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Topic 15 of 15

Diffraction & Polarization

Diffraction occurs when waves bend around obstacles or through openings comparable in size to their wavelength. Diffraction gratings disperse light into spectra with high resolution. Polarization describes the orientation of the electric field vector in transverse waves, governed by Malus's law.

Key Concepts

Single-Slit Diffraction

A slit of width

a

produces dark fringes at angles where

a\sin\theta = m\lambda

(

m = \pm1, \pm2, \ldots

). The central maximum spans twice the angular width of the other maxima.

Diffraction Grating

A grating with slit spacing

d

produces sharp bright maxima (principal maxima) at

d\sin\theta = m\lambda

. The grating equation is identical to Young's but applies to many slits, giving much sharper and brighter maxima.

Rayleigh Criterion

Two point sources are just resolved when the central maximum of one falls on the first minimum of the other. For a circular aperture of diameter

D

\theta_R = 1.22\lambda/D

Polarization

Light is linearly polarized when its electric field oscillates in a single plane. Unpolarized light has random orientations. A linear polarizer transmits only the component along its transmission axis.

Malus's Law

When polarized light of intensity

I_0

passes through a polarizer whose transmission axis makes angle

\theta

with the polarization direction, the transmitted intensity is

I = I_0\cos^2\theta

Brewster's Angle

\theta_B = \arctan(n_2/n_1)

, reflected light is completely polarized parallel to the surface (s-polarized component only).

Key Equations

Single-Slit Dark Fringes

a\sin\theta = m\lambda\quad(m = \pm1,\,\pm2,\ldots)

Condition for destructive interference; $a$ = slit width.

Diffraction Grating Bright Fringes

d\sin\theta = m\lambda\quad(m = 0,\,\pm1,\,\pm2,\ldots)

Principal maxima condition; $d$ = grating spacing (1/lines per meter).

Rayleigh Criterion

\theta_R = 1.22\,\dfrac{\lambda}{D}

Minimum angular separation for two sources to be resolved by a circular aperture of diameter $D$.

Malus's Law

I = I_0\cos^2\theta

Transmitted intensity through a polarizer; $\theta$ = angle between polarization and transmission axis.

Brewster's Angle

\tan\theta_B = \dfrac{n_2}{n_1}

Angle at which reflected light is completely polarized.

Worked Example

Diffraction Grating: First-Order Maximum

Problem

A diffraction grating has 500 lines/mm. Find the angle of the first-order maximum ( $m=1$ ) for light of wavelength $\lambda = 600\text{ nm}$ .

Solution

Find the grating spacing $d$ .

d = \frac{1}{500\text{ lines/mm}} = \frac{1}{500\times10^3\text{ lines/m}} = 2.00\times10^{-6}\text{ m}

Apply the grating equation for $m=1$ .

\sin\theta = \frac{m\lambda}{d} = \frac{(1)(600\times10^{-9})}{2.00\times10^{-6}} = 0.300

\theta = \arcsin(0.300) \approx 17.5°

Answer

\theta \approx 17.5°

Practice

Exercises

7 problems

1 of 7

A diffraction grating has 300 lines/mm. Find the angle (in degrees) of the first-order maximum for $\lambda = 550\text{ nm}$ .

Your answer

2 of 7

A grating with 400 lines/mm produces a first-order maximum at $\theta = 12.0°$ . What is the wavelength of the light in nm?

Your answer

3 of 7

A single slit of width $a = 0.040\text{ mm}$ is illuminated with $\lambda = 600\text{ nm}$ . What is the angle (in degrees) to the first dark fringe?

Your answer

4 of 7

The Hubble Space Telescope has a mirror diameter $D = 2.40\text{ m}$ . What is its angular resolution limit $\theta_R$ (in $\mu$ rad) for $\lambda = 550\text{ nm}$ ?

Your answer

μrad

5 of 7

Polarized light of intensity $I_0 = 120\text{ W/m}^2$ passes through a polarizer with its axis at $60°$ to the polarization direction. What is the transmitted intensity in W/m²?

Your answer

W/m²

6 of 7

Polarized light passes through a polarizer and the transmitted intensity is 25% of the incident intensity. What is the angle $\theta$ (in degrees) between the polarization direction and the polarizer axis?

Your answer

7 of 7

A diffraction grating has 500 lines/mm. What is the angle (in degrees) of the second-order maximum for $\lambda = 550\text{ nm}$ ?

Your answer

Key Takeaways

Single-slit diffraction dark fringes: $a\sin\theta = m\lambda$ ; diffraction grating bright maxima: $d\sin\theta = m\lambda$ (same form, very different sharpness).
The Rayleigh criterion $\theta_R = 1.22\lambda/D$ sets the fundamental resolution limit of circular optical instruments.
Malus's law $I = I_0\cos^2\theta$ describes intensity transmission through a polarizer; two polarizers at 90° block all light.