Reflection - College Physics

1. An object is placed at various distances from a concave mirror with focal length 12 cm. At which object distance will the reflected rays be parallel after reflection from the mirror?
   Answer: 12 cm from the mirror surface, since this equals the focal length When dealing with concave mirrors and ray behavior, you need to understand the relationship between object position and the resulting reflected rays. The key insight is that light rays are reversible - if parallel rays coming from infinity converge at the focal point, then rays emanating from the focal point will reflect as parallel rays. For reflected rays to be parallel after hitting a concave mirror, the object must be placed exactly at the focal point of the mirror. When an object is positioned at the focal length distance, each ray from the object hits the mirror and reflects parallel to the principal axis. Since the focal length is 12 cm, placing the object at 12 cm from the mirror surface produces parallel reflected rays, making answer D correct. Let's examine why the other options are incorrect: Option A suggests 6 cm because it's half the focal length, but this position would cause reflected rays to diverge significantly, not become parallel. Option B claims infinite distance is needed, but this reverses the cause and effect - parallel incident rays from infinity focus at the focal point, which is the opposite of what we want. Option C proposes 24 cm (twice the focal length), but at this distance, the reflected rays would converge to form a real image between the focal point and mirror, not emerge as parallel rays. Remember this principle: object at focal point equals parallel reflected rays. This reciprocal relationship between parallel rays and focal point behavior is fundamental to mirror optics and appears frequently on physics exams.
2. Two plane mirrors are arranged to form a corner reflector with an angle of 60° between them. A light ray enters parallel to one mirror and strikes the other mirror first. After reflecting from both mirrors, the ray emerges in what direction relative to its initial direction?
   Answer: At an angle of 120° to the initial direction For two plane mirrors at angle θ, a ray undergoes total deviation of 2θ regardless of the incident angle. With θ = 60°, the total deviation is 2×60° = 120°. The ray emerges at 120° to its initial direction. Choice A describes retroreflection (90° mirrors). Choice B incorrectly uses just the mirror angle. Choice D uses 360° - 120°, confusing angular directions.
3. A convex mirror used as a security mirror in a store has a focal length of -30 cm. An object 60 cm tall is positioned 90 cm in front of the mirror. What is the height of the image formed?
   Answer: 15 cm, upright When you encounter convex mirror problems, remember that these mirrors always produce virtual, upright, and diminished images. The key is applying the mirror equation and magnification formula systematically. Start with the mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$, where f = -30 cm (negative for convex mirrors), and $d_o$ = 90 cm. Solving for the image distance: $\frac{1}{-30} = \frac{1}{90} + \frac{1}{d_i}$. This gives $\frac{1}{d_i} = -\frac{1}{30} - \frac{1}{90} = -\frac{4}{90}$, so $d_i$ = -22.5 cm. The negative value confirms a virtual image. Next, use the magnification formula: $m = -\frac{d_i}{d_o} = -\frac{(-22.5)}{90} = 0.25$. The positive magnification indicates an upright image. The image height is: $h_i = m \times h_o = 0.25 \times 60 = 15$ cm. Choice A is correct: 15 cm, upright. Choice B (20 cm, upright) uses an incorrect calculation, possibly from rounding errors or algebraic mistakes. Choice C (30 cm, upright) might result from incorrectly using the focal length as the magnification factor. Choice D (15 cm, inverted) gets the height right but misunderstands that convex mirrors never produce inverted images. Remember this pattern: convex mirrors always create images that are virtual, upright, and smaller than the object. If you ever calculate an inverted image for a convex mirror, double-check your magnification sign.
4. Use the ray diagram above to determine the relationship. In the ray diagram shown, three rays from an object are traced to locate the image formed by a spherical mirror. Based on the ray paths, what can be concluded about the focal length and the object distance?
   Answer: The object distance equals twice the focal length of the spherical mirror When the object is placed at twice the focal length (at the center of curvature) of a concave mirror, the reflected rays converge to form an image also at twice the focal length on the opposite side. The ray diagram shows parallel rays reflecting through the focal point and rays through the focal point reflecting parallel, with the image forming at the same distance as the object from the mirror. Choice B would show no image formation. Choice C would show divergent rays. Choice D would show image closer than object. Choice E would show magnified image.
5. A student places a small object 20 cm in front of a concave spherical mirror with focal length 15 cm. She then covers the bottom half of the mirror with black tape. Compared to the uncovered mirror, the image formed will be:
   Answer: The same size and position, but with half the brightness and complete Using the mirror equation: 1/f = 1/do + 1/di gives 1/15 = 1/20 + 1/di, so di = 60 cm. Each point on the object sends rays to all parts of the mirror, so covering half the mirror reduces light intensity but doesn't eliminate any part of the image. The image remains complete, same size, same position, but dimmer. Choice A incorrectly assumes geometric correspondence. Choice C confuses aperture effects. Choice D incorrectly changes image distance.
6. A concave mirror has a radius of curvature of 24 cm. An object is placed at a distance equal to 1.5 times the focal length from the mirror. What is the magnification of the image?
   Answer: -3.0 When working with concave mirrors, you need to apply the mirror equation and magnification formula systematically. The key relationships are the mirror equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ and magnification $m = -\frac{d_i}{d_o}$. First, find the focal length. For any spherical mirror, $f = \frac{R}{2} = \frac{24\text{ cm}}{2} = 12\text{ cm}$. Since the object is placed at 1.5 times the focal length, $d_o = 1.5f = 1.5(12) = 18\text{ cm}$. Using the mirror equation: $\frac{1}{12} = \frac{1}{18} + \frac{1}{d_i}$ Solving for $d_i$: $\frac{1}{d_i} = \frac{1}{12} - \frac{1}{18} = \frac{3-2}{36} = \frac{1}{36}$ Therefore, $d_i = 36\text{ cm}$ The magnification is: $m = -\frac{d_i}{d_o} = -\frac{36}{18} = -3.0$ This confirms answer C is correct. Looking at the wrong answers: A) -0.5 would result from incorrectly inverting the magnification formula or making arithmetic errors. B) -2.0 might come from using the wrong object distance or focal length. D) +2.0 has the wrong sign—forgetting that real images formed by concave mirrors are always inverted, giving negative magnification. Study tip: Remember that when the object distance is between $f$ and $2f$ for a concave mirror, you always get a real, inverted, and magnified image. The negative magnification confirms the image is inverted, and $|m| > 1$ confirms it's magnified.
7. A light ray strikes a plane mirror at an angle of incidence of 35°. If the mirror is then rotated 15° clockwise about an axis perpendicular to the plane of incidence, what is the new angle of reflection?
   Answer: 35° When you encounter mirror rotation problems, the key principle is that the angle of reflection always equals the angle of incidence, regardless of the mirror's orientation. This is the fundamental law of reflection, and it holds true even when the mirror moves. Initially, the light ray hits the mirror at 35° and reflects at 35°. When the mirror rotates 15° clockwise, you might think this changes the reflection angle, but here's what actually happens: the mirror's surface normal (the perpendicular line to the mirror) also rotates 15°. However, the incoming light ray doesn't change direction. After rotation, you need to measure the new angle of incidence relative to the new normal. The geometry works out so that the angle between the incident ray and the rotated normal is still 35°. Since the angle of reflection must equal the angle of incidence, the reflected ray also makes a 35° angle with the new normal. Choice A) 20° incorrectly subtracts the rotation angle from the original reflection angle. Choice C) 50° mistakenly adds the rotation angle to the original angle. Choice D) 65° appears to double-count the rotation effect. These errors stem from confusing how angles are measured relative to the surface versus the normal. The correct answer is B) 35°. Study tip: Remember that reflection angles are always measured from the normal, not the mirror surface. The law of reflection (angle in = angle out) is invariant—it doesn't change when you rotate the mirror, only the direction of the reflected ray changes.
8. A dentist uses a concave mirror with a 15 cm focal length to examine a patient's teeth. To obtain an upright, magnified image of a tooth, at what distance should the tooth be positioned from the mirror?
   Answer: Less than 15 cm from the mirror When working with concave mirrors, you need to understand how object position affects image characteristics. Concave mirrors can produce both real and virtual images depending on where you place the object relative to the focal point. For a concave mirror with focal length $f = 15$ cm, the key relationship is object distance versus focal length. When an object is placed closer to the mirror than the focal point ($d_o < f$), the mirror produces a virtual image that is upright and magnified - exactly what the dentist needs to examine teeth clearly. Using the mirror equation $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$, when $d_o < 15$ cm, the image distance becomes negative, indicating a virtual image on the same side as the object. The magnification $m = -\frac{d_i}{d_o}$ becomes positive (upright) and greater than 1 (magnified). Answer B is correct because positioning the tooth less than 15 cm from the mirror creates the desired upright, magnified virtual image. Answer A is wrong because placing the object beyond the focal point produces a real, inverted image - useless for dental examination. Answer C is incorrect because at exactly the focal point, no image forms (rays emerge parallel). Answer D is wrong because at 30 cm (twice the focal length), you get a real, inverted image at the same distance on the opposite side. Study tip: Remember the concave mirror rule: object inside the focal point gives virtual, upright, and magnified images - perfect for magnifying glasses, makeup mirrors, and dental mirrors.
9. A beam of light undergoes total internal reflection at the boundary between two media. If the critical angle is 42°, and the beam initially travels at 50° to the normal in the denser medium, what happens when the angle is gradually decreased to 35°?
   Answer: Total internal reflection stops when the angle reaches 42°, and both reflection and refraction occur below this angle Total internal reflection occurs only when the incident angle exceeds the critical angle (42°). At 50°, total internal reflection occurs. When the angle decreases to 42°, the beam is exactly at the critical angle. Below 42°, the beam undergoes partial reflection and refraction into the second medium. Choice A ignores the critical angle condition. Choice B incorrectly suggests normal reflection only. Choice D incorrectly describes a gradual transition in total internal reflection intensity.
10. A convex spherical mirror has a focal length of magnitude 24 cm. An object is placed 36 cm in front of the mirror. What is the magnification of the image?
    Answer: $+0.4$ For a convex mirror, f = -24 cm (negative focal length). Using 1/f = 1/do + 1/di: 1/(-24) = 1/36 + 1/di, solving gives di = -14.4 cm. Magnification m = -di/do = -(-14.4)/36 = +0.4. The positive magnification indicates an upright virtual image. Choice B uses wrong sign convention. Choices C and D incorrectly calculate the reciprocal or use wrong focal length sign.
11. A laser beam strikes a plane mirror at an angle of 35° from the normal. The mirror is then rotated 15° clockwise about the point where the beam hits it. What is the angle between the initial reflected beam and the final reflected beam?
    Answer: 30° When a mirror rotates by angle θ, the reflected beam rotates by 2θ. Initially, the beam reflects at 35° from normal. When the mirror rotates 15° clockwise, the reflected beam rotates 2×15° = 30° from its original direction. Choice A incorrectly uses just the mirror rotation angle. Choice C adds the initial angle and twice the rotation (35° + 15°). Choice D adds the initial angle and the full beam rotation (35° + 30° + 5°).
12. A concave shaving mirror produces an upright image that is 3 times larger than the object when the object is placed 8 cm from the mirror. What is the focal length of the mirror?
    Answer: 12 cm When you encounter mirror problems involving magnification and object distance, you're working with the mirror equation and magnification relationships. For spherical mirrors, the key is understanding how object distance, image distance, and focal length relate. Given information: magnification = +3 (upright image), object distance = 8 cm. The positive magnification tells you the image is upright and virtual, which is characteristic of concave mirrors when the object is closer than the focal point. Using the magnification formula: $m = -\frac{d_i}{d_o}$, where $m = +3$ and $d_o = 8$ cm. This gives us: $3 = -\frac{d_i}{8}$, so $d_i = -24$ cm. The negative image distance confirms a virtual image. Now apply the mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{8} + \frac{1}{(-24)} = \frac{3-1}{24} = \frac{1}{12}$ Therefore, $f = 12$ cm, which is answer B. Looking at the wrong answers: A (6 cm) would result from incorrectly using $d_i = +24$ cm instead of negative. C (16 cm) might come from mistakenly adding the object and image distances directly. D (24 cm) could result from confusing the image distance with the focal length. Study tip: Always check the sign of your magnification first—it tells you whether the image is upright or inverted, which helps verify your setup is correct. Remember that for concave mirrors, virtual images occur when objects are closer than the focal point.
13. Two identical plane mirrors are arranged parallel to each other, separated by a distance of 1 meter. A point source of light is placed exactly halfway between the mirrors. How many images of the light source will be formed?
    Answer: Infinite images When you encounter parallel mirror problems, you're dealing with multiple reflections that create a series of images. The key insight is that each mirror reflects not only the original object, but also the images formed by the other mirror. With the light source positioned exactly halfway between two parallel mirrors, each mirror will form images at specific locations. The first mirror creates an image 0.5 meters behind it, and the second mirror creates an image 0.5 meters behind it on the opposite side. However, these aren't the only images formed. Each mirror also reflects the images created by the other mirror, generating additional images further back. This process continues theoretically forever, creating an infinite series of images that appear to recede into the distance on both sides. The images become progressively dimmer due to light loss with each reflection, but mathematically, infinite images are formed. Looking at the wrong answers: A) "Two images" only counts the primary reflection from each mirror, ignoring the reflections of reflections. B) "Four images" might account for the first two reflections from each mirror but misses the continuing series. C) "Six images" and D) "Eight images" represent attempts to count a finite number of reflections, but both underestimate the true situation. For parallel mirror problems, remember that when the object is positioned between the mirrors (not at a specific angle that would limit reflections), you always get infinite images. Watch for this pattern on physics exams—the "infinite images" answer often appears when dealing with parallel mirror arrangements.
14. A concave spherical mirror has a focal length of 20 cm. An object is placed 30 cm in front of the mirror. At what distance from the mirror will the image be formed?
    Answer: 60 cm in front of the mirror When you encounter spherical mirror problems, you need to apply the mirror equation and understand the sign conventions for concave mirrors. For any spherical mirror, use 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 concave mirrors, the focal length is positive, and when the image forms in front of the mirror (real image), the image distance is positive. Given: f = +20 cm and $d_o$ = +30 cm. Solving for $d_i$: $\frac{1}{20} = \frac{1}{30} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$ Therefore, $d_i$ = +60 cm Since $d_i$ is positive, the image forms 60 cm in front of the mirror, making D correct. Choice A (12 cm behind) results from incorrectly calculating $\frac{1}{d_i} = \frac{1}{20} + \frac{1}{30}$ and misapplying sign conventions. Choice B (12 cm in front) uses the same calculation error but correct location reasoning. Choice C (60 cm behind) applies the wrong sign convention—negative image distance indicates a virtual image behind the mirror, but this setup produces a real image. Remember: for concave mirrors, when the object is beyond the focal point, you'll always get a real image in front of the mirror. The key is careful algebra with the mirror equation and understanding that positive image distance means "in front" for concave mirrors.
15. A ray of light strikes a plane mirror at point P. The mirror is then moved parallel to itself by a distance d in the direction perpendicular to its surface, while the incident ray direction remains unchanged. How does the position of the reflected ray change?
    Answer: The reflected ray is displaced parallel to itself by distance 2d in the same direction as the mirror movement When light reflects off a mirror, you need to carefully track how both the incident and reflected rays behave when the mirror's position changes. This tests your understanding of geometric ray optics and the law of reflection. Let's trace what happens step by step. Initially, light hits the mirror at point P and reflects according to the law of reflection (angle of incidence equals angle of reflection). When you move the mirror perpendicular to its surface by distance d, the incident ray now hits the mirror at a new point that's distance d closer to the source along the incident ray's path. From this new intersection point, the light reflects at the same angle as before (since the mirror orientation hasn't changed). However, this new reflection point is shifted by distance d along the incident ray. When you draw the reflected ray from this new point, you'll find it's parallel to the original reflected ray but displaced by distance 2d in the same direction as the mirror movement. Why 2d? The displacement comes from two equal contributions: d from the mirror moving closer, and another d from the geometry of reflection. Answer A incorrectly gives displacement d instead of 2d. Answer B wrongly suggests the ray direction stays the same with just a shifted origin - the entire ray path changes. Answer C incorrectly describes rotation about the original point P, but P is no longer the reflection point. Remember this 2:1 ratio rule: when a mirror moves perpendicular to its surface by distance d, the reflected ray displaces by 2d. This geometric relationship appears frequently in optics problems.
16. Light traveling in water (n = 1.33) strikes a glass surface (n = 1.50) at an angle of incidence of 45°. What is the angle of reflection?
    Answer: 45° When light hits a surface between two media, you need to distinguish between reflection and refraction - they follow different laws and occur simultaneously. The law of reflection states that the angle of reflection always equals the angle of incidence, regardless of the materials involved. This is a fundamental principle that applies to any interface between two media. Since the incident ray strikes the glass surface at 45°, the reflected ray must also make a 45° angle with the normal to the surface. The refractive indices given (water n = 1.33, glass n = 1.50) are relevant for calculating the refraction angle using Snell's law, but they don't affect reflection at all. The angle of reflection depends only on the incident angle. Looking at the wrong answers: A) 30° and B) 37° likely come from attempting to use Snell's law incorrectly for reflection, or from trigonometric errors involving the given refractive indices. D) 53° might result from confusion about complementary angles or from misapplying the refractive indices in some calculation. The key insight is that reflection and refraction are separate phenomena. When light hits an interface, part of it reflects (following the law of reflection) and part refracts (following Snell's law). The correct answer is C) 45°. Remember this distinction: reflection angle always equals incidence angle, while refraction angle depends on the refractive indices. Don't let extra information about materials distract you when the question asks specifically about reflection.
17. A convex mirror in a parking garage produces images that are always 1/3 the height of the actual objects. A car that is actually 5 meters long appears in this mirror. What is the apparent length of the car's image in the mirror?
    Answer: 1.67 meters When you encounter mirror problems, the key relationship to remember is that the magnification factor applies uniformly to all dimensions of an object. Convex mirrors always produce diminished, upright, virtual images. Since this convex mirror produces images that are always 1/3 the height of actual objects, the magnification is $m = \frac{1}{3} = 0.33$. This same magnification applies to all linear dimensions—not just height, but also length, width, and any other measurement. For a car that is actually 5 meters long, the image length will be: $\text{Image length} = \text{magnification} \times \text{object length} = \frac{1}{3} \times 5 = 1.67 \text{ meters}$ Looking at the wrong answers: (A) 1.0 meters would result from incorrectly using $\frac{1}{5}$ as the magnification factor. (B) 1.5 meters might come from mistakenly using $\frac{3}{10}$ or making an arithmetic error. (D) 2.0 meters could result from using $\frac{2}{5}$ as the magnification, perhaps confusing the given ratio. The correct answer is (C) 1.67 meters. Remember that magnification in optics is a scaling factor that applies to all linear dimensions equally. When a problem gives you the magnification for one dimension (like height), you can confidently apply that same factor to any other linear measurement of the object. This principle holds for all types of mirrors and lenses.
18. A convex mirror produces an image that is 1/4 the size of the object. If the object is 40 cm from the mirror, what is the focal length of the mirror?
    Answer: -13.3 cm When dealing with curved mirrors, you need to understand the relationship between object distance, image distance, magnification, and focal length. Convex mirrors always produce virtual, upright, and diminished images, which is why the magnification is positive but less than 1. Start with the magnification equation: $m = -\frac{s_i}{s_o}$, where $s_i$ is image distance and $s_o$ is object distance. Since the image is 1/4 the size of the object and upright (typical for convex mirrors), $m = +0.25$. With $s_o = 40$ cm, you get: $0.25 = -\frac{s_i}{40}$, so $s_i = -10$ cm. The negative sign indicates a virtual image, which is correct for convex mirrors. Now apply the mirror equation: $\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} = \frac{1}{40} + \frac{1}{(-10)} = \frac{1}{40} - \frac{1}{10} = \frac{1-4}{40} = -\frac{3}{40}$. Therefore, $f = -\frac{40}{3} = -13.3$ cm. Choice A (-10 cm) incorrectly uses just the image distance as the focal length. Choice C (-20 cm) likely comes from incorrectly assuming $f = s_o/2$, which only applies to specific geometric situations. Choice D (-30 cm) might result from sign errors or incorrect application of the magnification formula. Remember: for convex mirrors, always expect negative focal lengths, and use both the magnification and mirror equations systematically. Don't confuse image distance with focal length.
19. A periscope uses two plane mirrors positioned at 45° to the horizontal. Light enters the periscope horizontally and exits horizontally in the same direction. What is the total number of reflections that occur in this optical system?
    Answer: Two reflections When analyzing periscope optics, you need to trace the light path step by step through the system to count each reflection event. In a standard periscope, light enters horizontally and must exit horizontally in the same direction after navigating around an obstacle. The two plane mirrors are positioned at 45° angles to accomplish this redirection. Here's how the light travels: First, the incoming horizontal light ray hits the upper mirror (angled at 45°), which reflects it downward at a 90° angle. This downward-traveling light then strikes the lower mirror (also at 45°), which reflects it horizontally again, sending it out in the same direction as the original incoming ray. Each time light hits a mirror surface and bounces off, that constitutes one reflection. In this system, there are exactly two such events. Looking at the wrong answers: (A) One reflection would only bend the light once, leaving it traveling vertically rather than horizontally as required. (C) Three reflections and (D) Four reflections would require additional mirrors beyond the two specified in the problem, and would likely redirect the light in unintended directions. The correct answer is (B) Two reflections. Study tip: For any mirror system problem, always trace the light path from entrance to exit, counting each mirror interaction. Remember that each mirror contact equals one reflection, regardless of the mirror's size or the complexity of the overall system. This methodical approach prevents miscounting in optical pathway problems.
20. Light traveling in a medium with refractive index 1.8 strikes the boundary with air at an angle of 25° from the normal. What percentage of the light intensity is reflected at this boundary, assuming the Fresnel reflection coefficient for this geometry is 0.15?
    Answer: 15% is reflected and 85% is transmitted, with no light loss First, check if total internal reflection occurs. The critical angle is θc = arcsin(1/1.8) ≈ 33.7°. Since 25° < 33.7°, the light can transmit into air, so total internal reflection does not occur. For angles below the critical angle, the Fresnel reflection coefficient determines what fraction reflects. Given coefficient 0.15 means 15% reflects and 85% transmits, with total energy conserved. Choice B reverses the percentages. Choice C incorrectly claims total internal reflection. Choice D incorrectly relates reflection percentage to incident angle.

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