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Wave Behaviour - The Wave Model

Grade 11IBPhysics

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

🔑Concepts

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Wavefronts and Rays: A wavefront is a surface joining points of the same phase. Rays are lines drawn perpendicular to wavefronts indicating the direction of energy transfer.

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Huygens' Principle: Every point on a wavefront acts as a source of secondary spherical wavelets that spread out in the forward direction at the speed of the wave. The new wavefront is the surface tangent to these wavelets.

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Reflection: The angle of incidence θi\theta_i is equal to the angle of reflection θr\theta_r. Both angles are measured relative to the normal.

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Refraction and Snell's Law: When a wave passes from one medium to another, its speed and wavelength change, causing it to bend. The frequency ff remains constant.

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Critical Angle and Total Internal Reflection (TIR): TIR occurs when a wave travels from a denser medium to a less dense medium and the angle of incidence exceeds the critical angle θc\theta_c.

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Superposition Principle: The resultant displacement of two or more interfering waves is the algebraic sum of their individual displacements at that point.

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Interference: Constructive interference occurs when path difference is an integer multiple of the wavelength (nλn\lambda), while destructive interference occurs at half-integer multiples ((n+12)λ(n + \frac{1}{2})\lambda).

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Diffraction: The spreading of waves as they pass through an aperture or around an edge. Significant diffraction occurs when the wavelength λ\lambda is comparable to the aperture size bb.

📐Formulae

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

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

n1n2=v2v1=λ2λ1\frac{n_1}{n_2} = \frac{v_2}{v_1} = \frac{\lambda_2}{\lambda_1}

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

s=λDds = \frac{\lambda D}{d}

ϕ=2πΔxλ\phi = \frac{2\pi \Delta x}{\lambda}

θ≈λb\theta \approx \frac{\lambda}{b}

💡Examples

Problem 1:

Light of wavelength 633 nm633\text{ nm} in air enters a glass block with a refractive index of 1.521.52. Calculate the speed and wavelength of the light inside the glass.

Solution:

v=cn=3.00×1081.52≈1.97×108 m s−1v = \frac{c}{n} = \frac{3.00 \times 10^8}{1.52} \approx 1.97 \times 10^8 \text{ m s}^{-1} λglass=λairn=633×10−91.52≈4.16×10−7 m=416 nm\lambda_{glass} = \frac{\lambda_{air}}{n} = \frac{633 \times 10^{-9}}{1.52} \approx 4.16 \times 10^{-7} \text{ m} = 416 \text{ nm}

Explanation:

The frequency of light remains constant across media. Using the definition of refractive index n=cvn = \frac{c}{v} and the wave equation v=fλv = f\lambda, the speed and wavelength decrease proportionally in the denser medium.

Problem 2:

In a Young’s double-slit experiment, the slit separation is 0.15 mm0.15\text{ mm} and the screen is placed 2.0 m2.0\text{ m} away. If the distance between the central maximum and the third-order bright fringe is 2.4 cm2.4\text{ cm}, find the wavelength of the light.

Solution:

s=2.4×10−23=8.0×10−3 ms = \frac{2.4 \times 10^{-2}}{3} = 8.0 \times 10^{-3} \text{ m} λ=s⋅dD=(8.0×10−3)(0.15×10−3)2.0=6.0×10−7 m\lambda = \frac{s \cdot d}{D} = \frac{(8.0 \times 10^{-3})(0.15 \times 10^{-3})}{2.0} = 6.0 \times 10^{-7} \text{ m}

Explanation:

First, calculate the fringe spacing ss by dividing the total distance to the third fringe by 33. Then, rearrange the double-slit formula s=λDds = \frac{\lambda D}{d} to solve for λ\lambda. The result is 600 nm600\text{ nm}.