Ch. 35 - Diffraction

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 35 - Diffraction

Problem 18

Chapter 34, Problem 18

Two 0.010-mm-wide slits are 0.030 mm apart (center to center). Determine (a) the spacing between interference fringes for 520-nm light on a screen 1.0 m away and (b) the distance between the two diffraction minima on either side of the central maximum of the envelope.

Verified step by step guidance

Step 1: Identify the given values for part (a). The slit separation (d) is 0.030 mm = 3.0 × 10⁻⁵ m, the wavelength of light (λ) is 520 nm = 5.2 × 10⁻⁷ m, and the distance to the screen (L) is 1.0 m. The formula for fringe spacing (Δy) in a double-slit interference pattern is Δy = (λ × L) / d.

Step 2: Substitute the given values into the formula for fringe spacing. Use Δy = (λ × L) / d, where λ = 5.2 × 10⁻⁷ m, L = 1.0 m, and d = 3.0 × 10⁻⁵ m. Simplify the expression to find Δy.

Step 3: For part (b), identify the given values for the diffraction envelope. The slit width (a) is 0.010 mm = 1.0 × 10⁻⁵ m, and the wavelength of light (λ) is 520 nm = 5.2 × 10⁻⁷ m. The formula for the angular position of the first diffraction minimum is sin(θ) = mλ / a, where m = ±1 for the first minima.

Step 4: Solve for the angular position (θ) of the first diffraction minima using sin(θ) = λ / a. Substitute λ = 5.2 × 10⁻⁷ m and a = 1.0 × 10⁻⁵ m into the equation. Then, calculate θ for both m = +1 and m = -1.

Step 5: To find the distance between the two diffraction minima on the screen, use the small-angle approximation, where y = L × tan(θ) ≈ L × sin(θ). Calculate the positions of the two minima (y₁ and y₂) on the screen and find the distance between them as |y₁ - y₂|.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Double-Slit Interference

Double-slit interference occurs when light waves pass through two closely spaced slits, creating an interference pattern on a screen. The pattern consists of alternating bright and dark fringes due to constructive and destructive interference of the light waves. The spacing between these fringes can be calculated using the formula: Δy = λL/d, where Δy is the fringe spacing, λ is the wavelength of light, L is the distance to the screen, and d is the distance between the slits.

Diffraction

Diffraction is the bending of waves around obstacles and the spreading of waves when they pass through narrow openings. In the context of the double-slit experiment, diffraction causes the light waves to spread out after passing through the slits, leading to a broader envelope of intensity that contains the interference pattern. The positions of the minima in the diffraction pattern can be determined using the formula: a sin(θ) = mλ, where a is the slit width, θ is the angle of the minima, m is the order of the minimum, and λ is the wavelength.

Interference and Diffraction Minima

In the context of the double-slit experiment, interference minima occur at points where the path difference between light from the two slits leads to destructive interference. Similarly, diffraction minima are points where the intensity of the light is significantly reduced due to the wave nature of light interacting with the slit width. The distance between these minima can be calculated based on the geometry of the setup and the wavelength of the light used.

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Diffraction

Related Practice

Textbook Question

(a) Derive an expression for the intensity in the interference pattern for three equally spaced slits. Express in terms of δ = 2πd sin θ / λ where d is the distance between adjacent slits and assume the slit width D ≈ λ.

(b) Show that there is only one secondary maximum between principal peaks.

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Textbook Question

Explain why the secondary maxima in the single-slit diffraction pattern do not occur precisely at β/2 = (m + 1/2)π where m = 1, 2, 3, ... Carefully and precisely plot the curves y = β/2 and y = tan β/2. From their intersections, determine the values of β for the first and second secondary maxima. What is the percent difference from β/2 = (m + 1/2)π?

Textbook Question

(III) Derive an expression for the intensity in the interference pattern for three equally spaced slits. Express in terms of δ = 2πd sin θ / λ where d is the distance between adjacent slits and assume the slit width D ≈ λ . Show that there is only one secondary maximum between principal peaks.

Textbook Question

In a double-slit experiment, let d = 5.00D = 40.0λ. Compare (as a ratio) the intensity of the third-order interference maximum with that of the zero-order maximum.

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Textbook Question

(a) Explain why the secondary maxima in the single-slit diffraction pattern do not occur precisely at β/2 = (m + 1/2)π where m = 1, 2, 3, ... .

(b) By differentiating Eq. 35–7 with respect to β show that the secondary maxima occur when β/2 satisfies the relation tan(β/2) = β/2.

(c) Carefully and precisely plot the curves y = β/2 and y = tan β/2. From their intersections, determine the values of β for the first and second secondary maxima. What is the percent difference from β/2 = (m + 1/2)π?

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Textbook Question

Monochromatic light of wavelength 633 nm falls on a slit. If the angle between the first bright fringes on either side of the central maximum is 32°, estimate the slit width.