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Ch 33: Wave Optics
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 33, Problem 31

Your artist friend is designing an exhibit inspired by circular-aperture diffraction. A pinhole in a red zone is going to be illuminated with a red laser beam of wavelength 670 nm, while a pinhole in a violet zone is going to be illuminated with a violet laser beam of wavelength 410 nm. She wants all the diffraction patterns seen on a distant screen to have the same size. For this to work, what must be the ratio of the red pinhole’s diameter to that of the violet pinhole?

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1
Understand the concept of circular-aperture diffraction: The size of the diffraction pattern is determined by the angular width of the central maximum, which depends on the wavelength of the light and the diameter of the aperture.
Recall the formula for the angular width of the central maximum in circular-aperture diffraction: \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of the light and \( D \) is the diameter of the aperture.
To ensure the diffraction patterns have the same size, the angular width \( \theta \) must be the same for both the red and violet beams. Set the angular widths equal: \( 1.22 \frac{\lambda_{\text{red}}}{D_{\text{red}}} = 1.22 \frac{\lambda_{\text{violet}}}{D_{\text{violet}}} \).
Simplify the equation by canceling the constant \( 1.22 \): \( \frac{\lambda_{\text{red}}}{D_{\text{red}}} = \frac{\lambda_{\text{violet}}}{D_{\text{violet}}} \). Rearrange to find the ratio of the diameters: \( \frac{D_{\text{red}}}{D_{\text{violet}}} = \frac{\lambda_{\text{red}}}{\lambda_{\text{violet}}} \).
Substitute the given wavelengths: \( \lambda_{\text{red}} = 670 \, \text{nm} \) and \( \lambda_{\text{violet}} = 410 \, \text{nm} \). The ratio of the diameters is \( \frac{D_{\text{red}}}{D_{\text{violet}}} = \frac{670}{410} \).

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Key Concepts

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

Diffraction

Diffraction is the bending of waves around obstacles and the spreading of waves when they pass through small openings. In the context of light, diffraction patterns arise when coherent light sources, like lasers, pass through apertures, creating interference patterns on a screen. The extent of diffraction is influenced by the wavelength of the light and the size of the aperture.
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Wavelength and Aperture Size

The wavelength of light is a critical factor in determining the diffraction pattern produced by an aperture. According to the principles of diffraction, a smaller aperture size relative to the wavelength results in more pronounced diffraction. To achieve similar diffraction patterns for different wavelengths, the ratio of the diameters of the apertures must be adjusted based on their respective wavelengths.
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Ratio of Aperture Diameters

To ensure that diffraction patterns from two different wavelengths appear the same size on a screen, the ratio of the diameters of the apertures must be inversely proportional to the ratio of the wavelengths. This means that if one wavelength is longer, its corresponding aperture must be smaller to maintain the same diffraction pattern size, allowing for a direct comparison of the two patterns.
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Related Practice
Textbook Question

Two 50-μm-wide slits spaced 0.25 mm apart are illuminated by blue laser light with a wavelength of 450 nm. The interference pattern is observed on a screen 2.0 m behind the slits. How many bright fringes are seen in the central maximum that spans the distance between the first missing order on one side and the first missing order on the other side?

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Textbook Question

Infrared light of wavelength 2.5 μm illuminates a 0.20-mm-diameter hole. What is the angle of the first dark fringe in radians? In degrees?

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Textbook Question

Light from a helium-neon laser (λ = 633 nm) passes through a circular aperture and is observed on a screen 4.0 m behind the aperture. The width of the central maximum is 2.5 cm. What is the diameter (in mm) of the hole?

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Textbook Question

Figure EX33.26 shows the light intensity on a screen behind a single slit. The wavelength of the light is 600 nm and the slit width is 0.15 mm. What is the distance from the slit to the screen?

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Textbook Question

Light of 600 nm wavelength passes through a single slit and creates a 2.0-cm-wide central maximum on a screen behind the slit. What wavelength of light will create a 3.0-cm-wide central maximum on a screen twice as far away?

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Textbook Question

You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon laser (λ=633 nm) and a 0.12-mm-diameter pinhole. How far behind the pinhole should you place the screen that's to be photographed?

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