C3-3. Slit Interference and Gratings

1. Double-Slit Interference

Background: Thomas Young's classic double-slit experiment demonstrates the wave nature of light. When coherent, monochromatic light passes through two narrow, closely spaced slits, the slits act as two in-phase point sources. The overlapping waves from these sources undergo superposition, creating an interference pattern of alternating bright (constructive) and dark (destructive) fringes on a distant screen.

Derivation from First Principles:

Consider two slits separated by a distance $d$ and a screen at a distance $D$ (where $D \gg d$). For a point on the screen at a vertical distance $y$ from the central axis, the light waves from the two slits travel slightly different distances.

  1. Path Difference: The extra distance the wave from the lower slit travels is the path difference, $\Delta L$. Using simple geometry and assuming the rays heading to the screen are nearly parallel, $\Delta L \approx d \sin(\theta)$, where $\theta$ is the angle from the central axis to the point on the screen.
  2. Small-Angle Approximation: Because $D \gg y$, the angle $\theta$ is very small. We can apply the small-angle approximation: $\sin(\theta) \approx \tan(\theta) = \dfrac{y}{D}$.
  3. Substitute and Equate: Replacing $\sin(\theta)$ with $\dfrac{y}{D}$, the geometric path difference becomes:
    $$\Delta L \approx \dfrac{d \cdot y}{D}$$
  4. Constructive Interference: For a bright fringe (maximum) to appear, the path difference must be an integer multiple of the wavelength: $\Delta L = m\lambda$ (where $m = 0, 1, 2, \dots$).
    $$\dfrac{d \cdot y_m}{D} = m\lambda \implies y_m = \dfrac{m\lambda D}{d}$$
  5. Fringe Spacing ($s$): The distance between two adjacent bright fringes (e.g., $m$ and $m+1$) is calculated by finding the difference in their vertical positions:
    $$s = y_{m+1} - y_m = \dfrac{(m+1)\lambda D}{d} - \dfrac{m\lambda D}{d} = \dfrac{\lambda D}{d}$$
$$s = \dfrac{\lambda D}{d}$$
  • $s$ = fringe spacing / distance between adjacent maxima ($\text{m}$)
  • $\lambda$ = wavelength ($\text{m}$)
  • $D$ = distance from slits to screen ($\text{m}$)
  • $d$ = slit separation ($\text{m}$)

Key idea: To increase the spread of the interference pattern (larger $s$), use a longer wavelength ($\lambda$), move the screen further away ($D$), or bring the slits closer together (decrease $d$).

S1 S2 Screen Intensity (I) n=2 n=1 n=0 n=1 n=2 Constructive Interference (Path difference nλ) s 2nd Max (Bright) 1st Max (Bright) Central Max 1st Max (Bright) 2nd Max (Bright) path to 1st maximum D d

Example 1: Calculating Fringe Spacing

Problem: A pair of slits separated by $1.0\text{ mm}$ are illuminated with light of wavelength $480\text{ nm}$. What is the separation of the fringes at a distance of $2.0\text{ m}$?

Solution: $s = \dfrac{(480 \times 10^{-9}\text{ m}) \times (2.0\text{ m})}{1.0 \times 10^{-3}\text{ m}} = 960 \times 10^{-6}\text{ m} = \mathbf{0.96 \text{ mm}}$.

Example 2

S₁ S₂ Screen d x y

Problem: In a double-slit experiment, a source of monochromatic red light is incident on slits $S_1$ and $S_2$ separated by a distance $d$. A screen is located at distance $x$ from the slits. A pattern with fringe spacing $y$ is observed on the screen.

Three changes are possible for this arrangement:

  1. increasing $x$
  2. increasing $d$
  3. using green monochromatic light instead of red.

Which changes will cause a decrease in fringe spacing $y$?

Solution:

First, recall the double-slit interference formula for fringe spacing. Using the variables provided in the problem, the formula is:

$$y = \dfrac{\lambda x}{d}$$

We want to find which changes cause $y$ to decrease. Let's evaluate each option:

  • I. Increasing $x$: Because $y$ is directly proportional to $x$, increasing $x$ will increase $y$. (Incorrect)
  • II. Increasing $d$: Because $y$ is inversely proportional to $d$, increasing $d$ will decrease $y$. (Correct)
  • III. Using green light instead of red: The wavelength of green light is shorter than that of red light ($\lambda_{\text{green}} < \lambda_{\text{red}}$). Since $y$ is directly proportional to $\lambda$, decreasing the wavelength decreases $y$. (Correct)

Conclusion: Changes II and III will cause a decrease in the fringe spacing $y$.

Example 3

Problem: Monochromatic light is used to produce double-slit interference fringes on a screen. The fringe separation on the screen is $y$. The distance from the slits to the screen and the separation of the slits are both doubled, and the light source is unchanged. What is the new fringe separation on the screen?

Solution:

Set up the initial condition using the standard fringe separation formula. Let $D$ be the distance to the screen and $d$ be the slit separation:

$$y = \dfrac{\lambda D}{d}$$

Next, define the new parameters based on the problem description:

  • New distance to screen: $D' = 2D$
  • New slit separation: $d' = 2d$
  • Wavelength remains unchanged: $\lambda' = \lambda$

Substitute these new values into the formula to find the new fringe separation, $y'$:

$$y' = \dfrac{\lambda (2D)}{(2d)}$$

The factors of 2 in the numerator and denominator cancel out:

$$y' = \dfrac{\lambda D}{d} = y$$

Conclusion: The new fringe separation on the screen remains $y$.

2. Single-Slit Diffraction

When coherent light passes through a single narrow slit of width $b$, it creates a diffraction pattern consisting of a wide, intense central maximum flanked by much fainter, narrower secondary maxima.

$$\theta = \dfrac{\lambda}{b}$$
  • $\theta$ = Angle from the central axis to the first diffraction minimum (radians).
  • $\lambda$ = Wavelength of the incident light (meters).
  • $b$ = Physical width of the slit opening (meters).
b θ Intensity Curve 1st Min 1st Min

Example 2: Macroscopic Width & Refractive Index

Problem: A single slit is illuminated by a helium-neon laser ($\lambda = 632.8\text{ nm}$). The central diffraction maximum observed on a screen located $2.50\text{ m}$ away is exactly $4.00\text{ cm}$ wide.

  1. Calculate the physical width of the slit $b$.
  2. If the entire apparatus (laser, slit, and screen) is completely submerged in water (refractive index $n = 1.33$), calculate the new width of the central maximum on the screen.

Solution:

(a) Slit Width: The total linear width $W$ of the central maximum on a distant screen ($L$) stretches from the first minimum above the axis to the first minimum below it, making $W = 2y$, where $y \approx L\theta$.

$$W = \dfrac{2\lambda L}{b} \implies b = \dfrac{2\lambda L}{W}$$

Substituting the values ($W = 0.040\text{ m}$):

$$b = \dfrac{2(632.8 \times 10^{-9}\text{ m})(2.50\text{ m})}{0.040\text{ m}} = 7.91 \times 10^{-5}\text{ m} \text{ (or } 79.1\text{ }\mu\text{m})$$

(b) Submerged Apparatus: When light enters a medium with a higher refractive index, its frequency remains constant, but its speed and wavelength decrease. The new wavelength $\lambda'$ is:

$$\lambda' = \dfrac{\lambda}{n}$$

Since the central maximum width $W$ is directly proportional to the wavelength ($W \propto \lambda$), the new width $W'$ will simply be scaled by the same factor:

$$W' = \dfrac{W}{n} = \dfrac{4.00\text{ cm}}{1.33} = \mathbf{3.01\text{ cm}}$$

3. Diffraction Gratings

A diffraction grating contains thousands of extremely closely spaced, parallel slits (lines). The interference of light from all these numerous slits produces highly intense, razor-sharp principal maxima at specific angles, making gratings ideal for spectral analysis.

$$n\lambda = d \sin\theta$$
  • $n$ = Order number of the bright fringe ($n = 0, 1, 2, \dots$).
  • $\lambda$ = Wavelength (meters).
  • $d$ = Grating spacing / distance between adjacent slits (meters). Calculated as $1/N$ where $N$ is lines per meter.
  • $\theta$ = Angle of diffraction for the $n$-th order maximum.
D d Screen y Grating θv θr Violet d θ Small ΔL Red d θ Large ΔL
White Light Beam Diffraction Grating n = 0 (Central Max) All colors overlap (White) Red (~700 nm) Violet (~400 nm) n = 1 Red (~700 nm) Violet (~400 nm) n = 1 Because λ (red) > λ (violet): The grating formula dictates that θ (red) must be > θ (violet).

Example 3: Overlapping Spectral Orders

Problem: A continuous spectrum of white light containing wavelengths from $400\text{ nm}$ (violet) to $700\text{ nm}$ (red) is incident normally on a diffraction grating ruled with $400\text{ lines per millimeter}$.

  1. Calculate the angular width (spread) of the entire first-order ($n=1$) visible spectrum.
  2. Determine mathematically whether the second-order ($n=2$) spectrum and the third-order ($n=3$) spectrum overlap.

Solution:

First, find the grating spacing $d$ in meters:

$$d = \dfrac{1}{N} = \dfrac{1\text{ mm}}{400} = 2.5 \times 10^{-3}\text{ mm} = 2.5 \times 10^{-6}\text{ m}$$

(a) Angular Width of First Order ($n=1$):
Find the diffraction angles for the extreme ends of the spectrum using $$ \sin\theta = \dfrac{n\lambda}{d} $$ For Violet ($400\text{ nm}$), $$ \sin\theta_v = \frac{(1)(400 \times 10^{-9})}{2.5 \times 10^{-6}} = 0.160 \implies \theta_v \approx 9.21^\circ $$ For Red ($700\text{ nm}$), $$ \sin\theta_r = \frac{(1)(700 \times 10^{-9})}{2.5 \times 10^{-6}} = 0.280 \implies \theta_r \approx 16.26^\circ $$ The angular width is $\Delta\theta = 16.26^\circ - 9.21^\circ = \mathbf{7.05^\circ}$.

(b) Checking for Spectral Overlap:
Overlap occurs if the shortest wavelength of the higher order ($n=3$, violet) bends at a smaller angle than the longest wavelength of the lower order ($n=2$, red) For $n=2$ Red, $$ \sin\theta_{2r} = \frac{2(700 \times 10^{-9})}{2.5 \times 10^{-6}} = 0.560 \implies \theta_{2r} \approx 34.06^\circ $$ For $n=3$ Violet, $$ \sin\theta_{3v} = \frac{3(400 \times 10^{-9})}{2.5 \times 10^{-6}} = 0.480 \implies \theta_{3v} \approx 28.69^\circ $$ Because $\theta_{3v} < \theta_{2r}$ ($28.69^\circ < 34.06^\circ$), the start of the third-order spectrum appears "inside" the end of the second-order spectrum. Therefore, yes, the second and third orders overlap.