is the only zone where concave mirrors behave like magnifying glasses.

Question 5

A physics student claims: 'For any thin lens system, if the object is real and the image is also real, then the lens must be converging (positive focal length). A diverging lens can never produce a real image of a real object.'

Which of the following best evaluates the student's claim from the perspective of ray tracing and the thin-lens equation?

  1. The claim is correct. For a real object (do>0d_o > 0) and a real image (di>0d_i > 0), the thin-lens equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} requires f>0f > 0, confirming that only a converging lens can produce a real image of a real object under these conditions.
  2. The claim is partially correct for isolated lenses but fails for lens combinations. A single diverging lens cannot produce a real image of a real object, but in a two-lens system, the diverging lens can intercept a converging beam (virtual object) and form a real image, though the student's original object is still technically real. (correct answer)
  3. The claim is incorrect because a diverging lens can produce a real image of a real object when the object distance equals the magnitude of the focal length, i.e., do=∣f∣d_o = |f|, which causes the refracted rays to converge on the far side of the lens.
  4. The claim is incorrect because the thin-lens equation places no restriction on the sign of ff when do>0d_o > 0 and di>0d_i > 0; both converging and diverging lenses can satisfy 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} with positive values of dod_o and did_i simultaneously.
Explanation: When evaluating claims about lens systems, you need to think carefully about whether the claim applies to isolated lenses or lens combinations — this distinction is what separates a partially correct statement from a fully correct one. The student's reasoning holds perfectly for a single lens in isolation. If you have a real object (do>0d_o > 0) and a real image (di>0d_i > 0), then 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} gives you the sum of two positive terms, forcing f>0f > 0. No single diverging lens can satisfy this with a real object. Ray tracing confirms this: a diverging lens spreads rays outward, so they never converge to form a real image on the far side. So far, the student is right. But the claim breaks down in multi-lens systems. If a converging lens begins focusing light toward a point, a diverging lens placed before that focal point intercepts a converging beam. That converging beam acts as a virtual object (do<0d_o < 0) for the diverging lens. Now the thin-lens equation can yield di>0d_i > 0 even with f<0f < 0. The original object from the student's perspective is still real, yet the diverging lens ultimately contributes to forming a real image. This makes B correct — the claim is partially right for isolated lenses but fails for combinations. A is wrong because it over-generalizes to all lens systems, ignoring multi-lens setups. C is wrong — setting do=∣f∣d_o = |f| for a diverging lens gives 1di=1∣f∣−1∣f∣⋅(−1)\frac{1}{d_i} = \frac{1}{|f|} - \frac{1}{|f|} \cdot(-1)... actually it produces a virtual image, not real. D is wrong because the math does restrict the sign of ff; you cannot get f<0f < 0 when both dod_o and did_i are positive in the thin-lens equation. Study tip: Whenever a question generalizes about "any lens system," ask yourself whether multi-lens configurations (with virtual objects) might create exceptions — they almost always do.

Question 6

In a ray diagram for a thin converging lens, a student draws a ray from the tip of an object that passes through the front focal point of the lens and then strikes the lens. After refraction, this ray must exit the lens in a specific direction. Separately, a second ray is drawn from the same object tip, traveling at a slight downward angle such that it passes through the optical center of the lens. Which of the following correctly describes the refracted directions of both rays, and which ray — if either — changes direction upon passing through the lens?

  1. Ray 1 (through the front focal point) exits toward the back focal point after refraction, because the front and back focal points are symmetric and any ray aimed at one focal point exits toward the other. Ray 2 (through the optical center) continues without bending in its original direction.
  2. Ray 1 (through the front focal point) exits parallel to the optical axis after refraction. Ray 2 (through the optical center) bends toward the axis after passing through the center, because the converging lens redirects all transmitted rays toward the back focal point.
  3. Ray 1 (through the front focal point) exits parallel to the optical axis after refraction, because any ray passing through the front focal point is redirected parallel to the axis by a converging lens. Ray 2 (through the optical center) continues in its original direction without bending, because the lens surfaces at the center are locally parallel and produce no net deviation. (correct answer)
  4. Both rays exit the lens without changing direction, because the three standard ray-tracing rules apply only to rays parallel to the axis or aimed at a focal point — a ray through the optical center and a ray through the front focal point are both special cases that undergo zero net deflection.
Explanation: When working with thin lens ray diagrams, you need to memorize the three principal rays — each follows a specific rule based on where it enters the lens, and these rules are not interchangeable. The first principal ray travels through the front focal point before hitting the lens. Because the front focal point is defined as the point from which rays emerge parallel after refraction, any ray passing through it exits the lens parallel to the optical axis. Think of it as the reverse of a ray coming in parallel (which converges to the back focal point) — reversibility of light makes this symmetric. The second principal ray passes through the optical center. At the very center of a thin lens, the two surfaces are locally parallel and essentially act like a thin flat glass slab, producing no net angular deviation. The ray continues in its original direction, unchanged. This makes C correct. Choice A is tempting but wrong — it invents a rule where rays aimed at the front focal point exit toward the back focal point. No such rule exists; the correct behavior is parallel exit, not redirection toward the opposite focal point. Choice B gets Ray 1 right but incorrectly claims Ray 2 bends toward the axis. The center of a thin lens does not redirect rays — that's the whole point of the central-ray rule. Choice D incorrectly claims Ray 1 also passes through unchanged. Only the central ray does this; the front-focal-point ray definitely refracts to become parallel. Study tip: Memorize all three principal rays as distinct rules — parallel→back focal point, front focal point→parallel, center→straight through. Mixing them up is one of the most common lens-diagram errors on physics exams.

Question 7

A student sets up a ray diagram for a thin diverging lens (focal length f=−15 cmf = -15 \text{ cm}) with an object placed 45 cm45 \text{ cm} to the left of the lens. The student draws the following three rays from the tip of the object: Ray 1 travels parallel to the optical axis and, after refraction, diverges as if coming from the near focal point. Ray 2 travels toward the far focal point and, after refraction, exits parallel to the axis. Ray 3 passes through the optical center without bending.

After correctly completing this ray diagram, the student notes that the three refracted rays diverge and do not meet on the right side of the lens. The student then extends the refracted rays backward (to the left). Which of the following correctly describes what the student finds and what it implies about the image?

  1. The backward extensions of the three rays meet 22.5 cm22.5 \text{ cm} to the left of the lens, forming a virtual, upright, and diminished image, because the image distance exceeds the focal length magnitude when the object is beyond twice the focal length.
  2. The backward extensions of the three rays meet 11.25 cm11.25 \text{ cm} to the left of the lens, forming a virtual, upright, and diminished image, because the diverging lens always produces a virtual image located between the lens and its near focal point on the object side. (correct answer)
  3. The backward extensions of the three rays meet 11.25 cm11.25 \text{ cm} to the left of the lens, forming a virtual, inverted, and diminished image, because the diverging lens inverts the image whenever the object distance is greater than the magnitude of the focal length.
  4. The backward extensions of the three rays do not meet at a single point, because the principal ray rules for a diverging lens become inconsistent when the object distance is more than three times the focal length magnitude, making the diagram indeterminate.
Explanation: Whenever you encounter a diverging lens problem, your first instinct should be to apply the thin lens equation and remember the golden rule: diverging lenses always produce virtual, upright, and diminished images on the same side as the object. Using the thin lens equation 1di=1f−1do\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}, plug in f=−15 cmf = -15\text{ cm} and do=45 cmd_o = 45\text{ cm}: 1di=1−15−145=−345−145=−445\frac{1}{d_i} = \frac{1}{-15} - \frac{1}{45} = \frac{-3}{45} - \frac{1}{45} = \frac{-4}{45} di=−11.25 cmd_i = -11.25\text{ cm} The negative sign confirms the image forms 11.25 cm to the left of the lens — virtual and on the same side as the object. The magnification m=−di/do=11.25/45=0.25m = -d_i/d_o = 11.25/45 = 0.25 confirms the image is upright and diminished. This makes B correct. A is wrong because it gives 22.5 cm22.5\text{ cm}, which doesn't come from a valid application of the lens equation — it's simply a miscalculation. The additional claim about image distance exceeding focal length magnitude is also irrelevant reasoning for a diverging lens. C gets the distance right (11.25 cm11.25\text{ cm}) but incorrectly states the image is inverted. Diverging lenses never produce inverted images for real objects; that's exclusive to converging lenses when the object is beyond the focal point. D is a fabricated distractor — there's no rule that makes principal rays inconsistent at any object distance. All three rays always converge to a single virtual image point. Study tip: For diverging lenses, memorize that did_i is always negative and the image is always virtual, upright, and smaller — no exceptions for real objects.

Question 8

A concave (converging) spherical mirror has a radius of curvature of R=30 cmR = 30 \text{ cm}. An object is placed 10 cm10 \text{ cm} in front of the mirror. A student performs a ray trace using the standard three principal rays. Which of the following correctly identifies where the extensions of the reflected rays converge and what this implies about the image?

  1. The reflected rays, when extended behind the mirror, converge at a point 30 cm30 \text{ cm} behind the mirror surface, forming a virtual, upright, and magnified image because the object lies between the mirror and its focal point. (correct answer)
  2. The reflected rays converge in front of the mirror at 30 cm30 \text{ cm}, forming a real, inverted, and magnified image, because the object is between the center of curvature and the focal point of the mirror.
  3. The reflected rays converge in front of the mirror at 15 cm15 \text{ cm}, forming a real, inverted image of the same size as the object, because the object is placed exactly at the focal length of the mirror.
  4. The reflected rays diverge after reflection and cannot be made to converge anywhere, so no image is formed, because an object placed inside the focal length of a concave mirror always produces an indeterminate ray diagram.
Explanation: When a concave mirror question involves an object placed inside the focal length, your first move should be to locate the focal point. With R=30 cmR = 30\text{ cm}, the focal length is f=R/2=15 cmf = R/2 = 15\text{ cm}. The object at 10 cm10\text{ cm} sits between the mirror and the focal point — this is the critical setup that determines everything. Applying the mirror equation: 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} gives 115=110+1di\frac{1}{15} = \frac{1}{10} + \frac{1}{d_i}, so 1di=115−110=−130\frac{1}{d_i} = \frac{1}{15} - \frac{1}{10} = -\frac{1}{30}, meaning di=−30 cmd_i = -30\text{ cm}. The negative sign means the image forms behind the mirror — it's virtual. When you trace the three principal rays, they diverge after reflection; extending them behind the mirror, their extensions converge at 30 cm, producing a virtual, upright, and magnified image. That's exactly what answer A describes. Answer B is wrong on two counts: the object isn't between the center of curvature and the focal point (it's inside the focal point), and a real image would require a positive did_i. Answer C incorrectly claims the object is at the focal length — it's not; at do=fd_o = f, reflected rays are parallel and no image forms at all. Answer D is a misconception: placing an object inside the focal length of a concave mirror absolutely produces an image — it just happens to be virtual. A reliable rule to memorize: for a concave mirror, whenever do<fd_o < f, the image is always virtual, upright, and magnified, formed behind the mirror. This is the same principle that makes concave mirrors useful as makeup or shaving mirrors.