These twelve questions cover reflection, refraction, total internal reflection and lens images from light and imaging. All data is invented for practice and is not taken from any exam paper. They go from easier to harder.
Draw every diagram on paper with a sharp pencil and a ruler, then work out your answer, before you open the worked answer. Use a calculator in degree mode. Some questions use the refractive index equation and the critical angle equation, which may apply only to the Extended route, so check the Cambridge page for your exam year.
Questions
1. A ray of light hits a plane mirror with an angle of incidence of 25°. State the angle of reflection and the angle between the incident and reflected rays.
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The angle of reflection equals the angle of incidence, so it is 25°. The angle between the rays is 25° + 25° = 50°.
2. A ray makes an angle of 30° with the surface of a plane mirror. Find the angle of reflection and the angle between the incident and reflected rays.
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Angles are measured from the normal. Angle of incidence = 90° − 30° = 60°. The angle of reflection is 60°. The angle between the rays is 60° + 60° = 120°.
3. A student stands 1.8 m from a plane mirror. How far is the student from the image? The student then steps back 0.4 m. How far is the student from the image now?
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The image is as far behind the mirror as the student is in front. At first, the distance is 1.8 + 1.8 = 3.6 m. After stepping back, the student is 1.8 + 0.4 = 2.2 m from the mirror, so the distance to the image is 2.2 + 2.2 = 4.4 m.
4. A ray of light in air hits a plastic block with i = 60° and r = 35°. Calculate the refractive index of the plastic.
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n = sin i ÷ sin r = sin 60° ÷ sin 35° = 0.8660 ÷ 0.5736 = 1.510. The refractive index is 1.5 (to 2 significant figures).
5. A ray in air hits a glass block of refractive index 1.5 with an angle of incidence of 40°. Find the angle of refraction and state which way the ray bends.
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sin r = sin i ÷ n = sin 40° ÷ 1.5 = 0.6428 ÷ 1.5 = 0.4285. r = sin⁻¹(0.4285) = 25.4°, so r ≈ 25°. The ray bends toward the normal, as the angle in the glass (25°) is smaller than the angle in air (40°).
6. A material has a refractive index of 2.4. Calculate its critical angle.
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sin c = 1 ÷ n = 1 ÷ 2.4 = 0.4167. c = sin⁻¹(0.4167) = 24.6°, so c ≈ 25°.
7. A ray inside glass (n = 1.5, critical angle 41.8°) meets the glass-to-air boundary with an angle of incidence of 38°. Does total internal reflection occur? Find the angle of refraction in the air.
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38° is less than 41.8°, so there is no total internal reflection. sin r = n × sin i = 1.5 × sin 38° = 1.5 × 0.6157 = 0.9235. r = sin⁻¹(0.9235) = 67.4°, so the ray emerges at r ≈ 67° from the normal, bending away from it.
8. An object 3.0 cm tall is 18.0 cm from a converging lens of focal length 6.0 cm. Draw the image with two rays. Find its distance from the lens and its height, and describe it.
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The object is beyond 2f (2f = 12 cm). Ray 1 (parallel to the axis) bends through the far F. Ray 2 (through the optical centre) goes straight on. They meet 9.0 cm from the lens on the far side, 1.5 cm below the axis. Magnification = 1.5 ÷ 3.0 = 0.5. The image is real, inverted and diminished, between f and 2f.
Check with the lens equation for interest: 1/v = 1/6 − 1/18 = 2/18, so v = 9 cm. This matches the drawing.
9. An object 2.0 cm tall is 9.0 cm from a converging lens of focal length 6.0 cm. Find the image distance and height from a scale drawing, and describe the image.
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The object is between f and 2f. The two standard rays meet 18 cm from the lens, on the far side, and the image is 4.0 cm tall, which is 2 times the object height (magnification = 18 ÷ 9 = 2). The image is real, inverted and magnified.
Check: 1/v = 1/6 − 1/9 = 3/18 − 2/18 = 1/18, so v = 18 cm.
10. An object 1.5 cm tall is 5.0 cm from a converging lens of focal length 8.0 cm. Describe the image, say where it is, and find its height.
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The object is inside f (5 cm is less than 8 cm). The two rays diverge after the lens, so you extend them back with dashed lines. They meet about 13 cm from the lens, on the same side as the object. The magnification is 13.3 ÷ 5.0 ≈ 2.7, so the height is about 1.5 × 2.7 = 4.0 cm. The image is virtual, upright and magnified.
Check: 1/v = 1/8 − 1/5 = 5/40 − 8/40 = −3/40, so v = −13.3 cm (negative means virtual, same side).
11. The glass core of an optical fibre has a refractive index of 1.5 and is surrounded by air. A ray inside the core meets the wall with an angle of incidence of 50°. Does the ray stay in the fibre? Explain.
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The critical angle is sin⁻¹(1 ÷ 1.5) = 41.8°. The ray is in the denser material heading for a less dense one, and 50° is greater than 41.8°. Both conditions hold, so the ray undergoes total internal reflection and stays in the fibre, reflecting at 50° from the normal.
12. A student says, “The image in the magnifying glass is real because I can see it, and it is bigger than the object.” Explain what is wrong and what should be said instead.
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Seeing an image with your eye does not show it is real, and being bigger does not either. A magnifying glass forms a virtual image: the object is inside the focal length, the rays spread out after the lens, and they only appear to come from a point behind the object. No screen could show the image. The correct statement is that the image is virtual, upright and magnified, on the same side as the object.
If you got these wrong
Match each slip to the lesson that repairs it.
- Wrong angle in reflection (questions 1 to 3): go to constructing a reflected ray and re-check that every angle is measured from the normal.
- Wrong refractive index or angle (questions 4 and 5): go to using a refraction relationship and check that you used sines, not the angles.
- Wrong critical angle or total internal reflection decision (questions 6, 7 and 11): go to total internal reflection conditions and check both conditions in order.
- Wrong image position or size from rays (questions 8 to 10): go to drawing a simple lens image and redraw with a ruler.
- Wrong word, real or virtual (questions 10 and 12): go to real and virtual image construction and test the rays, not the size.
Afterwards, write the slip in the mistake log and retest queue so it returns later as a fresh question.
If one type of question keeps failing even after you revisit the lesson, a teacher watching the diagram as you draw it can often see the cause quickly. That is what our online one-to-one Physics tuition is designed for.