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Wave Behaviour - Wave model

Grade 12IBPhysics

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 that are in phase. Rays are lines perpendicular to the wavefronts that indicate the direction of energy transfer.

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The Principle of Superposition: When two or more waves of the same type arrive at the same point, the resultant displacement is the algebraic sum of the individual displacements: y=y1+y2y = y_1 + y_2.

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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 with respect to the normal to the surface.

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Refraction: The change in direction of a wave as it transmits from one medium to another due to a change in its phase speed vv. The frequency ff remains constant.

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Snell's Law: Relates the angles of incidence and refraction to the refractive indices or wave speeds: n1n2=sin⁡θ2sin⁡θ1=v2v1\frac{n_1}{n_2} = \frac{\sin \theta_2}{\sin \theta_1} = \frac{v_2}{v_1}.

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Diffraction: The spreading of a wave as it passes through an aperture or around an obstacle. It is most significant when the wavelength λ\lambda is approximately equal to the size of the opening bb.

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Interference: The result of superposition. Constructive interference occurs when waves are in phase (path difference =nλ= n\lambda). Destructive interference occurs when waves are 180∘180^{\circ} out of phase (path difference =(n+12)λ= (n + \frac{1}{2})\lambda).

📐Formulae

v=fλv = f \lambda

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}

Path Difference=nλ (Constructive)\text{Path Difference} = n\lambda \text{ (Constructive)}

Path Difference=(n+12)λ (Destructive)\text{Path Difference} = (n + \frac{1}{2})\lambda \text{ (Destructive)}

💡Examples

Problem 1:

A light ray travels from air (n1=1.00n_1 = 1.00) into a diamond (n2=2.42n_2 = 2.42) at an angle of incidence of 30∘30^{\circ}. Calculate the angle of refraction θ2\theta_2.

Solution:

n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2 1.00×sin⁡(30∘)=2.42×sin⁡θ21.00 \times \sin(30^{\circ}) = 2.42 \times \sin \theta_2 sin⁡θ2=1.00×0.52.42\sin \theta_2 = \frac{1.00 \times 0.5}{2.42} sin⁡θ2≈0.2066\sin \theta_2 \approx 0.2066 θ2=arcsin⁡(0.2066)≈11.9∘\theta_2 = \arcsin(0.2066) \approx 11.9^{\circ}

Explanation:

We apply Snell's Law. Since the refractive index of diamond is much higher than air, the light ray slows down significantly and bends toward the normal, resulting in a smaller angle.

Problem 2:

Two coherent sources S1S_1 and S2S_2 produce waves with a wavelength λ=4.0 cm\lambda = 4.0\text{ cm}. A point PP is 22.0 cm22.0\text{ cm} from S1S_1 and 30.0 cm30.0\text{ cm} from S2S_2. Determine if the interference at point PP is constructive or destructive.

Solution:

Path Difference (ΔL)=∣S2P−S1P∣\text{Path Difference } (\Delta L) = |S_2 P - S_1 P| ΔL=30.0 cm−22.0 cm=8.0 cm\Delta L = 30.0\text{ cm} - 22.0\text{ cm} = 8.0\text{ cm} Ratio ΔLλ=8.0 cm4.0 cm=2\text{Ratio } \frac{\Delta L}{\lambda} = \frac{8.0\text{ cm}}{4.0\text{ cm}} = 2 ΔL=2λ\Delta L = 2\lambda

Explanation:

Since the path difference is an integer multiple of the wavelength (n=2n=2), the waves arrive at point PP in phase, resulting in constructive interference.