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Wave Behaviour - Wave Phenomena (Reflection, Refraction, Diffraction)

Grade 11IBPhysics

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

🔑Concepts

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Reflection: The change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. The Law of Reflection states θi=θr\theta_i = \theta_r.

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Phase Change in Reflection: When a wave reflects off a more dense medium (fixed boundary), it undergoes a phase change of π\pi radians (180∘180^\circ). No phase change occurs at a less dense medium (free boundary).

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Refraction: The change in direction of a wave passing from one medium to another, caused by its change in speed vv. The frequency ff remains constant during refraction.

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Refractive Index (nn): A dimensionless number that describes how fast light travels through a material. It is defined as n=cvn = \frac{c}{v}, where cc is the speed of light in a vacuum (3.00×108 m s−13.00 \times 10^8 \text{ m s}^{-1}).

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Snell's Law: Relates the angles of incidence θ1\theta_1 and refraction θ2\theta_2 to the refractive indices of the two media: n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2.

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Critical Angle (θc\theta_c): The angle of incidence that provides an angle of refraction of 90∘90^\circ. Total Internal Reflection (TIR) occurs when θi>θc\theta_i > \theta_c and the wave travels from a more dense to a less dense medium (n1>n2n_1 > n_2).

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Diffraction: The spreading of a wave as it passes through an aperture or around an edge. Diffraction effects are most significant when the wavelength λ\lambda is approximately equal to the size of the opening bb (i.e., λ≈b\lambda \approx b).

📐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

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}

💡Examples

Problem 1:

A light ray in air (n1=1.00n_1 = 1.00) strikes a glass block (n2=1.52n_2 = 1.52) at an angle of incidence of 40∘40^\circ. Calculate the angle of refraction within the glass.

Solution:

1.00×sin⁡(40∘)=1.52×sin⁡θ21.00 \times \sin(40^\circ) = 1.52 \times \sin \theta_2 sin⁡θ2=sin⁡(40∘)1.52≈0.64281.52≈0.4229\sin \theta_2 = \frac{\sin(40^\circ)}{1.52} \approx \frac{0.6428}{1.52} \approx 0.4229 θ2=arcsin⁡(0.4229)≈25.0∘\theta_2 = \arcsin(0.4229) \approx 25.0^\circ

Explanation:

Using Snell's Law, we substitute the known refractive indices and the incident angle to find the sine of the refractive angle, then take the inverse sine.

Problem 2:

Calculate the critical angle for light traveling from crown glass (n=1.52n = 1.52) into water (n=1.33n = 1.33).

Solution:

sin⁡θc=n2n1=1.331.52≈0.875\sin \theta_c = \frac{n_2}{n_1} = \frac{1.33}{1.52} \approx 0.875 θc=arcsin⁡(0.875)≈61.0∘\theta_c = \arcsin(0.875) \approx 61.0^\circ

Explanation:

The critical angle occurs when the angle of refraction is 90∘90^\circ. Since sin⁡(90∘)=1\sin(90^\circ) = 1, Snell's Law simplifies to sin⁡θc=n2n1\sin \theta_c = \frac{n_2}{n_1}.

Problem 3:

A sound wave of frequency 440 Hz440 \text{ Hz} passes from air where the speed is 340 m s−1340 \text{ m s}^{-1} into a wall where the speed is 1500 m s−11500 \text{ m s}^{-1}. Calculate the wavelength in the wall.

Solution:

f=440 Hz (remains constant)f = 440 \text{ Hz} \text{ (remains constant)} λwall=vwallf=1500440≈3.41 m\lambda_{wall} = \frac{v_{wall}}{f} = \frac{1500}{440} \approx 3.41 \text{ m}

Explanation:

When a wave is refracted, its frequency does not change. We use the wave equation v=fλv = f\lambda to find the new wavelength in the second medium.

Wave Phenomena (Reflection, Refraction, Diffraction) Revision - Grade 11 Physics IB