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Diffraction and Polarization
Single Slit Diffraction
Diffraction is the bending of light around obstacles and its spreading into geometrical shadow regions, fundamentally explained by Huygens' Principle.
$$\sin \theta \approx \frac{\lambda}{b}$$
Determines the angular position of the first minimum in a single-slit pattern.
$\theta$=Angle to the first dark fringe(radians)
$\lambda$=Wavelength of incident light(m)
$b$=Slit width (aperture size)(m)
$b \gg \lambda$
→Negligible diffraction (light travels in straight lines)
$b \approx \lambda$
→Significant spreading of the Central Maximum
Huygens' Principle: Every point on the wavefront within the slit acts as a source of secondary wavelets.
Central Maximum: The bright broad peak directly opposite the slit, containing most of the wave energy.
Secondary Maxima: Fainter bright fringes on either side of the center caused by partial constructive interference.
General Minima Condition: Dark fringes occur at $d \sin \theta = m\lambda$ where $m = \pm 1, \pm 2, \pm 3, ...$
Proportionality Shortcut: $\theta \propto \lambda / b$ — doubling wavelength doubles the spread; doubling slit width halves it.
Diffraction Prominence: The effect is significant only when $\lambda \sim b$. For $b \gg \lambda$, light essentially travels in straight lines.
Diffraction Gratings
A Diffraction Grating uses thousands of parallel slits to produce extremely sharp and localized interference maxima through Constructive Interference.
$$d \sin \theta = n\lambda$$
Predicts the angles of principal maxima for multi-slit interference.
$d$=Grating element (spacing between slits)(m)
$n$=Order of diffraction(integer)
$\theta$=Angular deviation of the nth order(degrees)
$n = 0$
→Zero-order maximum (central spot)
$n \lambda > d$
→Order not observable (sin θ cannot exceed 1)
Grating Element: The slit spacing $d$ is the reciprocal of line density: $d = 1/N$. Convert units carefully — 5000 lines/cm means $d = 1/500{,}000$ m.
Maximum Observable Order: Since $\sin \theta \leq 1$, the highest order is $n_{\text{max}} = \lfloor d/\lambda \rfloor$ (the greatest integer ≤ $d/\lambda$).
Angular Dispersion: Higher orders ($n$) separate different wavelengths more effectively, producing wider spectra.
White Light: Each wavelength diffracts at a different angle, producing a spectrum at each order. Higher orders show more spread-out spectra.
Bragg's Law — X-ray Diffraction by Crystals
X-rays have wavelengths (~$10^{-10}$ m) comparable to interplanar spacing in crystals, so crystal lattice planes act as a natural Diffraction Grating for X-rays.
$$2d \sin \theta = n\lambda$$
Bragg's equation: gives the condition for constructive interference of X-rays reflected from parallel crystal planes.
$d$=Interplanar spacing between crystal planes(m)
$\theta$=Bragg angle (angle of incidence measured from the crystal plane, not the normal)(degrees)
$n$=Order of reflection(integer)
$\lambda$=Wavelength of incident X-rays(m)
$n = 1$
→First-order reflection — strongest and most commonly observed
$\theta$ increases
→Higher-order reflections become possible, but intensity decreases
Path Difference: The extra distance traveled by the beam reflecting from the lower plane is $2d \sin \theta$, leading to the factor of 2.
Applications: Determining crystal structure (X-ray crystallography), used to find the structure of DNA and haemoglobin.
Grating vs Bragg: The grating equation $d \sin \theta = n\lambda$ has no factor of 2 because it involves transmission through slits, not reflection from parallel planes.
Polarization
Polarization proves light is a Transverse Wave by restricting oscillations to a single plane using a Polarizer.
$$I = I_0 \cos^2 \theta$$
Malus's Law: Relates transmitted intensity to the alignment of two polarizers.
$I$=Transmitted intensity(W/m²)
$I_0$=Intensity of polarized light entering the analyzer(W/m²)
$\theta$=Angle between polarizer and analyzer axes(degrees)
$\theta = 0^\circ$
→Parallel polarizers; full intensity $I_0$ transmitted
$\theta = 45^\circ$
→Half intensity transmitted ($\cos^2 45° = 1/2$)
$\theta = 90^\circ$
→Crossed polarizers; zero intensity transmitted
Unpolarized Light: Random oscillations in all planes perpendicular to propagation.
Selective Absorption: Polaroid sheets (dichroic materials) absorb E-field components perpendicular to the transmission axis.
Half-Rule: When unpolarized light passes through the first polarizer, transmitted intensity is always $I_0/2$, regardless of polarizer orientation.
Crossed Polarizers: Two polaroids at $90°$ block all light. Inserting a third at $45°$ between them allows light through — a classic demonstration.
Proof of Transverse Nature: Longitudinal waves cannot be polarized. The fact that light can be polarized proves it is a transverse wave.
Optical Rotation: Certain substances (quartz, sugar solutions) rotate the plane of polarization. This property is used to measure concentration of optically active solutions.
Brewster's Law
Light reflected from a surface is perfectly plane-polarized at Brewster's Angle, where the reflected and refracted rays are perpendicular.
$$n = \tan \theta_p$$
Calculates the specific angle of incidence for complete polarization by reflection.
$n$=Refractive index of the medium(dimensionless)
$\theta_p$=Brewster angle of incidence(degrees)
Water ($n=1.33$)
→$\theta_p \approx 53^\circ$
90° Rule: At Brewster's angle, the reflected and refracted rays are exactly perpendicular ($\theta_p + \theta_r = 90°$).
Glare Reduction: Polaroid sunglasses block horizontally polarized light reflected from roads, water, and glass surfaces.
Partial Polarization: At angles other than Brewster's, the reflected ray is only partially polarized.
Common Values: Glass ($n \approx 1.5$) → $\theta_p \approx 56.3°$; Water ($n = 1.33$) → $\theta_p \approx 53°$.
Application Strategy
Worked Examples
Problem Statement
Calculate the angular width of the central maximum for a single slit of width 0.02 mm illuminated by light of wavelength 500 nm.
Single slit diffractionAngular width calculation
Strategic Toolkit
$\theta \approx \frac{\lambda}{b}$
Step 1
Identify Variables
Convert units to meters: $b = 0.02 \text{ mm} = 2 \times 10^{-5} \text{ m}$ and $\lambda = 500 \text{ nm} = 5 \times 10^{-7} \text{ m}$.
$b = 2 \times 10^{-5} \text{ m}, \lambda = 5 \times 10^{-7} \text{ m}$
Step 2
Find Angular Position of First Minimum
The first minimum occurs at $\theta = \lambda / b$.
$\theta = \frac{5 \times 10^{-7}}{2 \times 10^{-5}} = 0.025 \text{ rad}$
Step 3
Calculate Total Width
The central maximum spans from the first minimum on the left to the first on the right ($2\theta$).
$\text{Width} = 2 \times 0.025 = 0.050 \text{ rad}$
Problem Statement
Unpolarized light of $100 \text{ W/m}^2$ passes through three polarizers: 1st is vertical, 2nd is at $30^\circ$ to vertical, and 3rd is horizontal ($90^\circ$ to vertical). Find final intensity.
Malus's LawInitial polarization loss
Strategic Toolkit
$I = I_0 \cos^2 \theta$
Step 1
First Polarizer
Unpolarized light loses half its intensity upon entering the first polarizer.
$I_1 = \frac{100}{2} = 50 \text{ W/m}^2$
Step 2
Second Polarizer
The light is now polarized vertically. The angle to the 2nd polarizer is $30^\circ$.
$I_2 = 50 \cdot \cos^2(30^\circ) = 50 \cdot (0.866)^2 = 37.5 \text{ W/m}^2$
Step 3
Third Polarizer
The angle between the 2nd ($30^\circ$) and 3rd ($90^\circ$) is $60^\circ$.
$I_3 = 37.5 \cdot \cos^2(60^\circ) = 37.5 \cdot (0.5)^2 = 9.375 \text{ W/m}^2$
Problem Statement
Light of wavelength 450 nm is incident on a diffraction grating with 5000 lines/cm. How many orders of spectra can be observed?
Grating equationMaximum order limit
Strategic Toolkit
$$d \sin \theta = n\lambda$$
Step 1
Find Grating Element
Convert line density to spacing: $d = 1/5000$ cm $= 1/500{,}000$ m $= 2 \times 10^{-6}$ m.
$d = 2 \times 10^{-6} \text{ m}$
Step 2
Apply Maximum Condition
The maximum order occurs when $\sin \theta = 1$, so $n_{\text{max}} = d / \lambda$.
$n_{\text{max}} = \frac{2 \times 10^{-6}}{450 \times 10^{-9}} = 4.44$
Step 3
Round Down
Since $n$ must be an integer and $\sin \theta$ cannot exceed 1, the maximum observable order is 4.
$n_{\text{max}} = 4$
Problem Statement
X-rays of wavelength 0.150 nm undergo first-order reflection at a Bragg angle of 13.3° from a quartz crystal. Find the interplanar spacing.
Bragg's lawInterplanar spacing
Strategic Toolkit
$$2d \sin \theta = n\lambda$$
Step 1
Identify Variables
$\lambda = 0.150$ nm, $\theta = 13.3°$, $n = 1$ (first order).
$\lambda = 0.150 \times 10^{-9} \text{ m}$
Step 2
Rearrange Bragg's Equation
Solve for $d$: $d = n\lambda / (2 \sin \theta)$.
$d = \frac{1 \times 0.150 \times 10^{-9}}{2 \times \sin 13.3°}$
Step 3
Calculate
$\sin 13.3° = 0.2300$. Substitute and compute.
$d = \frac{0.150 \times 10^{-9}}{2 \times 0.2300} = 0.326 \times 10^{-9} \text{ m} = 0.326 \text{ nm}$
Memorize formulas
Single Slit Minimum
Location of the first dark fringe (where wavelets cancel out).
Formula
$\sin \theta = \frac{\lambda}{b}$
Grating Maxima
The 'Bragg-like' condition for bright spots in a multi-slit grating.
Formula
$d \sin \theta = n\lambda$
Bragg's Law
X-ray reflection from crystal planes; note the factor of 2 and that θ is from the surface.
Formula
$2d \sin \theta = n\lambda$
Malus's Law
Remember: intensity depends on the square of the cosine of the angle.
Formula
$I = I_0 \cos^2 \theta$
Brewster's Tangent
Perfect polarization happens when the tangent of the angle equals the index.
Formula
$n = \tan \theta_p$
Review
1
A diffraction grating has 4000 lines per cm. Find the angular position of the first-order maximum for light of wavelength 500 nm.
Ans
$11.5°$
Review
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Find the grating spacing
Convert the line density to slit spacing using the reciprocal of 4000 lines per cm. This gives the distance between adjacent lines in meters.
$$d = \frac{0.01}{4000} = 2.5 \times 10^{-6}\ \text{m}$$
Apply the first-order grating equation
For the first bright fringe, use $d\sin\theta = n\lambda$ with $n=1$ and substitute the given wavelength.
$$\sin\theta = \frac{\lambda}{d} = \frac{500 \times 10^{-9}}{2.5 \times 10^{-6}} = 0.200$$
Evaluate the angle
Take the inverse sine of the result to obtain the angular position of the first-order maximum.
$$\theta = \arcsin(0.200) = 11.5^\circ$$
2
Two stars have an angular separation of $1.2 \times 10^{-6}$ radians. What is the minimum telescope diameter required to resolve them using light of wavelength 500 nm?
Ans
0.51 m
Review
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Use the Rayleigh criterion
The minimum resolvable angle for a circular aperture is given by $\theta = 1.22\lambda/D$. Rearrange this to solve for the diameter.
$$D = \frac{1.22\lambda}{\theta}$$
Substitute the data
Insert the wavelength and angular separation with consistent SI units.
$$D = \frac{1.22(500 \times 10^{-9})}{1.2 \times 10^{-6}} = 0.508\ \text{m}$$
3
Plane-polarized light of intensity $I_0$ passes through an analyzer. If the transmitted intensity is $I_0/4$, the angle between the polarization direction and the analyzer's transmission axis is:
Ans
$60°$
Review
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Apply Malus's law
For already polarized light, the transmitted intensity through an analyzer is $I = I_0\cos^2\theta$. Use the given intensity fraction directly.
$$\frac{I_0}{4} = I_0\cos^2\theta$$
Solve for the angle
Cancel $I_0$, take the square root, and read the angle from the cosine value.
$$\cos^2\theta = \frac14 \Rightarrow \cos\theta = \frac12 \Rightarrow \theta = 60^\circ$$
4
Two crossed polarizers transmit zero light. A third polarizer is now inserted between them at $45°$ to both. What fraction of the original unpolarized intensity $I_0$ emerges?
Ans
$I_0/8$
Review
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Transmit through the first polarizer
Unpolarized light loses half its intensity after the first polarizer.
$$I_1 = \frac{I_0}{2}$$
Pass through the middle polarizer
The middle polarizer is at $45^\circ$ to the first, so apply Malus's law with $\cos^2 45^\circ = 1/2$.
$$I_2 = I_1\cos^2 45^\circ = \frac{I_0}{2}\cdot\frac12 = \frac{I_0}{4}$$
Pass through the final polarizer
The last polarizer is again $45^\circ$ to the middle one, so the same factor applies once more.
$$I_3 = I_2\cos^2 45^\circ = \frac{I_0}{4}\cdot\frac12 = \frac{I_0}{8}$$
5
Light is incident on a glass block (n=1.5) at the Brewster angle. What is the angle between the reflected and refracted rays?
Ans
$90^\circ$
Review
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Use the Brewster-angle condition
At Brewster's angle, the reflected and refracted rays are perpendicular to one another by definition.
$$\angle(\text{reflected},\text{refracted}) = 90^\circ$$
6
X-rays of wavelength 0.071 nm undergo second-order Bragg reflection at an angle of $14.8°$. What is the interplanar spacing of the crystal?
Ans
$0.278 \text{ nm}$
Review
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Start from Bragg's law
Use the Bragg condition $2d\sin\theta = n\lambda$ and rearrange it for the spacing $d$.
$$d = \frac{n\lambda}{2\sin\theta}$$
Substitute the second-order values
Insert $n=2$, $\lambda=0.071$ nm, and $\theta=14.8^\circ$.
$$d = \frac{2(0.071)}{2\sin 14.8^\circ} = \frac{0.142}{2\times 0.2554} \approx 0.278\ \text{nm}$$
7
In a diffraction grating spectrum, the third-order image of wavelength $\lambda_1$ coincides with the fourth-order image of wavelength $\lambda_2$. The ratio $\lambda_1 / \lambda_2$ is:
Ans
$4/3$
Review
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Match the same diffraction angle
If two spectral lines coincide, they have the same angle, so the grating equation gives equal path differences for both.
$$3\lambda_1 = 4\lambda_2$$
Form the wavelength ratio
Divide both sides by $3\lambda_2$ to isolate the required ratio.
$$\frac{\lambda_1}{\lambda_2} = \frac{4}{3}$$
8
A single slit of width $d$ produces a diffraction pattern. If the slit width is halved and the wavelength is doubled, the angular width of the central maximum:
Ans
Becomes 4 times wider
Review
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Use the central-maxima scaling
For single-slit diffraction, the angular width of the central maximum is proportional to $\lambda/d$.
$$\text{width} \propto \frac{\lambda}{d}$$
Apply both changes together
Doubling the wavelength doubles the width, and halving the slit width doubles it again, so the total factor is four.
$$\frac{2\lambda}{d/2} = 4\,\frac{\lambda}{d}$$
9
A second-order spectrum is formed at $38.0°$ when light falls normally on a grating with 5400 lines/cm. The wavelength of light is approximately:
Ans
570 nm
Review
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Find the grating spacing
Convert the line density into slit spacing by taking its reciprocal in SI units.
$$d = \frac{0.01}{5400} = 1.852 \times 10^{-6}\ \text{m}$$
Use the grating equation
Apply $d\sin\theta = n\lambda$ with $n=2$ and solve for the wavelength.
$$\lambda = \frac{d\sin 38^\circ}{2} = \frac{(1.852 \times 10^{-6})(0.6157)}{2} \approx 5.70 \times 10^{-7}\ \text{m}$$
Convert to nanometers
Rewrite the wavelength in a more standard optical unit.
$$5.70 \times 10^{-7}\ \text{m} = 570\ \text{nm}$$
10
In Bragg's law $2d \sin \theta = n\lambda$, the angle $\theta$ is measured from:
Ans
The crystal surface (glancing angle)
Review
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Recall the Bragg-angle convention
In Bragg diffraction, $\theta$ is the glancing angle measured between the beam and the crystal plane, not the surface normal.
$$\theta\text{ is measured from the crystal surface}$$
11
Polaroid sunglasses reduce glare from a wet road because:
Ans
Reflected light is partially polarized horizontally, and the sunglasses block horizontal polarization
Review
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Identify the polarization of reflected glare
Light reflected from a horizontal surface is partially polarized, with the electric field tending to vibrate horizontally.
$$\text{reflected glare is mainly horizontally polarized}$$
Match the sunglasses transmission axis
Polaroid sunglasses are typically oriented to block horizontal polarization, so they suppress the glare while still allowing much of the remaining ambient light through.
$$\text{horizontal glare} \rightarrow \text{blocked by the polaroid}$$
Question 1 of 11
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1
A diffraction grating has 4000 lines per cm. Find the angular position of the first-order maximum for light of wavelength 500 nm.