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In this lesson, you learned what happens to light when it is reflected and refracted.
In this assignment, you will differentiate between reflection and refraction, use the law of reflection to make predictions, apply Snell’s law to solve problems, and analyze and interpret ray diagrams.
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Below is a complete assignment you can give students (with an answer key). It covers distinguishing reflection vs refraction, using the law of reflection, applying Snell’s law and the critical angle, and interpreting ray diagrams.
Instructions for students
- Show all work and label all angles relative to the normal unless stated otherwise.
- Use nair = 1.00, nwater = 1.33, nglass = 1.50 unless told otherwise.
- Use Snell’s law: n1 sin θ1 = n2 sin θ2.
- Critical angle (for light going from medium 1 into medium 2, with n1 > n2): sin θc = n2 / n1.
PART A — Conceptual (short answers)
1. Define reflection and refraction. Give one everyday example of each.
2. Explain the difference between specular (regular) reflection and diffuse reflection.
3. State the law of reflection in words and write its formula.
PART B — Law of reflection (calculate and sketch)
4. A light ray strikes a plane mirror so that the angle between the incident ray and the normal is 30°. What is the angle between the reflected ray and the normal? Draw a simple sketch showing the incident ray, normal, and reflected ray.
5. A beam hits a plane mirror at 20° to the normal. If the mirror is rotated by 10° about an axis perpendicular to the plane of the paper, by how many degrees does the reflected ray rotate? Explain briefly.
PART C — Snell’s law and critical angle (calculations)
6. Light goes from air into water. If θ1 = 40° (in air), find θ2 in the water. (Give answer to 1 decimal place.)
7. Light in glass (n = 1.50) strikes the glass–air interface at θ1 = 60°. Determine whether the light is refracted into air or totally internally reflected. If TIR occurs, explain why. If refracted, find θ2.
8. Calculate the critical angle for the glass–air interface (nglass = 1.50 to nair = 1.00). Give answer to 1 decimal place.
9. A laser beam in air strikes a glass slab (n = 1.50) at θ = 25°. Find: (a) the angle inside the glass, (b) the angle of the beam when it emerges back into air after passing through a parallel-faced slab.
10. A swimming pool has depth 3.00 m. What depth does the bottom appear to a person viewing from directly above (approximately normal incidence)? (Use nwater = 1.33.)
PART D — Ray diagrams and interpretation (short tasks with sketches/calculations)
11. A ray in air hits a flat piece of glass at point P with incident angle 50° to the normal. Determine whether the ray is refracted into the glass or totally internally reflected. If it is refracted, calculate the angle inside the glass. Sketch the ray path and label angles.
12. A ray inside water strikes the water–air surface with angle 60° to the normal. Will it be refracted into air or totally internally reflected? Explain and sketch the ray(s).
13. Draw a ray diagram (qualitative) showing why an object under water appears closer to the surface when viewed from above. Label the real depth and apparent depth and indicate the refracted rays.
Extension (optional)
14. Two plane mirrors are placed at 90° to each other. A ray hits one mirror and reflects to the second mirror and then to an observer. Using the law of reflection, describe qualitatively the final direction relative to the original ray (hint: two reflections across perpendicular planes reverse direction).
Answer key (worked solutions)
PART A
1. Reflection: light bounces off a surface (e.g., a mirror). Refraction: light changes direction when it passes from one medium to another because its speed changes (e.g., a straw looks bent in a glass of water).
2. Specular reflection: light reflects from a smooth surface with a single reflected direction per incoming direction (clear images form). Diffuse reflection: light reflects from a rough surface in many directions (no clear image).
3. Law of reflection: the angle of incidence equals the angle of reflection. Formula: θi = θr (angles measured from the normal).
PART B
4. Angle of reflection = 30° (because θi = θr). Sketch: incident ray at 30° to normal; reflected ray on the other side of normal at 30°.
5. If the mirror rotates by 10°, the reflected ray rotates by 20°. Explanation: rotating the mirror by Δ changes the normal by Δ, so the reflected direction (which is symmetric about the normal) changes by 2Δ.
PART C
6. Use Snell: n1 sin θ1 = n2 sin θ2 → 1.00 · sin 40° = 1.33 · sin θ2 → sin θ2 = sin40° / 1.33.
sin40° = 0.6428 → sin θ2 = 0.6428/1.33 = 0.4837 → θ2 = arcsin(0.4837) ≈ 28.9°.
Answer: θ2 ≈ 28.9°.
7. n1 = 1.50 (glass), n2 = 1.00 (air). Compute sin θ2 = (n1/n2) sin θ1 = (1.50/1.00) · sin60° = 1.50 · 0.8660 = 1.299 > 1. Since sin θ2 would be > 1, refraction into air is impossible — total internal reflection occurs. (Alternatively check θ1 vs critical angle from Q8.)
8. Critical angle θc = arcsin(n2/n1) = arcsin(1.00/1.50) = arcsin(0.6667) ≈ 41.8°.
Answer: θc ≈ 41.8°.
9. (a) Inside the glass: sin θglass = (nair/nglass) sin 25° = (1/1.50)·sin25°.
sin25° = 0.4226 → sin θglass = 0.4226/1.50 = 0.2818 → θglass = arcsin(0.2818) ≈ 16.4°.
(b) For a parallel slab, the emergent ray is parallel to the incident ray (though laterally displaced), so the emergent angle in air = 25°.
10. Apparent depth for viewing from air into water at near-normal incidence:
apparent depth = real depth × (n_air / n_water) = 3.00 m × (1.00 / 1.33) = 3.00 / 1.33 ≈ 2.26 m.
Answer: ≈ 2.26 m.
PART D
11. n1 = 1.00 (air), n2 = 1.50 (glass). sin θglass = (1.00/1.50) · sin 50°. sin50° = 0.7660 → sin θglass = 0.7660/1.50 = 0.5107 → θglass = arcsin(0.5107) ≈ 30.7°. Since sin θglass < 1, the ray refracts into the glass at ≈ 30.7°. Sketch: incident ray making 50° with normal in air, refracted ray bent toward the normal in glass at 30.7°.
12. Light inside water to air: critical angle (water → air) = arcsin(1.00/1.33) ≈ 48.8° (from earlier). Since 60° > 48.8°, total internal reflection occurs — the ray is reflected back into the water. Sketch: incident ray at 60° to the normal; reflected ray leaving at 60° on the other side of the normal.
13. Sketch explanation: Draw an object point on the bottom. Draw two rays from the point: one near the normal, one at an angle. At the water–air interface both rays refract away from the normal (bend toward the observer), making them appear to originate from a shallower point along a backward extension. Label real depth d and apparent depth d' (d' < d). The apparent depth can be estimated by d' ≈ d / n for near-normal viewing.
Extension answer (brief)
14. Two perpendicular mirrors produce two reflections that reverse the direction of the original ray (a 180° change); the final ray is antiparallel to the original direction (it goes back along a line parallel to the incoming ray but opposite in direction).
Notes for grading
- Full marks require correct formulas, substitution, numerical work, and sketches where requested.
- Allow small rounding differences (±0.1° or ±0.01 m where appropriate).
If you want, I can convert this into a printable worksheet (PDF-style) or produce diagram images for the sketch questions. Which would you prefer?
Instructions for students
- Show all work and label all angles relative to the normal unless stated otherwise.
- Use nair = 1.00, nwater = 1.33, nglass = 1.50 unless told otherwise.
- Use Snell’s law: n1 sin θ1 = n2 sin θ2.
- Critical angle (for light going from medium 1 into medium 2, with n1 > n2): sin θc = n2 / n1.
PART A — Conceptual (short answers)
1. Define reflection and refraction. Give one everyday example of each.
2. Explain the difference between specular (regular) reflection and diffuse reflection.
3. State the law of reflection in words and write its formula.
PART B — Law of reflection (calculate and sketch)
4. A light ray strikes a plane mirror so that the angle between the incident ray and the normal is 30°. What is the angle between the reflected ray and the normal? Draw a simple sketch showing the incident ray, normal, and reflected ray.
5. A beam hits a plane mirror at 20° to the normal. If the mirror is rotated by 10° about an axis perpendicular to the plane of the paper, by how many degrees does the reflected ray rotate? Explain briefly.
PART C — Snell’s law and critical angle (calculations)
6. Light goes from air into water. If θ1 = 40° (in air), find θ2 in the water. (Give answer to 1 decimal place.)
7. Light in glass (n = 1.50) strikes the glass–air interface at θ1 = 60°. Determine whether the light is refracted into air or totally internally reflected. If TIR occurs, explain why. If refracted, find θ2.
8. Calculate the critical angle for the glass–air interface (nglass = 1.50 to nair = 1.00). Give answer to 1 decimal place.
9. A laser beam in air strikes a glass slab (n = 1.50) at θ = 25°. Find: (a) the angle inside the glass, (b) the angle of the beam when it emerges back into air after passing through a parallel-faced slab.
10. A swimming pool has depth 3.00 m. What depth does the bottom appear to a person viewing from directly above (approximately normal incidence)? (Use nwater = 1.33.)
PART D — Ray diagrams and interpretation (short tasks with sketches/calculations)
11. A ray in air hits a flat piece of glass at point P with incident angle 50° to the normal. Determine whether the ray is refracted into the glass or totally internally reflected. If it is refracted, calculate the angle inside the glass. Sketch the ray path and label angles.
12. A ray inside water strikes the water–air surface with angle 60° to the normal. Will it be refracted into air or totally internally reflected? Explain and sketch the ray(s).
13. Draw a ray diagram (qualitative) showing why an object under water appears closer to the surface when viewed from above. Label the real depth and apparent depth and indicate the refracted rays.
Extension (optional)
14. Two plane mirrors are placed at 90° to each other. A ray hits one mirror and reflects to the second mirror and then to an observer. Using the law of reflection, describe qualitatively the final direction relative to the original ray (hint: two reflections across perpendicular planes reverse direction).
Answer key (worked solutions)
PART A
1. Reflection: light bounces off a surface (e.g., a mirror). Refraction: light changes direction when it passes from one medium to another because its speed changes (e.g., a straw looks bent in a glass of water).
2. Specular reflection: light reflects from a smooth surface with a single reflected direction per incoming direction (clear images form). Diffuse reflection: light reflects from a rough surface in many directions (no clear image).
3. Law of reflection: the angle of incidence equals the angle of reflection. Formula: θi = θr (angles measured from the normal).
PART B
4. Angle of reflection = 30° (because θi = θr). Sketch: incident ray at 30° to normal; reflected ray on the other side of normal at 30°.
5. If the mirror rotates by 10°, the reflected ray rotates by 20°. Explanation: rotating the mirror by Δ changes the normal by Δ, so the reflected direction (which is symmetric about the normal) changes by 2Δ.
PART C
6. Use Snell: n1 sin θ1 = n2 sin θ2 → 1.00 · sin 40° = 1.33 · sin θ2 → sin θ2 = sin40° / 1.33.
sin40° = 0.6428 → sin θ2 = 0.6428/1.33 = 0.4837 → θ2 = arcsin(0.4837) ≈ 28.9°.
Answer: θ2 ≈ 28.9°.
7. n1 = 1.50 (glass), n2 = 1.00 (air). Compute sin θ2 = (n1/n2) sin θ1 = (1.50/1.00) · sin60° = 1.50 · 0.8660 = 1.299 > 1. Since sin θ2 would be > 1, refraction into air is impossible — total internal reflection occurs. (Alternatively check θ1 vs critical angle from Q8.)
8. Critical angle θc = arcsin(n2/n1) = arcsin(1.00/1.50) = arcsin(0.6667) ≈ 41.8°.
Answer: θc ≈ 41.8°.
9. (a) Inside the glass: sin θglass = (nair/nglass) sin 25° = (1/1.50)·sin25°.
sin25° = 0.4226 → sin θglass = 0.4226/1.50 = 0.2818 → θglass = arcsin(0.2818) ≈ 16.4°.
(b) For a parallel slab, the emergent ray is parallel to the incident ray (though laterally displaced), so the emergent angle in air = 25°.
10. Apparent depth for viewing from air into water at near-normal incidence:
apparent depth = real depth × (n_air / n_water) = 3.00 m × (1.00 / 1.33) = 3.00 / 1.33 ≈ 2.26 m.
Answer: ≈ 2.26 m.
PART D
11. n1 = 1.00 (air), n2 = 1.50 (glass). sin θglass = (1.00/1.50) · sin 50°. sin50° = 0.7660 → sin θglass = 0.7660/1.50 = 0.5107 → θglass = arcsin(0.5107) ≈ 30.7°. Since sin θglass < 1, the ray refracts into the glass at ≈ 30.7°. Sketch: incident ray making 50° with normal in air, refracted ray bent toward the normal in glass at 30.7°.
12. Light inside water to air: critical angle (water → air) = arcsin(1.00/1.33) ≈ 48.8° (from earlier). Since 60° > 48.8°, total internal reflection occurs — the ray is reflected back into the water. Sketch: incident ray at 60° to the normal; reflected ray leaving at 60° on the other side of the normal.
13. Sketch explanation: Draw an object point on the bottom. Draw two rays from the point: one near the normal, one at an angle. At the water–air interface both rays refract away from the normal (bend toward the observer), making them appear to originate from a shallower point along a backward extension. Label real depth d and apparent depth d' (d' < d). The apparent depth can be estimated by d' ≈ d / n for near-normal viewing.
Extension answer (brief)
14. Two perpendicular mirrors produce two reflections that reverse the direction of the original ray (a 180° change); the final ray is antiparallel to the original direction (it goes back along a line parallel to the incoming ray but opposite in direction).
Notes for grading
- Full marks require correct formulas, substitution, numerical work, and sketches where requested.
- Allow small rounding differences (±0.1° or ±0.01 m where appropriate).
If you want, I can convert this into a printable worksheet (PDF-style) or produce diagram images for the sketch questions. Which would you prefer?
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