light interference diffraction and polarization

Light is an electromagnetic wave, so wave laws apply to it: interference arises when coherent waves superpose, and diffraction arises when the wavefront is limited. In Young's experiment the path difference $\Delta = d\sin\theta$ gives maxima when $\Delta = m\lambda$ and minima when $\Delta = (m+\tfrac12)\lambda$. For a diffraction grating, the condition for principal maxima $d\sin\theta = m\lambda$ gives the angles at which different wavelengths are separated. Single-slit diffraction is described by minima $a\sin\theta = m\lambda$, where $a$ is the slit width. Polarization reveals the transverse nature of light waves: after a polarizer, the intensity obeys Malus's law $I = I_0\cos^2\theta$ if the incident light is linearly polarized. These phenomena share a common wave language: amplitude, phase difference, and projection of the electric-field vector.

In Young's experiment, light with wavelength $\lambda = 600$ nm falls on two slits separated by $d = 0.20$ mm. The screen is placed at a distance $L = 2.0$ m. Find the fringe spacing, that is, the distance between adjacent bright fringes.

Unpolarized light of intensity $I_0$ passes through two polarizers. The axis of the first is vertical, and the axis of the second makes an angle $60^\circ$ with the vertical. Find the intensity of the light after the second polarizer.

Linearly polarized light with wavelength $\lambda = 600$ nm illuminates two slits separated by $d = 0.15$ mm. The screen is $L = 1.5$ m from the slits. An analyzer is placed in front of the screen, and its axis makes an angle $30^\circ$ with the polarization direction of the incident light. Find the distance from the central maximum to the first side maximum and the factor by which the intensity at this maximum is smaller than the intensity at the same maximum without the analyzer.

Carry out an independent investigation. Open the PhET simulation 'Wave Interference' (or a similar one) and select the light two-slit mode. Fix $\lambda = 600$ nm, $d = 0.20$ mm, and vary $L$ from 1 to 3 m; record the positions of the first three maxima and test the formula $\Delta y = \lambda L/d$. Then switch to one slit and measure the angles of the minima, comparing them with $a\sin\theta = m\lambda$. For polarization, safely use two polarizing filters (for example, from sunglasses) and a laptop screen: rotating the second filter, verify that at angles $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ the brightness qualitatively follows $I \propto \cos^2\theta$. Do not look directly at the Sun and do not use powerful sources; if in doubt, restrict yourself to the simulation.

Show how path difference creates interference fringes, how limiting the wavefront gives diffraction minima, and how a polarizer selects a projection of the electric field.

A labeled three-panel sketch: left — two slits, rays, path difference $d\sin\theta$, and fringes on a screen; middle — one slit with minima $a\sin\theta = m\lambda$; right — two polarizers with axes at angle $\theta$ and a graph of Malus's law $I/I_0 = \cos^2\theta$.

In a kitchen sink, a soap bubble is lit by white light. The soap film has thickness $t = 300\,\text{nm}$ and refractive index $n = 1.33$. Light reflecting from the outer and inner surfaces of the film interferes. The optical path difference is $2nt = 2 \cdot 1.33 \cdot 300\,\text{nm} = 798\,\text{nm}$. Because of the $\pi$ phase shift on reflection at the outer surface, the constructive condition is $2nt = (m+\tfrac12)\lambda$. For $m=1$, $\lambda = 2nt/1.5 = 532\,\text{nm}$, which is green. Thus the bubble appears green in reflected light.

This scenario uses interference in thin films: the optical path difference $\Delta = 2nt$ between reflections from the top and bottom surfaces plus a $\pi$ phase shift on one reflection determines which wavelengths are enhanced. It tests the condition $\Delta = (m+\tfrac12)\lambda$ for constructive interference.

A laser pointer with $\lambda = 650\,\text{nm}$ (red) shines on a CD, which acts as a reflection diffraction grating. The track spacing is $d = 1.6\,\mu\text{m} = 1600\,\text{nm}$. The first-order diffraction angle $\theta_1$ satisfies $\sin\theta_1 = \lambda/d = 650/1600 = 0.406$, so $\theta_1 \approx 24^\circ$. A second-order spot would require $\sin\theta_2 = 2\lambda/d = 1.30 > 1$, so it is not visible. The red spot appears at $24^\circ$ from the reflected beam.

This uses diffraction from a periodic grating: the condition $d\sin\theta = m\lambda$ for constructive interference of waves from adjacent tracks. It exercises the concept of diffraction orders and the limit $|\sin\theta|\le 1$.

A phone screen emits unpolarized light of intensity $I_0 = 10\,\text{W/m}^2$. The screen's built-in polarizer makes the light linearly polarized and halves the intensity: $0.5 I_0 = 5\,\text{W/m}^2$. Sunglasses with a polarizing filter are at $60^\circ$ to the screen's polarization. By Malus's law the transmitted intensity is $I = 5\,\text{W/m}^2 \cdot \cos^2 60^\circ = 1.25\,\text{W/m}^2$. Rotating the sunglasses to $90^\circ$ makes the screen nearly black.

This scenario applies polarization and Malus's law: unpolarized light becomes linearly polarized after passing through a polarizer, and the intensity after an analyzer at angle $\theta$ is $I = I_{\text{pol}}\cos^2\theta$. It tests the transverse nature of light and the cosine-squared dependence.