krit.club logo

Wave Behaviour - Wave Phenomena

Grade 12IBPhysics

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

🔑Concepts

•

Wavefronts and Rays: Wavefronts are surfaces of constant phase perpendicular to the direction of energy transfer (rays). The distance between consecutive wavefronts is one wavelength λ\lambda.

•

Reflection and Refraction: At a boundary, the angle of incidence θi\theta_i equals the angle of reflection θr\theta_r. Refraction is governed by Snell's Law, where light changes speed and direction passing between media of different refractive indices nn.

•

Total Internal Reflection: Occurs when a wave travels from a denser medium 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}.

•

Diffraction: The spreading of waves as they pass through an aperture or around an obstacle. It is most significant when the wavelength λ\lambda is comparable to the aperture size bb. For a single slit, the first minimum occurs at θ=λb\theta = \frac{\lambda}{b}.

•

Superposition and Interference: When two waves meet, the resultant displacement is the vector sum of individual displacements. Constructive interference occurs when path difference ΔL=nλ\Delta L = n\lambda, and destructive when ΔL=(n+12)λ\Delta L = (n + \frac{1}{2})\lambda.

•

Young's Double Slit: Produces an interference pattern of equally spaced bright and dark fringes. The fringe spacing ss is proportional to the distance to the screen DD and wavelength λ\lambda, and inversely proportional to slit separation dd.

•

Polarization: Transverse waves can be polarized. Malus's Law states that the intensity II of polarized light passing through an analyzer is I=I0cos⁡2θI = I_0 \cos^2 \theta, where θ\theta is the angle between the polarizer and analyzer axes.

•

Standing Waves: Formed by the superposition of two counter-propagating waves of the same frequency and amplitude. They feature nodes (zero displacement) and antinodes (maximum displacement), and do not transfer energy.

📐Formulae

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

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

s=λDds = \frac{\lambda D}{d}

θ=λb\theta = \frac{\lambda}{b}

dsin⁡θ=nλd \sin \theta = n \lambda

I=I0cos⁡2θI = I_0 \cos^2 \theta

θ≈1.22λb\theta \approx 1.22 \frac{\lambda}{b}

💡Examples

Problem 1:

Monochromatic light of wavelength λ=633 nm\lambda = 633 \text{ nm} is incident on a double slit with a separation d=0.50 mmd = 0.50 \text{ mm}. If the interference pattern is observed on a screen D=2.0 mD = 2.0 \text{ m} away, calculate the distance between adjacent bright fringes.

Solution:

s=λDd=(633×10−9 m)(2.0 m)0.50×10−3 m=2.53×10−3 ms = \frac{\lambda D}{d} = \frac{(633 \times 10^{-9} \text{ m})(2.0 \text{ m})}{0.50 \times 10^{-3} \text{ m}} = 2.53 \times 10^{-3} \text{ m} or 2.53 mm2.53 \text{ mm}.

Explanation:

We use the fringe spacing formula for Young's double-slit experiment. Convert all units to meters (SI) before calculation to ensure the result is in meters.

Problem 2:

Unpolarized light of intensity IinI_{in} is incident on a polarizer. The resulting polarized light then passes through an analyzer whose transmission axis is at an angle of 30∘30^\circ to the polarizer. Calculate the final intensity II in terms of IinI_{in}.

Solution:

Step 1: After the first polarizer, I0=12IinI_0 = \frac{1}{2} I_{in}. Step 2: Apply Malus's Law: I=I0cos⁡2(30∘)=12Iin×(32)2=12Iin×34=38Iin=0.375IinI = I_0 \cos^2(30^\circ) = \frac{1}{2} I_{in} \times (\frac{\sqrt{3}}{2})^2 = \frac{1}{2} I_{in} \times \frac{3}{4} = \frac{3}{8} I_{in} = 0.375 I_{in}.

Explanation:

An ideal polarizer reduces the intensity of unpolarized light by half. The second stage follows Malus's Law based on the relative angle between the two polarizing filters.

Problem 3:

Calculate the critical angle θc\theta_c for a boundary between glass (n=1.52n = 1.52) and air (n=1.00n = 1.00).

Solution:

sin⁡θc=nairnglass=1.001.52≈0.6579\sin \theta_c = \frac{n_{air}}{n_{glass}} = \frac{1.00}{1.52} \approx 0.6579. Therefore, θc=arcsin⁡(0.6579)≈41.1∘\theta_c = \arcsin(0.6579) \approx 41.1^\circ.

Explanation:

Total internal reflection occurs only when moving from a high refractive index to a lower one. The critical angle is found when the angle of refraction is 90∘90^\circ.

Wave Phenomena - Revision Notes & Key Formulas | IB Grade 12 Physics