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Physics - Waves (Light, Sound, and Electromagnetic Spectrum)

Grade 10IB

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

🔑Concepts

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Waves are oscillations that transfer energy and information from one point to another without the transfer of matter.

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Transverse waves oscillate perpendicular to the direction of energy transfer (e.g., light, water waves), while longitudinal waves oscillate parallel to the direction of energy transfer (e.g., sound).

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Key wave properties include amplitude (AA), wavelength (λ\lambda), frequency (ff), and period (TT), where T=1fT = \frac{1}{f}.

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The Electromagnetic (EM) Spectrum consists of transverse waves that travel at the speed of light (c≈3.00×108 m s−1c \approx 3.00 \times 10^8 \text{ m s}^{-1}) in a vacuum. In order of increasing frequency: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, and Gamma rays.

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The Law of Reflection states that the angle of incidence (θi\theta_i) is equal to the angle of reflection (θr\theta_r) relative to the normal.

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Refraction occurs when a wave changes speed and direction as it enters a medium with a different optical density. This is described by Snell's Law: 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 to a less dense medium at an angle of incidence greater than the critical angle (θc\theta_c), where sin⁡θc=n2n1\sin \theta_c = \frac{n_2}{n_1}.

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Sound is a longitudinal mechanical wave requiring a medium. Its pitch depends on frequency, and its loudness depends on the square of the amplitude (A2A^2).

📐Formulae

v=fλv = f \lambda

T=1fT = \frac{1}{f}

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

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

sin⁡θc=1n\sin \theta_c = \frac{1}{n}

v=2dt (for echo calculations)v = \frac{2d}{t} \text{ (for echo calculations)}

💡Examples

Problem 1:

A light ray travels from air (n1=1.00n_1 = 1.00) into a glass block with a refractive index of n2=1.50n_2 = 1.50. If the angle of incidence is 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. Substituting values: 1.00×sin⁡(30∘)=1.50×sin⁡θ21.00 \times \sin(30^\circ) = 1.50 \times \sin \theta_2. Since sin⁡(30∘)=0.5\sin(30^\circ) = 0.5, we have 0.5=1.50sin⁡θ20.5 = 1.50 \sin \theta_2. Thus, sin⁡θ2=0.51.5=0.333\sin \theta_2 = \frac{0.5}{1.5} = 0.333. Therefore, θ2=arcsin⁡(0.333)≈19.5∘\theta_2 = \arcsin(0.333) \approx 19.5^\circ.

Explanation:

Light slows down as it enters the glass (1.50>1.001.50 > 1.00), causing it to bend toward the normal, resulting in a smaller angle of refraction compared to the angle of incidence.

Problem 2:

A sound wave has a frequency of 440 Hz440 \text{ Hz} and travels at 340 m s−1340 \text{ m s}^{-1} in air. Determine its wavelength.

Solution:

Use the wave equation v=fλv = f \lambda. Rearranging for wavelength: λ=vf\lambda = \frac{v}{f}. Substituting the values: λ=340440≈0.773 m\lambda = \frac{340}{440} \approx 0.773 \text{ m}.

Explanation:

The wavelength is the distance between consecutive compressions or rarefactions in the longitudinal sound wave.

Problem 3:

A ship uses sonar to find the depth of the ocean. It sends a sound pulse that reflects off the seabed and returns to the ship 1.2 s1.2 \text{ s} later. If the speed of sound in water is 1500 m s−11500 \text{ m s}^{-1}, find the depth.

Solution:

The total distance traveled by the pulse is dtotal=v×t=1500×1.2=1800 md_{total} = v \times t = 1500 \times 1.2 = 1800 \text{ m}. Since the pulse travels to the bottom and back, the depth is half the total distance: depth=18002=900 m\text{depth} = \frac{1800}{2} = 900 \text{ m}.

Explanation:

Echo calculations must account for the 'there and back' nature of the sound signal by dividing the total distance by 2.

Waves (Light, Sound, and Electromagnetic Spectrum) Revision - Grade 10 Science IB