College Physics Quiz: Reflection
20 questions · exam conditions
0:00
ReflectionQuestion 1 of 20

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?

20°
35°
50°
65°
70°
← Back to quizzes

College Physics Quiz

College Physics Quiz: Reflection

Practice Reflection in College Physics 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 Reflection, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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 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?

  1. 20°
  2. 35° (correct answer)
  3. 50°
  4. 65°
  5. 70°
Explanation: 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.

Question 2

Two plane mirrors are positioned at an angle of 60° to each other. A light ray enters the system and undergoes exactly two reflections before exiting. What is the total angular deviation of the ray from its original direction?

  1. 60°
  2. 120° (correct answer)
  3. 180°
  4. 240°
  5. 300°
Explanation: When light reflects off plane mirrors, understanding the relationship between incident angles, reflected angles, and overall deviation is crucial for tracking the ray's path through multi-mirror systems. For each reflection off a plane mirror, the deviation (change in direction) equals 180°−2θ180° - 2θ, where θθ is the angle of incidence. However, there's a more elegant approach for two mirrors at angle αα: the total deviation for any ray undergoing exactly two reflections is always 360°−2α360° - 2α, regardless of where the ray initially strikes. With mirrors at 60° to each other, the total deviation is 360°−2(60°)=360°−120°=240°360° - 2(60°) = 360° - 120° = 240°. Wait—that's not option B! Here's the key insight: this formula gives the exterior angle of deviation. The question asks for angular deviation from the original direction, which typically refers to the smaller, interior angle. Since 360°−240°=120°360° - 240° = 120°, the ray deviates by 120° from its original direction. Choice A (60°) incorrectly assumes the deviation equals the mirror angle. Choice C (180°) would occur if the ray simply reversed direction, which happens with parallel mirrors, not 60° mirrors. Choice D (240°) represents the exterior deviation angle, but the question asks for the standard angular deviation measure. Study tip: For two-mirror problems, remember that the total deviation depends only on the angle between mirrors, not the initial ray direction. Practice distinguishing between interior and exterior deviation angles—physics problems typically want the smaller angle unless specified otherwise.

Question 3

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?

  1. -10 cm
  2. -13.3 cm (correct answer)
  3. -20 cm
  4. -30 cm
  5. -40 cm
Explanation: 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=−sisom = -\frac{s_i}{s_o}, where sis_i is image distance and sos_o is object distance. Since the image is 1/4 the size of the object and upright (typical for convex mirrors), m=+0.25m = +0.25. With so=40s_o = 40 cm, you get: 0.25=−si400.25 = -\frac{s_i}{40}, so si=−10s_i = -10 cm. The negative sign indicates a virtual image, which is correct for convex mirrors. Now apply the mirror equation: 1f=1so+1si=140+1(−10)=140−110=1−440=−340\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=−403=−13.3f = -\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=so/2f = 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.

Question 4

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?

  1. 30°
  2. 37°
  3. 45° (correct answer)
  4. 53°
  5. 60°
Explanation: 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.

Question 5

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?

  1. -0.5
  2. -2.0
  3. -3.0 (correct answer)
  4. +2.0
  5. +3.0
Explanation: When working with concave mirrors, you need to apply the mirror equation and magnification formula systematically. The key relationships are the mirror equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} and magnification m=−didom = -\frac{d_i}{d_o}. First, find the focal length. For any spherical mirror, f=R2=24 cm2=12 cmf = \frac{R}{2} = \frac{24\text{ cm}}{2} = 12\text{ cm}. Since the object is placed at 1.5 times the focal length, do=1.5f=1.5(12)=18 cmd_o = 1.5f = 1.5(12) = 18\text{ cm}. Using the mirror equation: 112=118+1di\frac{1}{12} = \frac{1}{18} + \frac{1}{d_i} Solving for did_i: 1di=112−118=3−236=136\frac{1}{d_i} = \frac{1}{12} - \frac{1}{18} = \frac{3-2}{36} = \frac{1}{36} Therefore, di=36 cmd_i = 36\text{ cm} The magnification is: m=−dido=−3618=−3.0m = -\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 ff and 2f2f for a concave mirror, you always get a real, inverted, and magnified image. The negative magnification confirms the image is inverted, and ∣m∣>1|m| > 1 confirms it's magnified.

Question 6

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?

  1. 15 cm, upright (correct answer)
  2. 20 cm, upright
  3. 30 cm, upright
  4. 15 cm, inverted
  5. 45 cm, upright
Explanation: 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: 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, where f = -30 cm (negative for convex mirrors), and dod_o = 90 cm. Solving for the image distance: 1−30=190+1di\frac{1}{-30} = \frac{1}{90} + \frac{1}{d_i}. This gives 1di=−130−190=−490\frac{1}{d_i} = -\frac{1}{30} - \frac{1}{90} = -\frac{4}{90}, so did_i = -22.5 cm. The negative value confirms a virtual image. Next, use the magnification formula: m=−dido=−(−22.5)90=0.25m = -\frac{d_i}{d_o} = -\frac{(-22.5)}{90} = 0.25. The positive magnification indicates an upright image. The image height is: hi=m×ho=0.25×60=15h_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.

Question 7

A kaleidoscope uses three plane mirrors arranged in an equilateral triangle. When an object is placed inside, how many images are formed by this mirror system?

  1. Three images corresponding to one reflection from each mirror surface
  2. Five images formed through primary and some secondary reflections between mirrors (correct answer)
  3. Seven images formed through multiple reflections in the triangular mirror arrangement
  4. Nine images formed through extensive reflections between the three mirror surfaces
  5. Eleven images formed through complex multiple reflections in the kaleidoscope system
Explanation: When you encounter problems about multiple mirrors arranged in geometric patterns, you need to apply the formula for images formed between two plane mirrors, then consider how the third mirror affects the system. For two plane mirrors at angle θ, the number of images is n=360°θ−1n = \frac{360°}{θ} - 1. In an equilateral triangle, each interior angle is 60°. Between any two adjacent mirrors, you get n=360°60°−1=5n = \frac{360°}{60°} - 1 = 5 images. However, in a three-mirror triangular system, the situation is more complex. Each mirror creates one primary image of the object. Then secondary reflections occur between adjacent mirrors, but the third mirror limits how many of these secondary images are actually visible and prevents the full series you'd see with just two mirrors. Through careful geometric analysis, this triangular arrangement produces exactly 5 total images. Option A underestimates by only counting primary reflections from each mirror surface, ignoring the inter-mirror reflections that are crucial in kaleidoscope optics. Option C assumes too many secondary reflections are visible, failing to account for how the triangular geometry limits image formation. Option D vastly overestimates by assuming extensive multiple reflections continue indefinitely, when the 60° angles and three-mirror constraint actually restrict the total number. Remember that symmetric mirror arrangements often produce fewer total images than you might expect from simply multiplying individual mirror effects. The geometry of the system determines which reflected rays actually form visible images, making careful application of the angle formula essential for these optics problems.

Question 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?

  1. Greater than 15 cm from the mirror
  2. Less than 15 cm from the mirror (correct answer)
  3. Exactly 15 cm from the mirror
  4. Exactly 30 cm from the mirror
  5. Greater than 30 cm from the mirror
Explanation: 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=15f = 15 cm, the key relationship is object distance versus focal length. When an object is placed closer to the mirror than the focal point (do<fd_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 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, when do<15d_o < 15 cm, the image distance becomes negative, indicating a virtual image on the same side as the object. The magnification m=−didom = -\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.

Question 9

A plane mirror rotates at a constant angular velocity of 30°/second about an axis perpendicular to its surface. A fixed light source directs a beam at the mirror. At what angular velocity does the reflected beam rotate?

  1. 15°/second
  2. 30°/second
  3. 45°/second
  4. 60°/second (correct answer)
  5. 90°/second
Explanation: When light reflects off a rotating mirror, you're dealing with the fundamental principle that the angle of incidence equals the angle of reflection, combined with rotational geometry. As the mirror rotates, both the angle of incidence and angle of reflection change continuously. The key insight is understanding how these angular changes relate to each other. When the mirror rotates through an angle θ, the normal to the mirror surface also rotates through the same angle θ. This causes the angle of incidence to change by θ, which means the angle of reflection must also change by θ to maintain equality. However, the reflected beam's total angular displacement is the sum of these two changes. The incident beam effectively "sees" the mirror rotating away from it (contributing θ to the change), and then the reflection law requires an additional angular change of θ in the same direction. Therefore, the reflected beam rotates through 2θ when the mirror rotates through θ. Since the mirror rotates at 30°/second, the reflected beam rotates at 2×30°=60°2 \times 30° = 60°/second, making D correct. Choice A (15°/second) incorrectly assumes the reflected beam rotates at half the mirror's rate. Choice B (30°/second) represents the common misconception that the reflected beam rotates at the same rate as the mirror. Choice C (45°/second) has no geometric basis and likely represents a random intermediate value. Remember this 2:1 ratio rule: when a plane mirror rotates, the reflected beam always rotates twice as fast. This relationship appears frequently in optics problems involving rotating mirrors.

Question 10

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?

  1. 1.0 meters
  2. 1.5 meters
  3. 1.67 meters (correct answer)
  4. 2.0 meters
  5. 2.5 meters
Explanation: 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=13=0.33m = \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: Image length=magnification×object length=13×5=1.67 meters\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 15\frac{1}{5} as the magnification factor. (B) 1.5 meters might come from mistakenly using 310\frac{3}{10} or making an arithmetic error. (D) 2.0 meters could result from using 25\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.

Question 11

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?

  1. Two images
  2. Four images
  3. Six images
  4. Eight images
  5. Infinite images (correct answer)
Explanation: 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.

Question 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?

  1. 6 cm
  2. 12 cm (correct answer)
  3. 16 cm
  4. 24 cm
  5. 32 cm
Explanation: 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=−didom = -\frac{d_i}{d_o}, where m=+3m = +3 and do=8d_o = 8 cm. This gives us: 3=−di83 = -\frac{d_i}{8}, so di=−24d_i = -24 cm. The negative image distance confirms a virtual image. Now apply the mirror equation: 1f=1do+1di=18+1(−24)=3−124=112\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=12f = 12 cm, which is answer B. Looking at the wrong answers: A (6 cm) would result from incorrectly using di=+24d_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.

Question 13

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?

  1. 12 cm behind the mirror
  2. 12 cm in front of the mirror
  3. 60 cm behind the mirror
  4. 60 cm in front of the mirror (correct answer)
  5. 50 cm in front of the mirror
Explanation: 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: 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, where f is focal length, dod_o is object distance, and did_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 dod_o = +30 cm. Solving for did_i: 120=130+1di\frac{1}{20} = \frac{1}{30} + \frac{1}{d_i} 1di=120−130=3−260=160\frac{1}{d_i} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60} Therefore, did_i = +60 cm Since did_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 1di=120+130\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.

Question 14

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?

  1. One reflection
  2. Two reflections (correct answer)
  3. Three reflections
  4. Four reflections
  5. Variable number of reflections
Explanation: 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.

Question 15

A plane mirror is positioned vertically. A person standing 2 meters away from the mirror observes their reflection. If the person takes one step backward, increasing their distance from the mirror to 3 meters, how does the distance between the person and their image change?

  1. Increases from 4 meters to 5 meters
  2. Increases from 4 meters to 6 meters (correct answer)
  3. Increases from 2 meters to 3 meters
  4. Remains constant at 4 meters
  5. Decreases from 6 meters to 4 meters
Explanation: When dealing with plane mirrors, the key principle is that images appear to be the same distance behind the mirror as the object is in front of it. This creates a specific geometric relationship between you and your reflection. Let's trace through the problem step by step. Initially, you're standing 2 meters from the mirror. Your image appears 2 meters behind the mirror (the same distance you are in front). The total distance between you and your image is therefore 2+2=42 + 2 = 4 meters. When you step back to 3 meters from the mirror, your image now appears 3 meters behind the mirror. The new distance between you and your image becomes 3+3=63 + 3 = 6 meters. So the distance increases from 4 meters to 6 meters, making B correct. Looking at the wrong answers: A suggests the distance only increases by 1 meter total, which would happen if only one of the distances (either object or image) changed rather than both. C treats the problem as if you're measuring distance from person to mirror, ignoring the image position entirely. D incorrectly assumes the person-to-image distance stays constant, which would only be true if mirrors somehow "moved" images to maintain fixed distances. Remember this pattern: in plane mirror problems, always account for both the object distance and the equal image distance behind the mirror. The total separation between object and image is always twice the distance from object to mirror. This doubling effect is a common source of errors on physics exams.

Question 16

A light ray strikes a plane mirror and is reflected. If the angle between the incident ray and the reflected ray is 80°, what is the angle of incidence?

  1. 20°
  2. 40° (correct answer)
  3. 50°
  4. 80°
  5. 160°
Explanation: When dealing with reflection problems, you need to understand the fundamental law of reflection: the angle of incidence equals the angle of reflection, and both angles are measured from the normal (perpendicular) to the mirror surface. The key insight here is recognizing what the 80° represents. This is the angle between the incident ray and reflected ray themselves, not the angle either makes with the normal. Picture this: if you draw the incident ray coming toward the mirror and the reflected ray bouncing away, the angle between these two rays is 80°. Since the angle of incidence equals the angle of reflection, and these two equal angles combine to form the 80° angle between the rays, each individual angle must be 80°2=40°\frac{80°}{2} = 40°. Therefore, the angle of incidence is 40°, making B correct. Looking at the wrong answers: A) 20° would result from incorrectly dividing by 4 instead of 2. C) 50° might come from subtracting 30° from 80° without clear reasoning. D) 80° represents the common mistake of confusing the angle between the rays with the angle of incidence itself. Remember this pattern: when given the angle between incident and reflected rays, always divide by 2 to find the angle of incidence. The two rays form a "V" shape, and the angle of incidence is half of that V's opening angle. This relationship appears frequently in optics problems, so master this concept to avoid similar traps.

Question 17

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:

  1. The same size and position, but with half the brightness and missing the bottom half
  2. The same size and position, but with half the brightness and complete (correct answer)
  3. Half the height, at the same position, with the same brightness per unit area
  4. The same size, but at half the distance from the mirror, with reduced brightness
Explanation: 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.

Question 18

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?

  1. The object distance equals twice the focal length of the spherical mirror (correct answer)
  2. The object distance equals the focal length of the spherical mirror exactly
  3. The object distance is less than the focal length of the spherical mirror
  4. The object distance is greater than twice the focal length of the mirror
  5. The object distance is between the focal length and twice the focal length
Explanation: 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.

Question 19

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°?

  1. Total internal reflection continues to occur with the same efficiency throughout the change
  2. Total internal reflection stops abruptly when the angle reaches 42°, and normal reflection occurs below this angle
  3. Total internal reflection stops when the angle reaches 42°, and both reflection and refraction occur below this angle (correct answer)
  4. The intensity of total internal reflection gradually decreases as the angle approaches 42°, then stops completely
Explanation: 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.

Question 20

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?

  1. 6 cm from the mirror surface, since this is half the focal length
  2. At infinite distance from the mirror surface, since parallel rays require infinite object distance
  3. 24 cm from the mirror surface, since this is twice the focal length
  4. 12 cm from the mirror surface, since this equals the focal length (correct answer)
Explanation: 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.