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Waves - Reflection and Refraction

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Reflection: When a wave strikes a boundary and bounces back into the same medium. The Law of Reflection states that the angle of incidence θi\theta_i is equal to the angle of reflection θr\theta_r.

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Refraction: The change in direction of a wave as it passes from one medium to another. This is caused by a change in the wave's speed vv and wavelength λ\lambda, while the frequency ff remains constant.

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Refractive Index (nn): A dimensionless number that describes how fast light travels in a medium relative to a vacuum. It is defined as n=cvn = \frac{c}{v}, where c≈3.00×108 m/sc \approx 3.00 \times 10^8 \text{ m/s}.

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Snell's Law: Describes the relationship between the angles of incidence and refraction for a wave crossing the boundary between two different isotropic media: n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2.

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Total Internal Reflection (TIR): Occurs when light travels from a denser medium (n1n_1) to a less dense medium (n2n_2) at an angle of incidence greater than the critical angle θc\theta_c.

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Critical Angle (θc\theta_c): The specific angle of incidence that results in an angle of refraction of 90∘90^\circ. It can be calculated as sin⁡θc=n2n1\sin \theta_c = \frac{n_2}{n_1}.

📐Formulae

θi=θr\theta_i = \theta_r

n=cvn = \frac{c}{v}

n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2

n=λvacuumλmediumn = \frac{\lambda_{\text{vacuum}}}{\lambda_{\text{medium}}}

sin⁡θc=n2n1\sin \theta_c = \frac{n_2}{n_1}

💡Examples

Problem 1:

A ray of light travels from air (n1=1.00n_1 = 1.00) into a glass block (n2=1.50n_2 = 1.50) at an angle of incidence of 30∘30^\circ. Calculate the angle of refraction.

Solution:

Using Snell's Law: n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2 1.00⋅sin⁡(30∘)=1.50⋅sin⁡θ21.00 \cdot \sin(30^\circ) = 1.50 \cdot \sin \theta_2 0.50=1.50⋅sin⁡θ20.50 = 1.50 \cdot \sin \theta_2 sin⁡θ2=0.501.50=0.333\sin \theta_2 = \frac{0.50}{1.50} = 0.333 θ2=arcsin⁡(0.333)≈19.47∘\theta_2 = \arcsin(0.333) \approx 19.47^\circ

Explanation:

Since the light moves from a less dense medium (air) to a more dense medium (glass), the ray bends toward the normal, resulting in an angle of refraction smaller than the angle of incidence.

Problem 2:

Calculate the speed of light in a diamond which has a refractive index of 2.422.42. (Take c=3.00×108 m/sc = 3.00 \times 10^8 \text{ m/s})

Solution:

Using the formula for refractive index: n=cvn = \frac{c}{v} 2.42=3.00×108 m/sv2.42 = \frac{3.00 \times 10^8 \text{ m/s}}{v} v=3.00×108 m/s2.42v = \frac{3.00 \times 10^8 \text{ m/s}}{2.42} v≈1.24×108 m/sv \approx 1.24 \times 10^8 \text{ m/s}

Explanation:

The high refractive index of diamond causes the speed of light to slow down significantly to approximately 1.24×108 m/s1.24 \times 10^8 \text{ m/s}.

Problem 3:

Determine the critical angle for light traveling from water (n=1.33n = 1.33) into air (n=1.00n = 1.00).

Solution:

sin⁡θc=n2n1\sin \theta_c = \frac{n_2}{n_1} sin⁡θc=1.001.33\sin \theta_c = \frac{1.00}{1.33} sin⁡θc≈0.7519\sin \theta_c \approx 0.7519 θc=arcsin⁡(0.7519)≈48.75∘\theta_c = \arcsin(0.7519) \approx 48.75^\circ

Explanation:

Any light ray hitting the water-air interface from the water side at an angle greater than 48.75∘48.75^\circ will undergo total internal reflection and will not escape into the air.