Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Define and apply the laws of reflection and refraction.
• Calculate the refractive index of a material using Snell's Law.
• Sketch ray diagrams for lenses.
• Explain the conditions and applications of total internal reflection.
• Solve quantitative problems involving these concepts.

Lesson summary

This week, we delve into the fascinating world of geometrical optics, a branch of physics that deals with the behavior of light as it travels in straight lines (rays). Understanding geometrical optics is crucial because it explains how lenses and mirrors work, which are essential components in many devices we use daily. From eyeglasses and cameras to telescopes and microscopes, geometrical optics underpins a large part of our technological landscape.

Lesson notes

2.1 Reflection Reflection occurs when light bounces off a surface.

The law of reflection states: The angle of incidence (θᵢ) is equal to the angle of reflection (θᵣ). Both angles are measured relative to the normal, which is a line perpendicular to the reflecting surface at the point where the light ray hits the surface. The incident ray, the reflected ray, and the normal all lie in the same plane. Mirrors are designed to reflect light efficiently. Plane mirrors produce virtual, upright, and same-size images. Concave mirrors can produce real or virtual images depending on the object's position relative to the focal point, while convex mirrors always produce virtual, upright, and diminished images.

Example 1: A ray of light strikes a plane mirror at an angle of incidence of 30°. What is the angle of reflection?

Solution: According to the law of reflection, the angle of incidence equals the angle of reflection.

Therefore, the angle of reflection is also 30°. 2.2 Refraction Refraction is the bending of light as it passes from one medium to another (e.g., from air to water). This bending occurs because the speed of light changes in different media. The refractive index (n) of a medium is a measure of how much the speed of light is reduced in that medium compared to its speed in a vacuum (c = 3 x 10⁸ m/s). n = c / v where v is the speed of light in the medium. Snell's Law describes the relationship between the angles of incidence and refraction at the interface between two media: n₁sinθ₁ = n₂sinθ₂ where: n₁ is the refractive index of the first medium. θ₁ is the angle of incidence in the first medium. n₂ is the refractive index of the second medium. θ₂ is the angle of refraction in the second medium.

Example 2: A ray of light travels from air (n₁ = 1.00) into water (n₂ = 1.33) at an angle of incidence of 45°. What is the angle of refraction?

Solution: Apply Snell's Law: n₁sinθ₁ = n₂sinθ₂ Substitute the given values: (1.00)sin(45°) = (1.33)sin(θ₂) Solve for sin(θ₂): sin(θ₂) = (sin(45°)) / 1.33 ≈ 0.532 Find θ₂: θ₂ = arcsin(0.532) ≈ 32.1° Therefore, the angle of refraction is approximately 32.1°. 2.3 Lenses Lenses are transparent objects that refract light. Convex (converging) lenses are thicker in the middle and converge parallel light rays to a focal point. Concave (diverging) lenses are thinner in the middle and diverge parallel light rays. The focal length (f) is the distance from the lens to the focal point. The lens equation relates the object distance (u), image distance (v), and focal length (f): 1/f = 1/u + 1/v Magnification (M) is the ratio of the image height (hᵢ) to the object height (h₀): M = hᵢ/h₀ = -v/u (The negative sign indicates an inverted image).

Example 3: An object is placed 20 cm from a convex lens with a focal length of 10 cm. Determine the image distance and magnification.

Solution: Apply the lens equation: 1/f = 1/u + 1/v Substitute the given values: 1/10 = 1/20 + 1/v Solve for 1/v: 1/v = 1/10 - 1/20 = 1/20 Find v: v = 20 cm The image distance is 20 cm.

Now calculate the magnification: M = -v/u = -20/20 = -1 The magnification is -1, indicating that the image is the same size as the object and inverted. 2.4 Total Internal Reflection (TIR) Total internal reflection occurs when light traveling from a medium with a higher refractive index (n₁) to a medium with a lower refractive index (n₂) is incident at an angle greater than the critical angle (θc). At the critical angle, the angle of refraction is 90°. For angles greater than θc, all the light is reflected back into the original medium. The critical angle can be calculated using: sinθc = n₂/n₁ Conditions for TIR: Light must travel from a medium of higher refractive index to a medium of lower refractive index. The angle of incidence must be greater than the critical angle.

Applications of TIR: Optical fibers (used in telecommunications and medical endoscopes) utilize TIR to transmit light over long distances with minimal loss. In South Africa, fibre optic cables are crucial for improving internet connectivity, particularly in rural areas.

Example 4: Calculate the critical angle for light traveling from water (n = 1.33) to air (n = 1.00).

Solution: Apply the formula: sinθc = n₂/n₁ Substitute the values: sinθc = 1.00 / 1.33 ≈ 0.752 Find θc: θc = arcsin(0.752) ≈ 48.8° The critical angle is approximately 48.8°. Guided Practice (With Solutions)

Question 1: A ray of light is incident on a glass block (n = 1.50) at an angle of 60° to the normal. Calculate the angle of refraction inside the glass block.

Solution: Apply Snell's Law: n₁sinθ₁ = n₂sinθ₂ Here, n₁ = 1 (air), θ₁ = 60°, n₂ = 1.50. (1)sin(60°) = (1.50)sin(θ₂) sin(θ₂) = sin(60°) / 1.50 ≈ 0.577 θ₂ = arcsin(0.577) ≈ 35.3° The angle of refraction is approximately 35.3°.

Commentary: This question tests the direct application of Snell's Law. Pay careful attention to which medium has which refractive index and which angle is being used.

Question 2: A convex lens has a focal length of 15 cm.