Multiple Slit Diffraction

Strategy

For a diffraction grating, maxima occur at angles satisfying $d \sin \theta = m\lambda$, where d is the spacing between adjacent slits (grating constant), m is the order, and λ is the wavelength. First, calculate d from the line density, then solve for θ with m = 1.

Solution

Given:

• Line density: 2000 lines/cm
• Wavelength: $\lambda = 520 \text{ nm} = 520 \times 10^{-9} \text{ m}$
• Order: $m = 1$ (first-order maximum)

Calculate the grating constant (slit separation):

For the first-order maximum:

The first-order maximum for 520-nm green light occurs at 5.97°.

Discussion

The small angle (5.97°) for the first-order maximum indicates that the grating constant (5.00 μm) is about 10 times the wavelength (520 nm). Diffraction gratings with 2000 lines/cm are relatively coarse gratings. Finer gratings (e.g., 10,000 lines/cm) would produce larger diffraction angles and better spectral separation. The green wavelength (520 nm) falls in the middle of the visible spectrum. Higher orders (m = 2, 3, …) would appear at larger angles: the second order at about 11.96° and the third at about 18.05°.

Strategy

For a diffraction grating, we use $d \sin \theta = m\lambda$ where d is the distance between lines. First, find d from the line density, then solve for θ with m = 3.

Solution

Given:

• $$ \lambda = 580 \text{ nm} = 580 \times 10^{-9} \text{ m} $$
• Line density = 1500 lines/cm
• $m = 3$ (third-order maximum)

Step 1: Find the distance between lines

Step 2: Find the angle

Using $d \sin \theta = m\lambda$:

Discussion

The third-order maximum appears at 15.1°. With a grating spacing of 6.67 μm (about 11.5 wavelengths), this grating can produce several orders of maxima. We can check: the maximum order would be m_max = d/λ = 6.67/0.580 ≈ 11.5, so up to 11 orders are theoretically possible (though higher orders become very dim and widely separated).

Strategy

Use $d \sin \theta = m\lambda$ to find the grating spacing d, then calculate the number of lines per centimeter as N = 1/d.

Solution

Given:

• $$ \lambda = 470 \text{ nm} = 470 \times 10^{-9} \text{ m} $$
• $$ \theta = 25.0° $$
• $m = 1$ (first-order maximum)

Step 1: Find the grating spacing

Using $d \sin \theta = m\lambda$:

Step 2: Find the number of lines per centimeter

Discussion

A grating with approximately 9000 lines/cm is a moderately fine diffraction grating. This spacing of about 1.11 μm (about 2.4 wavelengths of blue light) produces the first-order maximum at 25.0°, which is a convenient angle for observation. This is close to commercial gratings often available with 5000-10,000 lines/cm. Higher orders would appear at larger angles: the second order at about 57.1° and the third order at about 80.8°, with a maximum possible order of m_max = d/λ ≈ 2.4, meaning only orders 0, 1, and 2 would be fully observable.

Strategy

Use $d \sin \theta = m\lambda$ with m = 2 (second-order) to solve for the line spacing d.

Solution

Given:

• $$ \lambda = 760 \text{ nm} = 760 \times 10^{-9} \text{ m} $$
• $$ \theta = 60.0° $$
• $m = 2$ (second-order maximum)

Using $d \sin \theta = m\lambda$:

Discussion

The line spacing of 1.76 μm is very small - just over twice the wavelength of red light. This close spacing is necessary to produce the second-order maximum at the relatively large angle of 60°.

We can verify this makes sense: if the second-order maximum is at 60°, the first-order would be at θ₁ where sin θ₁ = (1)(760)/(1.76 × 10³) = 0.432, giving θ₁ = 25.6°. This seems reasonable.

For this grating, the maximum order observable would be: m_max = d/λ = 1.76/0.76 ≈ 2.3, so only second-order maxima (and possibly a dim third-order) would be observable before reaching 90°.

Strategy

Use $d \sin \theta = m\lambda$ with m = 2 (second-order) to solve for the wavelength λ. First, find the grating spacing d from the line density.

Solution

Given:

• Line density: 5000 lines/cm
• $$ \theta = 45.0° $$
• $m = 2$ (second-order maximum)

Step 1: Find the grating spacing

Step 2: Find the wavelength

Using $d \sin \theta = m\lambda$:

Discussion

The wavelength of 707 nm falls in the red region of the visible spectrum (near the long-wavelength end). This is close to deep red light. The second-order maximum appears at 45.0°, which is a reasonably large angle, making it easy to observe and measure.

We can verify this is reasonable: the first-order maximum would appear at an angle where sin θ₁ = λ/d = 707/2000 = 0.354, giving θ₁ = 20.7°. The third-order would be at sin θ₃ = 3λ/d = 2121/2000 = 1.06 > 1, which is impossible, so no third-order maximum exists. This grating can show at most the zeroth, first, and second orders for this wavelength.

Strategy

Following the Problem-Solving Strategies for Wave Optics:

Step 1: This involves a diffraction grating. Step 2: We use $d \sin \theta = m\lambda$ for constructive interference. Step 3: We need to find the wavelengths λ for each angle. Step 4: Given: 10,000 lines/cm, m = 1, four angles. Step 5: First find d, then solve $\lambda = d \sin \theta / m$ for each angle.

Solution

Find the grating spacing:

For each angle, calculate λ using $\lambda = \frac{d \sin \theta}{m}$ with m = 1:

Angle 1: θ = 24.2°

Angle 2: θ = 25.7°

Angle 3: θ = 29.1°

Angle 4: θ = 41.0°

Step 6: All results are in the visible range (380-760 nm), which is reasonable for hydrogen emission lines.

Discussion

These wavelengths correspond to prominent lines in the hydrogen Balmer series:

• 410 nm (H-delta): violet
• 434 nm (H-gamma): violet-blue
• 486 nm (H-beta): blue-green
• 656 nm (H-alpha): red

The H-alpha line at 656 nm is the brightest and most famous line in the visible hydrogen spectrum. These emission lines result from electrons transitioning from higher energy levels (n = 3, 4, 5, 6) down to n = 2. This spectral “fingerprint” is characteristic of hydrogen and is observed in stars, nebulae, and laboratory hydrogen discharge tubes. The precision of diffraction gratings makes them invaluable tools for spectroscopy and identifying elements.

Strategy

Use the four hydrogen wavelengths from the previous problem (410, 434, 486, and 656 nm) with the new grating. For part (a), find first-order angles with the 5000 lines/cm grating. For part (b), find second-order angles. For part (c), compare how changing the grating spacing by a factor affects the angles.

Solution

The four hydrogen wavelengths are: 410 nm, 434 nm, 486 nm, and 656 nm.

For the 5000-line-per-centimeter grating:

(a) First-order angles with 5000 lines/cm grating:

Using $\sin \theta = \frac{m\lambda}{d}$ with m = 1:

For λ = 410 nm:

For λ = 434 nm:

For λ = 486 nm:

For λ = 656 nm:

First-order angles: 11.8°, 12.5°, 14.1°, 19.2°

(b) Second-order angles with 5000 lines/cm grating:

Using $\sin \theta = \frac{m\lambda}{d}$ with m = 2:

For λ = 410 nm:

For λ = 434 nm:

For λ = 486 nm:

For λ = 656 nm:

Second-order angles: 24.2°, 25.7°, 29.1°, 41.0°

(c) Relationship between line density and angles:

Notice that the second-order angles for the 5000 lines/cm grating (24.2°, 25.7°, 29.1°, 41.0°) are exactly the same as the first-order angles for the 10,000 lines/cm grating from the previous problem!

This demonstrates a general principle: When you decrease the number of lines per centimeter by a factor of n (thereby increasing d by a factor of n), the angles for the n-th order maximum with the new grating equal the angles for the first-order maximum with the original grating.

Mathematically: If $d_2 = n \cdot d_1$, then $\sin \theta_{m=1, d_1} = \frac{\lambda}{d_1}$ equals $\sin \theta_{m=n, d_2} = \frac{n\lambda}{n \cdot d_1} = \frac{\lambda}{d_1}$.

In this case, halving the line density (10,000 → 5000 lines/cm) means doubling d, so the second-order (n=2) angles with the coarser grating equal the first-order angles with the finer grating.

Discussion

This scaling relationship is very useful in diffraction grating design. It shows that using a coarser grating with higher-order maxima can produce the same angular positions as using a finer grating with lower-order maxima. However, higher orders are generally dimmer and may suffer from overlapping spectra from different orders. The 5000 lines/cm grating produces more widely spaced orders, making them easier to observe separately, while the 10,000 lines/cm grating provides better resolution within each order.

Strategy

For a complete first-order spectrum, both the shortest (violet, ~380 nm) and longest (red, ~760 nm) wavelengths of visible light must have first-order maxima at angles less than 90°. The limiting condition is for the longest wavelength (red) at θ = 90°.

Solution

For the complete visible spectrum to be observable in first order, the red end (longest wavelength) must appear before 90°. Using $d \sin \theta = m\lambda$ with m = 1 and $\theta = 90°$ (where $\sin 90° = 1$):

This is the minimum spacing needed. The maximum number of lines per centimeter is:

Rounding to appropriate significant figures:

Discussion

If the grating had more than ~13,200 lines/cm, the line spacing would be smaller than 760 nm, and the first-order maximum for red light would require $\sin \theta > 1$, which is impossible. The red end of the spectrum would not be observable in first order.

Let’s verify: with d = 760 nm:

• Red (760 nm): sin θ = (1)(760)/760 = 1.00 → θ = 90° (just barely observable)
• Violet (380 nm): sin θ = (1)(380)/760 = 0.50 → θ = 30° (easily observable)

This grating would spread the complete visible spectrum from 30° to 90° in first order. Commercial gratings often have fewer lines/cm (~5000-10,000) to ensure good visibility of the complete spectrum at more convenient angles.

Strategy

Use $d \sin \theta = m\lambda$ to find each wavelength from its angle. First, calculate the grating spacing d from the line density, then solve for λ at each angle.

Solution

Given:

• Line density: 10,000 lines/cm
• $$ \theta_1 = 36.093° $$
• $$ \theta_2 = 36.129° $$
• $m = 1$ (first-order maxima)

Step 1: Find the grating spacing

Step 2: Find the first wavelength

Using $\lambda = d \sin \theta / m$:

Step 3: Find the second wavelength

Discussion

These two wavelengths, 589.1 nm and 589.6 nm, form the famous sodium D-line doublet. The two lines are separated by only 0.5 nm (or about 6 nm), which is why sodium light appears to be pure yellow to the eye - the two wavelengths are too close for the eye to distinguish.

The doublet arises from the fine structure of sodium’s electron energy levels: transitions from two closely spaced upper states (²P₃/₂ and ²P₁/₂) to the same ground state (²S₁/₂) produce the D₂ line (589.0 nm) and D₁ line (589.6 nm).

This problem demonstrates the high resolving power of diffraction gratings. The angular separation is only 0.036°, yet the grating cleanly separates the two lines. The ability to resolve such closely spaced wavelengths is crucial for spectroscopic analysis in astronomy, chemistry, and physics.

Strategy

For a reflection grating, we use the same equation as for a transmission grating: $d \sin \theta = m\lambda$. We solve for d with m = 1.

Solution

Given:

• $$ \lambda = 525 \text{ nm} = 525 \times 10^{-9} \text{ m} $$
• $$ \theta = 30.0° $$
• $m = 1$ (first-order maximum)

Using $d \sin \theta = m\lambda$:

Discussion

The spacing of 1.05 μm (1050 nm) is about twice the wavelength of the green light being diffracted. This is typical for structural coloration in bird feathers and insect wings.

The iridescent colors in peacock feathers, hummingbird feathers, and butterfly wings are produced by such natural diffraction gratings. The spacing is created by regular arrangements of microscopic structures like barbules (in feathers) or scales (in butterfly wings). The color changes with viewing angle because different angles produce constructive interference for different wavelengths according to $d \sin \theta = m\lambda$.

This biological “engineering” produces brilliant colors without pigments - the colors are purely structural and result from wave interference. These colors often don’t fade over time like pigment-based colors do, which is why museum specimens of iridescent birds and butterflies remain colorful for centuries.

Strategy

Use $d \sin \theta = m\lambda$ with m = 1 to find the angle. First, calculate the spacing d from the line density, then solve for θ.

Solution

Given:

• Line density: 8000 lines/cm
• $$ \lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m} $$
• $m = 1$ (first-order maximum)

Step 1: Find the structure spacing

Step 2: Find the angle

Using $d \sin \theta = m\lambda$:

Discussion

The first-order maximum for orange-red light (600 nm) appears at 28.7° from the normal. This angle is typical for structural coloration in bird feathers. The spacing of 1.25 μm is roughly twice the wavelength, which is ideal for producing visible diffraction patterns.

Birds with iridescent feathers (like hummingbirds, peacocks, and starlings) have microscopic structures on their feather barbules that act as natural diffraction gratings. As the bird moves and the viewing angle changes, different wavelengths satisfy the constructive interference condition, causing the brilliant color shifts characteristic of iridescence. For this feather, shorter wavelengths (blue, violet) would appear at smaller angles, while longer wavelengths (red) would appear at larger angles, creating a rainbow effect with changing perspective.

Strategy

Use $d \sin \theta = m\lambda$ for the first-order maximum (m = 1) with d = 8 μm. Use λ ≈ 700 nm for red and λ ≈ 450 nm for blue.

Solution

Given:

• $$ d = 8 \text{ μm} = 8 \times 10^{-6} \text{ m} $$
• $m = 1$ (first-order, typically the brightest)

(a) Red light (λ ≈ 700 nm):

(b) Blue light (λ ≈ 450 nm):

Discussion

The diffraction angles are quite small (3.2° for blue, 5.0° for red) because the grating spacing (8 μm) is much larger than the wavelengths of visible light. Red light appears at a larger angle than blue light, which is why when you rotate an opal, you see different colors from different angles - this is the source of opal’s famous “play of color” or iridescence.

The complete visible spectrum would spread from about 3° (violet) to 5° (red). Higher orders (m = 2, 3, etc.) would appear at larger angles but with decreasing intensity. The maximum order possible is m_max = d/λ_max ≈ 8000/700 ≈ 11, so many orders are theoretically possible, though in practice only the first few are bright enough to see.

Opals contain tiny silica spheres arranged in regular patterns, creating these natural diffraction gratings. The spacing of 8 μm is roughly 10-20 wavelengths of visible light, perfect for producing brilliant iridescent colors.

Strategy

Use $d \sin \theta = m\lambda$ for both orders. From the first-order angle, find the ratio λ/d, then use this to find the second-order angle.

Solution

Given:

• $\theta_1 = 20.0°$ (first-order maximum)
• Find: $\theta_2$ (second-order maximum)

Step 1: Find the ratio λ/d from the first-order condition

For the first-order maximum (m = 1):

Step 2: Find the second-order angle

For the second-order maximum (m = 2):

Discussion

The second-order maximum appears at 43.2°, more than twice the first-order angle (20.0°). This is because the sine function is nonlinear - while sin θ doubles from first to second order, the angle itself more than doubles.

This result is general: for any diffraction grating, if the first-order maximum is at angle θ₁, the second-order maximum is at θ₂ = sin⁻¹(2 sin θ₁). Notice that the second-order angle exists only if 2 sin θ₁ ≤ 1, which requires θ₁ ≤ 30°. Since 20° < 30°, the second-order maximum is observable. If θ₁ were greater than 30°, no second-order maximum would exist (consistent with our earlier proof).

Strategy

We use $d \sin \theta = m\lambda$ for both first and second orders. For the second-order to exist, it must occur at θ < 90°.

Solution

For the first-order maximum (m = 1):

For the second-order maximum (m = 2) to be observable, it must occur at some angle θ₂ < 90°. At the limiting case where θ₂ = 90°:

With $\sin \theta_2 = \sin 90° = 1$:

Now substitute this back into the first-order equation:

Therefore: If θ₁ = 30.0°, the second-order maximum just barely appears at θ₂ = 90°. If θ₁ > 30.0°, then d < 2λ, which would require $\sin \theta_2 > 1$ for the second-order maximum - which is impossible.

Conclusion: A second-order maximum can only exist if the first-order maximum appears at θ₁ < 30.0°. Q.E.D.

Discussion

This result has important practical implications for diffraction grating design. If you observe that a first-order maximum appears at an angle greater than 30°, you immediately know that:

1. No second-order maximum exists for that wavelength
2. The grating spacing is less than twice the wavelength (d < 2λ)
3. At most, only first-order maxima will be observable (plus the zero-order central maximum)

For example, if violet light (380 nm) produces its first maximum at 35°, the grating spacing is only about 600 nm, and no higher orders can be observed for visible light.

Strategy

Use $d \sin \theta = m\lambda$ for both wavelengths. Find d from the violet condition, then use it to find the angle for red light.

Solution

Given:

• Shortest visible wavelength: $\lambda_V = 380 \text{ nm}$ (violet)
• Longest visible wavelength: $\lambda_R = 760 \text{ nm}$ (red)
• $\theta_V = 30.0°$ for violet light
• $m = 1$ (first-order maximum)

Step 1: Find the grating spacing from the violet condition

Step 2: Find the angle for red light

Discussion

The remarkable result is that the red end of the visible spectrum appears at exactly 90°! This is the limiting case where the first-order maximum just barely exists.

Notice that the grating spacing d = 760 nm exactly equals the wavelength of red light. This is the critical condition: when d = λ_max, the longest wavelength appears at 90°. If the grating had even slightly finer spacing (d < 760 nm), red light would not produce a first-order maximum at all.

This grating with d = 760 nm produces a complete visible spectrum spreading from 30° (violet) to 90° (red) in first order. The complete visible spectrum is compressed into a 60° angular range. This is actually the finest-spacing grating that can show the complete visible spectrum in first order - any finer and the red end would be cut off.

Strategy

For part (a), the smallest visible wavelength (violet, ~380 nm) must produce at least a first-order maximum before θ = 90°. For parts (b) and (c), we check if UV and IR wavelengths can produce maxima with this grating spacing.

Solution

(a) Maximum lines per centimeter:

The smallest visible wavelength is approximately λ = 380 nm (violet). For a first-order maximum to just barely exist at θ = 90°:

Maximum number of lines per centimeter:

Rounding: N ≈ 26,300 lines/cm

(b) Useful for ultraviolet spectra?

UV wavelengths are shorter than visible light (λ < 380 nm). For example, UV-A is around 315-380 nm.

Since d = 380 nm and λ_UV < 380 nm, we have d > λ_UV, so:

For first order (m = 1), sin θ < 1, so maxima will exist for UV light.

Yes, this grating would be useful for UV spectra. In fact, UV wavelengths would produce maxima at smaller angles than violet light, making them easier to observe.

(c) Useful for infrared spectra?

IR wavelengths are longer than visible light. For example, near-IR is around 760 nm - 2500 nm.

For λ_IR = 760 nm with d = 380 nm:

Since sin θ cannot exceed 1, no first-order maximum exists for 760 nm IR light.

No, this grating would NOT be useful for infrared spectra because d < λ for all IR wavelengths, making it impossible to produce even first-order maxima.

Discussion

This problem illustrates an important principle: gratings with very fine spacing (high line density) are excellent for short wavelengths (UV, violet) but useless for long wavelengths (IR, red). The ideal grating depends on the wavelength range of interest. For IR spectroscopy, gratings with much wider spacing (fewer lines/cm) are needed.

Strategy

For part (a), find the grating spacing d, then determine the maximum wavelength that can produce a first-order maximum (at θ = 90°). For part (b), this maximum wavelength is the answer. For part (c), the longest visible wavelength (760 nm) must produce a second-order maximum at θ < 90°.

Solution

(a) Show that 30,000 lines/cm won’t produce visible light maxima:

First, find the grating spacing:

For a first-order maximum to exist, we need θ ≤ 90°. At the limiting case (θ = 90°):

Since the visible spectrum ranges from 380 nm (violet) to 760 nm (red), and the maximum wavelength this grating can diffract is 333.3 nm (which is in the ultraviolet range), no visible light wavelengths can produce even a first-order maximum. All visible wavelengths exceed the grating spacing, making sin θ > 1, which is impossible.

(b) Longest wavelength for first-order maximum:

From part (a), the longest wavelength that can produce a first-order maximum is:

This is in the ultraviolet (UV) range.

(c) Greatest number of lines/cm for complete second-order visible spectrum:

For a complete second-order spectrum, the longest visible wavelength (760 nm, red) must have its second-order maximum at θ ≤ 90°. At the limiting case:

With m = 2 and λ_red = 760 nm:

Maximum number of lines per centimeter:

Rounding to three significant figures: N ≈ 6.58 × 10³ lines/cm

Discussion

Part (a) demonstrates an important limitation: extremely fine gratings (like 30,000 lines/cm with d = 333 nm) cannot diffract visible light at all because the wavelengths exceed the grating spacing. Such fine gratings are useful only for short-wavelength radiation like UV or X-rays.

For part (c), the result shows that to observe a complete second-order visible spectrum, the grating cannot have more than about 6,580 lines/cm. If it had more lines, the red end of the spectrum wouldn’t appear in second order. This is roughly half the maximum for a first-order spectrum (~13,200 lines/cm), which makes sense because second-order requires twice the path difference (2λ instead of λ).

Practical commercial gratings typically have 300-10,000 lines/cm, comfortably within the range to show complete visible spectra in both first and second orders.

Strategy

The CD acts as a reflection grating. We use $d \sin \theta = m\lambda$ with m = 1 for the first fringe. First, find the angle θ from the geometry, then solve for d. He-Ne laser wavelength is 633 nm.

Solution

Given:

• Distance to wall: L = 1.50 m
• Position of first fringe: y = 0.600 m
• λ = 633 nm (He-Ne laser)
• m = 1 (first-order maximum)

Step 1: Find the angle

From geometry:

Step 2: Find the groove spacing

Using $d \sin \theta = m\lambda$:

Discussion

The groove spacing of 1.71 μm is typical for CDs. The actual standard spacing for CDs is 1.6 μm (corresponding to about 625 grooves per mm). Our calculated value of 1.71 μm is quite close, with the small difference possibly due to:

• Measurement uncertainties in the fringe position
• The angle of incidence affecting the effective spacing
• Rounding in the given measurements

This demonstrates how CDs can be used as inexpensive diffraction gratings for educational demonstrations. The regular spacing of pits and lands on a CD creates a periodic structure that diffracts laser light, producing the colorful iridescent patterns you see when white light reflects from a CD surface.

Strategy

Use the formula for fringe spacing: $\Delta y = x\lambda/d$, where x is the distance to the screen. First, find d from the line density, then calculate Δy.

Solution

Given:

• Line density: 125 lines/cm
• $$ \lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m} $$
• Distance to screen: $x = 1.50 \text{ m}$

Step 1: Find the grating spacing

Step 2: Calculate the fringe spacing

Using $\Delta y = \frac{x\lambda}{d}$:

Discussion

The fringe spacing of 1.13 cm is quite large and easily observable. This is because the grating has a relatively coarse spacing (125 lines/cm = 80 μm between lines), which is about 133 wavelengths. The formula Δy = xλ/d shows that:

• Larger screen distance (x) increases fringe spacing
• Larger wavelength (λ) increases fringe spacing
• Smaller slit separation (d) increases fringe spacing

With adjacent fringes separated by over 1 cm at a distance of 1.5 m, this pattern would be very easy to observe and measure. Finer gratings (more lines/cm) would produce smaller fringe spacings. For example, a 1000 lines/cm grating would have fringes separated by only 1.4 mm at the same distance.

Strategy

Use $d \sin \theta = m\lambda$ for double-slit interference with m = 1. Examine whether the result is physically possible.

Solution

(a) Angle for first-order maximum:

Given:

• $$ \lambda = 700 \text{ nm} = 700 \times 10^{-9} \text{ m} $$
• $$ d = 400 \text{ nm} = 400 \times 10^{-9} \text{ m} $$
• $$ m = 1 $$

Using $d \sin \theta = m\lambda$:

This requires $\theta = \sin^{-1}(1.75)$, which is undefined (no solution exists).

(b) What is unreasonable about this result?

The sine function can only have values between -1 and +1 for real angles. A value of sin θ = 1.75 is impossible - there is no real angle whose sine equals 1.75. This means no first-order maximum can exist for this configuration.

(c) Which assumptions are unreasonable or inconsistent?

The unreasonable assumption is that a slit separation of 400 nm can produce an interference pattern for 700-nm light.

For a double-slit interference pattern to show at least one maximum (besides the central maximum), we need:

In this problem, d = 400 nm < λ = 700 nm. When the slit separation is smaller than the wavelength, no off-axis maxima can occur. The only observable feature would be the central maximum (m = 0), which exists for all wavelengths.

The inconsistency is assuming that red light (700 nm) with a relatively long wavelength can create a normal double-slit pattern with slits separated by only 400 nm. Either:

• The wavelength should be shorter (≤ 400 nm, like UV or violet light), OR
• The slit separation should be larger (≥ 700 nm)

Discussion

This problem illustrates an important physical limitation: you cannot create interference patterns of a certain order if the wavelength exceeds the slit separation. In practical terms, attempting this experiment would result in only seeing a broad central diffraction maximum with no side fringes - essentially no interference pattern. This is why diffraction gratings and double-slit apparatus must have appropriate spacing for the wavelength being studied.

Strategy

Use $d \sin \theta = m\lambda$ to find the wavelength. First, calculate the grating spacing d from the line density, then solve for λ with m = 4. Examine whether the result makes sense.

Solution

(a) Calculate the wavelength:

Given:

• Line density: 25,000 lines/cm
• $$ \theta = 25.0° $$
• $m = 4$ (fourth-order maximum)

Step 1: Find the grating spacing

Step 2: Find the wavelength

Using $d \sin \theta = m\lambda$:

(b) What is unreasonable about this result?

The calculated wavelength of 42.3 nm is NOT visible light. The visible spectrum ranges from approximately 380 nm (violet) to 760 nm (red). A wavelength of 42.3 nm falls in the extreme ultraviolet (EUV) range, which is far beyond what the human eye can detect. The problem asks for a “visible wavelength,” but the result is not visible at all.

(c) Which assumptions are unreasonable or inconsistent?

There are two unreasonable aspects to this problem:

1. The grating is too fine for visible light diffraction:

The grating spacing is only 400 nm, which is barely larger than the shortest visible wavelength (380 nm). For a fourth-order maximum to exist for visible light, we would need:

For the shortest visible wavelength (380 nm):

No visible wavelength can produce even a first-order maximum with a 400 nm grating spacing at any reasonable angle, let alone a fourth-order maximum at 25°.

2. Practical manufacturing limitations:

A grating with 25,000 lines/cm (d = 400 nm) is at the very edge of what can be manufactured using advanced nanofabrication techniques. While modern electron-beam lithography can achieve features around 50 nm, creating uniform, large-area diffraction gratings with such fine spacing is extremely challenging and expensive. Such gratings would be more suitable for X-ray or EUV spectroscopy, not visible light.

The inconsistent assumptions are:

• Expecting visible light to produce a fourth-order maximum with such fine grating spacing
• Assuming 25,000 lines/cm is reasonable for a visible-light spectrometer (it’s far too fine)
• For visible light spectroscopy, gratings typically have 300-10,000 lines/cm

Discussion

This problem highlights the importance of matching grating spacing to wavelength. The grating spacing should be several times larger than the wavelength of interest to produce observable higher-order maxima. A grating with d = 400 nm is appropriate for ultraviolet or extreme ultraviolet radiation (wavelengths < 100 nm), not for visible light. For visible light and fourth-order maxima, a grating with d ≥ 4 × 760 nm = 3040 nm (corresponding to ~3300 lines/cm or fewer) would be more appropriate.