Reflection, Refraction & Optics — Drills · Grade 12 Science

Chapter: Waves, Sound & Optics

What these drills covers

By the end of Reflection, Refraction & Optics you should be able to:

6 questions, multiple choice, numeric and written answer, each with a worked explanation.

Vocabulary used: reflection, normal, refraction, index of refraction (n), total internal reflection, focal length (f), real vs. virtual image.

Problem 1

According to the law of reflection, the angle of incidence always equals the angle of…
  1. refraction
  2. reflection
  3. the normal
  4. incidence doubled
Answer and explanation

reflection

The angle of incidence equals the angle of reflection, both measured from the normal.

Problem 2

Light travels at 1.5 × 10⁸ m/s inside a crystal. Using c = 3 × 10⁸ m/s, find its index of refraction.
Answer and explanation

2

n = c ÷ v = (3 × 10⁸) ÷ (1.5 × 10⁸) = 2.0.

Problem 3

Total internal reflection can occur only when light travels…
  1. from air into glass
  2. between any two media at any angle
  3. from a denser to a lower-index medium beyond the critical angle
  4. from a less dense to a higher-index medium
Answer and explanation

from a denser to a lower-index medium beyond the critical angle

Total internal reflection requires light going from a higher-index (denser) medium toward a lower-index one, at an angle greater than the critical angle.

Problem 4

An object is placed 60 cm from a converging lens with a focal length of 20 cm. Find the image distance.
Answer and explanation

30 cm

1 ÷ dᵢ = 1 ÷ f − 1 ÷ dₒ = 1 ÷ 20 − 1 ÷ 60 = 3 ÷ 60 − 1 ÷ 60 = 2 ÷ 60 = 1 ÷ 30, so dᵢ = 30 cm.

Problem 5

Explain how a single converging (convex) lens can form either a real image or a virtual image, depending on where the object is placed.
Answer and explanation

When the object is farther from the lens than the focal length, the rays converge on the far side and form a real, inverted image that can be projected on a screen (as in a camera or the eye). When the object is closer than the focal length, the rays diverge after the lens and only appear to come from a point behind it, forming an upright, enlarged virtual image — the way a magnifying glass works.

Problem 6

An object 3 cm tall is placed 15 cm from a lens, and its real image forms 45 cm away on the other side. How tall is the image? (magnification magnitude = image distance ÷ object distance)
Answer and explanation

9 cm

Magnification magnitude = dᵢ ÷ dₒ = 45 ÷ 15 = 3, so image height = 3 × 3 cm = 9 cm.

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