Interference and diffraction of light

Light has a wave nature: in two-slit or grating experiments it produces alternating bright and dark fringes.

Interference occurs when coherent waves superpose, and their path difference $\Delta$ determines the phase difference. If $\Delta = m\lambda$ ($m = 0, 1, 2, \ldots$), the waves arrive in phase and form a maximum; if $\Delta = (m + 1/2)\lambda$, they cancel and form a minimum.

In Young's experiment with slit separation $d$ and screen distance $L$, for small angles the fringe spacing is $\Delta y = \lambda L / d$, which allows the wavelength to be measured.

Diffraction is the bending and spreading of a wave when it passes an aperture or obstacle whose size is comparable to $\lambda$; for a single slit of width $a$, minima occur at $a\sin\theta = m\lambda$ ($m = 1, 2, \ldots$).

A diffraction grating enhances these effects: its maxima satisfy $d\sin\theta = m\lambda$, so different wavelengths are deflected by different angles and form a spectrum.

In Young's experiment, the slit separation is $d = 0.20$ mm, the screen distance is $L = 1.5$ m, and the wavelength is $\lambda = 600$ nm. Find the distance between adjacent bright fringes on the screen.

A diffraction grating has $N = 500$ lines per millimeter. Light with wavelength $\lambda = 500$ nm falls normally on the grating. Find the angle at which the second-order maximum ($m = 2$) is observed.

White light with wavelengths from $400$ to $700$ nm falls normally on a diffraction grating with $600$ lines per millimeter. Find the angular width of the first-order spectrum, i.e., the difference in angles between the extreme red and violet rays.

Use a safe interactive simulation such as PhET 'Wave Interference' (or similar); do not shine a laser into eyes and do not use a real laser without safety goggles.

In the simulation, select the two-slit mode and set wavelength $\lambda = 600$ nm, slit separation $d = 0.20$ mm, and screen distance $L = 1.5$ m. Measure the distance between adjacent bright fringes.

Then change $\lambda$ by a factor of 1.5 and $d$ by a factor of 2, predict the new $\Delta y$ using $\Delta y = \lambda L / d$, and check the result. Record a conclusion about how fringe spacing depends on $\lambda$, $L$, and $d$.

Show the relation between path difference and the positions of bright and dark fringes, and the difference between an interference pattern and a diffraction pattern.

Diagram: two coherent slits $S_1$ and $S_2$, rays reaching point $P$ on a screen, path difference marked $\Delta = d\sin\theta$; next to it a graph of intensity $I$ versus coordinate $y$ with maxima at $d\sin\theta = m\lambda$ and minima at $d\sin\theta = (m+1/2)\lambda$. Additionally, a single-slit pattern of width $a$ with a central maximum and minima $a\sin\theta = m\lambda$.

A smartphone glass often has a thin transparent film about $100$ nm thick with refractive index $n=1.38$. White light from a lamp falls almost perpendicularly. For green light with wavelength $\lambda=550$ nm, reflections from the top and bottom boundaries of the film cancel each other, so glare is weaker.

This is thin-film interference: the optical path difference of the two reflected waves is $2nt$. In a coating on glass both reflections give a $\lambda/2$ shift, which cancels, so cancellation occurs when $2nt=(m+\tfrac12)\lambda$. For $m=0$: $2\cdot1.38\cdot100\text{ nm}=276\text{ nm}\approx\tfrac12\cdot550\text{ nm}$.

On a frosty night a bright ring, a corona, is visible around the Moon. It is produced by millions of small water droplets in a cloud with diameter about $d=10$ µm. For yellow-green light $\lambda=550$ nm, the first dark minimum is observed at an angle of about $\theta=1.22\lambda/d=1.22\cdot550\text{ nm}/10\text{ µm}=0.067$ rad, that is $3.8^\circ$.

This is Fraunhofer diffraction by circular apertures: waves bend around the droplets and add with different phases. The angular radius of the first dark ring $\theta\approx1.22\lambda/d$ uses the dependence of the diffraction pattern on wavelength and obstacle size.

At a kitchen sink, a soap bubble has a film about $t=340$ nm thick with refractive index $n=1.33$. Under white light at near-normal incidence, red light with $\lambda=600$ nm is enhanced because $2nt=2\cdot1.33\cdot340\text{ nm}=904\text{ nm}\approx1.5\cdot600\text{ nm}$.

Thin-film interference is at work here: the path difference $2nt$ and the extra $\lambda/2$ phase shift on reflection from a denser medium make the enhancement condition $2nt=(m+\tfrac12)\lambda$. For red light with $m=1$, constructive addition occurs.