Images Formed by Mirrors - College Physics

A spherical mirror produces an image that is located at the same distance from the mirror as the object, but on the opposite side. Which statement correctly describes this situation?
Answer: The mirror must be concave, the object is at the center of curvature, and the magnification is $-1$ When an object and its image are equidistant from a spherical mirror but on opposite sides, we have $s = |s'|$ with $s' > 0$ (real image). This means $s = s'$ , so from the mirror equation $\frac{1}{f} = \frac{1}{s} + \frac{1}{s} = \frac{2}{s}$ , giving $s = 2f = R$ (radius of curvature). This occurs only for concave mirrors when the object is at the center of curvature, producing $m = -s'/s = -1$ . Choice B incorrectly describes convex mirrors (which never form real images), choice C incorrectly suggests both mirror types work, and choice D incorrectly states that magnification varies.
An object is placed in front of a convex mirror. As the object is moved from very far away to very close to the mirror surface, the image formed by the mirror:
Answer: Remains virtual and upright, with size increasing from nearly zero to nearly full object size When analyzing convex mirrors, remember that they always produce virtual, upright, and diminished images regardless of object position. This is fundamentally different from concave mirrors, which can produce both real and virtual images depending on object placement. As you move an object from infinity toward a convex mirror, the image behavior follows a predictable pattern. When the object is very far away, the virtual image forms very close to the focal point and appears extremely small—nearly a point. As you bring the object closer to the mirror surface, the virtual image moves away from the mirror and grows larger, but it never exceeds the object's actual size. At the closest approach, the image reaches nearly the same size as the object but remains virtual and upright. Choice A is incorrect because convex mirrors never produce real images—the focal point is virtual (behind the mirror), so objects cannot actually be placed there. Choice C makes the same error, suggesting a transition to real images that's impossible with convex mirrors. This confusion likely stems from mixing up convex and concave mirror properties. Choice D describes behavior typical of concave mirrors when objects are beyond the focal point, not convex mirrors. The key insight is that convex mirrors have consistent behavior: they always diverge light rays, creating virtual images that appear behind the mirror surface. Study tip: Remember "convex = consistent"—convex mirrors always produce virtual, upright, and diminished images. If you see answer choices suggesting real or inverted images from a convex mirror, eliminate them immediately.
Two concave mirrors with the same focal length are placed facing each other. An object is placed between them, closer to mirror A than to mirror B. The image formed by mirror A serves as the object for mirror B. Which statement best describes the final image characteristics?
Answer: The final image characteristics depend on the object's position and the mirror separation distance When you encounter problems involving multiple mirrors, remember that each mirror creates its own image, and that image can serve as the object for subsequent reflections. This creates a complex system where the final result depends heavily on the specific geometry. For two facing concave mirrors, the behavior is remarkably sensitive to both the object's initial position and the distance separating the mirrors. Let's trace what happens: Mirror A forms an image of the original object using the mirror equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ . This image then acts as the object for Mirror B, which forms its own image. The characteristics of each successive image (real vs. virtual, upright vs. inverted, magnified vs. diminished) depend on whether the object distance is greater than, less than, or equal to the focal length. Answer C correctly recognizes that the final image characteristics vary dramatically based on the setup geometry. Small changes in object position or mirror separation can shift the final image from real to virtual, change its orientation, or alter its size significantly. Answer A is wrong because the final image can be real and located outside the mirror system under certain conditions. Answer B is incorrect because the image can be virtual or upright depending on the geometry. Answer D fails because the size and orientation vary with the specific setup. For mirror system problems, always work step-by-step through each reflection rather than trying to predict the outcome. The complexity of multiple reflections means the answer almost always depends on the specific geometric parameters given.
An object is placed in front of a convex spherical mirror. The image formed has a height that is exactly one-third the height of the object. If the focal length of the mirror is $-24$ cm, what is the object distance?
Answer: $48$ cm from the mirror, and the image is located $16$ cm behind the mirror surface For a convex mirror, $|m| = \frac{1}{3}$ means $m = +\frac{1}{3}$ (virtual, upright). Since $m = -\frac{s'}{s}$ , we have $\frac{1}{3} = -\frac{s'}{s}$ , so $s' = -\frac{s}{3}$ . Using the mirror equation: $\frac{1}{-24} = \frac{1}{s} + \frac{1}{-s/3} = \frac{1}{s} - \frac{3}{s} = -\frac{2}{s}$ . Solving gives $s = 48$ cm and $s' = -16$ cm. The other choices result from sign errors or incorrect application of the magnification formula.
A concave mirror is used to project an image of a light bulb onto a screen. The bulb is $50$ cm from the mirror and the screen is $25$ cm from the mirror. If the screen is moved to $50$ cm from the mirror, where should the bulb be repositioned to maintain a focused image on the screen?
Answer: $25$ cm from the mirror When working with concave mirrors, you'll always use the mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ , where $f$ is the focal length, $d_o$ is the object distance, and $d_i$ is the image distance. First, find the focal length using the initial setup. With the bulb at 50 cm and screen at 25 cm: $\frac{1}{f} = \frac{1}{50} + \frac{1}{25} = \frac{1}{50} + \frac{2}{50} = \frac{3}{50}$ . Therefore, $f = \frac{50}{3} = 16.7$ cm. Now for the new setup where the screen is at 50 cm, you need to find where to place the bulb. Using the same focal length: $\frac{1}{16.7} = \frac{1}{d_o} + \frac{1}{50}$ . Solving: $\frac{1}{d_o} = \frac{1}{16.7} - \frac{1}{50} = \frac{3}{50} - \frac{1}{50} = \frac{2}{50}$ . Therefore, $d_o = 25$ cm. Looking at the wrong answers: A) 12.5 cm would create an image much farther than 50 cm from the mirror. B) 16.7 cm is actually the focal length itself—placing an object at the focal point creates no real image. D) 33.3 cm would produce an image closer than 50 cm to the mirror. The key insight here is recognizing that you've essentially swapped the object and image distances from the original setup (50 cm and 25 cm), which is a fundamental property of optical systems—they're reciprocal. Remember: Always find the focal length first from given information, then apply it to the new situation. The mirror's focal length never changes.
A concave shaving mirror has a focal length of $20$ cm. A person's face is positioned $15$ cm from the mirror. What type of image does the person see, and what happens to this image as the person slowly moves away from the mirror?
Answer: Virtual and magnified; the image becomes real and inverted when the person passes the focal point When analyzing concave mirror problems, you need to determine the image type based on object position relative to the focal point, then predict how the image changes as the object moves. First, let's find the initial image characteristics. With the face at 15 cm (closer than the 20 cm focal length), the object is between the focal point and mirror surface. For concave mirrors, objects positioned here always produce virtual, upright, and magnified images that appear behind the mirror surface. Now consider what happens as the person moves away. While the object remains closer than the focal point, the image stays virtual and magnified but becomes smaller. However, once the person passes the focal point (moves beyond 20 cm), the physics changes dramatically. Objects beyond the focal point of a concave mirror create real, inverted images that form in front of the mirror. Answer A incorrectly describes the image movement direction - virtual images actually move away from the mirror surface as objects move away, not toward it. Answer B wrongly claims the initial image is real and inverted, but objects inside the focal length never produce real images in concave mirrors. Answer D incorrectly states the initial image is real and diminished, which contradicts the basic properties of concave mirrors for nearby objects. Answer C correctly identifies the initial virtual, magnified image and accurately predicts the transition to real and inverted when crossing the focal point. Study tip: Remember the focal point as the "dividing line" for concave mirrors - closer objects give virtual images, farther objects give real images.
A concave mirror with focal length $f = 15$ cm forms an image of an object placed $30$ cm from the mirror. If the object is then moved to a position $10$ cm from the mirror, how does the image distance change?
Answer: The image distance changes from $30$ cm to $-30$ cm When you encounter concave mirror problems, you need to apply the mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ , where $f$ is focal length, $d_o$ is object distance, and $d_i$ is image distance. For the initial position ( $d_o = 30$ cm, $f = 15$ cm): $\frac{1}{15} = \frac{1}{30} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{15} - \frac{1}{30} = \frac{2-1}{30} = \frac{1}{30}$ So $d_i = 30$ cm (real image). When the object moves to $d_o = 10$ cm: $\frac{1}{15} = \frac{1}{10} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{15} - \frac{1}{10} = \frac{2-3}{30} = -\frac{1}{30}$ So $d_i = -30$ cm (virtual image). The image distance changes from $30$ cm to $-30$ cm, making answer B correct. Answer A incorrectly calculates the final image distance as $60$ cm instead of $-30$ cm. Answer C uses wrong initial calculations, showing $60$ cm initially instead of $30$ cm. Answer D completely miscalculates both positions—the initial image distance isn't $15$ cm. The key insight here is recognizing that moving the object inside the focal length (from $30$ cm to $10$ cm) transforms a real image into a virtual one. Remember: when $d_o < f$ for concave mirrors, you always get virtual images with negative image distances. Watch for this sign change—it's crucial for understanding image formation.
A convex mirror in a store has a focal length of magnitude $30$ cm. A customer standing $90$ cm from the mirror appears to be what distance from the mirror in the image, and what is the magnification?
Answer: Image distance $= 22.5$ cm behind the mirror; magnification $= +0.25$ When working with convex mirrors, remember that they always produce virtual, upright, and diminished images. The key is applying the mirror equation with the correct sign conventions: focal length is negative for convex mirrors, object distance is positive when in front of the mirror, and image distance is negative when behind the mirror. Given: focal length magnitude = 30 cm (so $f = -30$ cm for convex), object distance $d_o = +90$ cm. Using the mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ Substituting: $\frac{1}{-30} = \frac{1}{90} + \frac{1}{d_i}$ Solving: $\frac{1}{d_i} = -\frac{1}{30} - \frac{1}{90} = -\frac{3}{90} - \frac{1}{90} = -\frac{4}{90} = -\frac{2}{45}$ Therefore: $d_i = -22.5$ cm (negative means behind the mirror) The magnification is: $m = -\frac{d_i}{d_o} = -\frac{(-22.5)}{90} = +0.25$ (positive confirms upright image) Answer A correctly states 22.5 cm behind the mirror with +0.25 magnification. Answer B uses incorrect focal length calculations, giving 45 cm distance. Answer C places the image in front of the mirror (wrong side) and uses negative magnification, which would indicate an inverted image that convex mirrors cannot produce. Answer D significantly overestimates the image distance at 67.5 cm, likely from sign errors in the calculation. Remember: convex mirrors always create virtual images behind the mirror with positive magnification less than 1, making objects appear smaller and closer than they actually are.
An object placed $40$ cm from a concave mirror produces a real image $20$ cm from the mirror. If a plane mirror is now placed $10$ cm behind the concave mirror (on the side opposite to the object), where will the final image appear after reflection from both mirrors?
Answer: $30$ cm behind the plane mirror When dealing with multiple mirror systems, you need to trace the light path step by step, treating each mirror separately and using the image from one mirror as the object for the next. First, let's find the focal length of the concave mirror using the mirror equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ . With object distance $d_o = 40$ cm and image distance $d_i = 20$ cm: $\frac{1}{f} = \frac{1}{40} + \frac{1}{20} = \frac{3}{40}$ , so $f = 13.33$ cm. The concave mirror creates a real image 20 cm in front of it. This image now acts as a virtual object for the plane mirror located 10 cm behind the concave mirror. The distance from this real image to the plane mirror is $20 + 10 = 30$ cm. For a plane mirror, the image appears the same distance behind the mirror as the object is in front of it. Since the real image from the concave mirror is 30 cm in front of the plane mirror, the plane mirror creates an image 30 cm behind itself. Looking at the incorrect options: Choice A (30 cm in front of concave mirror) incorrectly assumes the plane mirror somehow affects the concave mirror's focal properties. Choice B (10 cm behind plane mirror) confuses the plane mirror's position with the image distance. Choice C (50 cm in front of concave mirror) appears to add distances incorrectly without considering the plane mirror's behavior. The correct answer is D: 30 cm behind the plane mirror. Study tip: Always work through multi-mirror problems sequentially—the image from mirror 1 becomes the object for mirror 2.
A dentist uses a concave mirror with a $25$ cm focal length to examine a patient's tooth. If the tooth is positioned $20$ cm from the mirror, what are the characteristics of the image the dentist sees?
Answer: Virtual, upright, and located $100$ cm behind the mirror with magnification $+5$ When you encounter concave mirror problems, you need to apply the mirror equation and understand how object position relative to the focal point determines image characteristics. Using the mirror equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ , where $f = 25$ cm and $d_o = 20$ cm: $\frac{1}{25} = \frac{1}{20} + \frac{1}{d_i}$ Solving: $\frac{1}{d_i} = \frac{1}{25} - \frac{1}{20} = \frac{4-5}{100} = -\frac{1}{100}$ Therefore $d_i = -100$ cm. The negative sign indicates a virtual image located 100 cm behind the mirror. The magnification is $m = -\frac{d_i}{d_o} = -\frac{(-100)}{20} = +5$ . The positive magnification confirms the image is upright and enlarged. Since the object is inside the focal length (20 cm < 25 cm), concave mirrors always produce virtual, upright, and magnified images - exactly what dentists need for examination. Answer A incorrectly shows a real, inverted image with negative magnification. This would occur if the object were beyond the focal point. Answer C has the wrong image distance (50 cm instead of 100 cm) and incorrect magnification, suggesting computational errors. Answer D has the correct image type and location but wrong magnification (2.5 instead of 5), indicating a calculation mistake. Remember: when objects are inside a concave mirror's focal length, you always get virtual, upright, magnified images. This is why concave mirrors are used in makeup mirrors and dental examinations - they provide enlarged views of nearby objects.
A concave mirror produces an image that is $3$ times larger than the object and located $45$ cm from the mirror. If this is a real image, what is the object distance?
Answer: $15$ cm When working with concave mirrors, you need to understand the relationship between object distance, image distance, and magnification. The key equations are the mirror equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ and the magnification formula $m = -\frac{d_i}{d_o}$ , where negative magnification indicates a real, inverted image. Given that the image is 3 times larger and real, the magnification is $m = -3$ (negative because real images are inverted). The image distance is $d_i = 45$ cm. Using the magnification formula: $-3 = -\frac{45}{d_o}$ , which gives us $d_o = \frac{45}{3} = 15$ cm. Let's examine each option: Choice A (15 cm) is correct based on our calculation. Choice B (30 cm) would result from incorrectly using $m = +3$ instead of $m = -3$ , forgetting that real images have negative magnification. Choice C (45 cm) represents the image distance, not the object distance—a common mix-up when students don't carefully track which quantity they're solving for. Choice D (60 cm) comes from incorrectly adding the magnification factor to the image distance ( $45 + 15 = 60$ ), which has no physical basis in mirror optics. Remember that for concave mirrors producing real images, the magnification is always negative, and larger magnification means the object is closer to the focal point. Always double-check whether you're solving for object distance or image distance, as these are easily confused.
A makeup mirror (concave) has a focal length of $40$ cm. A person holds the mirror $30$ cm from their face. What is the apparent distance of the person's face from the mirror surface as seen in the image?
Answer: $120$ cm behind the mirror surface When you encounter concave mirror problems, you're working with the mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ , where $f$ is focal length, $d_o$ is object distance, and $d_i$ is image distance. Given: $f = 40$ cm and $d_o = 30$ cm (the person's face is 30 cm from the mirror). Substituting into the mirror equation: $\frac{1}{40} = \frac{1}{30} + \frac{1}{d_i}$ Solving for $d_i$ : $\frac{1}{d_i} = \frac{1}{40} - \frac{1}{30} = \frac{3-4}{120} = -\frac{1}{120}$ Therefore, $d_i = -120$ cm. The negative sign indicates the image forms behind the mirror surface, appearing 120 cm away. Looking at the wrong answers: A) suggests 60 cm, which would result from incorrectly using $d_i = 2f - d_o$ , a formula that doesn't exist. C) gives 90 cm, which might come from averaging the object distance and some multiple of the focal length. D) shows 45 cm, which could result from mistakenly using $d_i = f + \frac{f \cdot d_o}{2}$ or similar arithmetic errors. The correct answer is B) 120 cm behind the mirror surface. Study tip: For concave mirrors, when the object is closer than the focal point (as here, 30 cm < 40 cm), the image is always virtual (behind the mirror) and magnified. Always check your sign conventions—negative image distance means virtual image behind the mirror.
A concave shaving mirror has a radius of curvature of $40$ cm. When an object is placed $15$ cm in front of the mirror, what is the lateral magnification of the image, and where should a screen be placed to capture the image?
Answer: Magnification is $+4$ , and no screen position will work because the image is virtual With $R = 40$ cm, the focal length is $f = R/2 = 20$ cm. Since the object distance ( $15$ cm) is less than the focal length, the image is virtual. Using $\frac{1}{20} = \frac{1}{15} + \frac{1}{s'}$ : $\frac{1}{s'} = \frac{1}{20} - \frac{1}{15} = \frac{3-4}{60} = -\frac{1}{60}$ , so $s' = -60$ cm. The magnification is $m = -s'/s = -(-60)/15 = +4$ . Virtual images cannot be captured on a screen. Choice B incorrectly assumes a real image, choices C and D have calculation errors.
A spherical mirror forms an image with magnification $m = -0.5$ when an object is placed $30$ cm from it. What type of mirror is this, and what is its focal length?
Answer: Concave mirror with focal length $10$ cm When you encounter spherical mirror problems, you need to use the mirror equation and magnification formula together to determine both the mirror type and focal length. Given that magnification $m = -0.5$ and object distance $d_o = 30$ cm, start with the magnification formula: $m = -\frac{d_i}{d_o}$ . Substituting: $-0.5 = -\frac{d_i}{30}$ , so $d_i = 15$ cm. The positive image distance tells us this is a real image. Since magnification is negative, the image is inverted. Real, inverted images are formed only by concave mirrors when the object is beyond the focal point. Now apply the mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{30} + \frac{1}{15} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$ Therefore, $f = 10$ cm. The positive focal length confirms this is a concave mirror. Answer A gives the wrong focal length (20 cm instead of 10 cm). Answer B incorrectly identifies this as a convex mirror—convex mirrors always produce virtual, upright images with positive magnification, never real inverted images. Answer D makes the same convex mirror error and has the wrong sign for focal length. Remember this key pattern: negative magnification with the object outside the focal point always indicates a concave mirror forming a real, inverted image. Calculate image distance first from magnification, then use the mirror equation to find focal length.
A concave mirror has a focal length of $30$ cm. An object is moved from a position $60$ cm from the mirror to a position $20$ cm from the mirror. How do the image distance and magnification change?
Answer: Image distance changes from $60$ cm to $-60$ cm; magnification changes from $-1$ to $+3$ When working with concave mirrors, you need to master two key equations: the mirror equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ and the magnification formula $m = -\frac{d_i}{d_o}$ . The sign conventions are crucial: positive image distances mean real images (behind the mirror), while negative distances mean virtual images (in front). Let's solve both positions systematically. For the initial position at $d_o = 60$ cm: $\frac{1}{30} = \frac{1}{60} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{30} - \frac{1}{60} = \frac{1}{60}$ So $d_i = 60$ cm, and $m = -\frac{60}{60} = -1$ For the final position at $d_o = 20$ cm: $\frac{1}{30} = \frac{1}{20} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{30} - \frac{1}{20} = -\frac{1}{60}$ So $d_i = -60$ cm, and $m = -\frac{(-60)}{20} = +3$ Answer A correctly shows this transition from $60$ cm to $-60$ cm and magnification from $-1$ to $+3$ . Answer B uses incorrect calculations throughout. Answer C gets the initial conditions right but miscalculates the final image distance as $-30$ cm instead of $-60$ cm. Answer D appears to use completely different object distances. Remember: when an object moves inside the focal length of a concave mirror, the image switches from real and inverted to virtual and upright, which you can identify by the sign change in both image distance and magnification.

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