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

Knight Calc 5th Edition

Ch 33: Wave Optics

Problem 22

Chapter 33, Problem 22

Light of 630 nm wavelength illuminates a single slit of width 0.15 mm. FIGURE EX33.22 shows the intensity pattern seen on a screen behind the slit. What is the distance to the screen?

Verified step by step guidance

Step 1: Understand the problem. The question involves diffraction through a single slit, and we are tasked with finding the distance to the screen. The intensity pattern shown in the graph is a result of the diffraction of light of wavelength 630 nm through a slit of width 0.15 mm.

Step 2: Recall the formula for the angular position of minima in single-slit diffraction: \( \sin \theta = \frac{m \lambda}{a} \), where \( m \) is the order of the minima (e.g., \( m = 1 \) for the first minimum), \( \lambda \) is the wavelength of light, and \( a \) is the slit width.

Step 3: Relate the angular position \( \theta \) to the linear position \( x \) on the screen using the small angle approximation: \( \tan \theta \approx \sin \theta \approx \frac{x}{L} \), where \( L \) is the distance to the screen and \( x \) is the distance from the central maximum to the first minimum on the screen.

Step 4: From the graph, observe that the distance \( x \) from the central maximum to the first minimum is approximately 0.5 cm (or 0.005 m). Substitute \( \sin \theta \approx \frac{m \lambda}{a} \) and \( \sin \theta \approx \frac{x}{L} \) into the equation to solve for \( L \): \( L = \frac{x a}{m \lambda} \). Use \( m = 1 \), \( \lambda = 630 \times 10^{-9} \) m, \( a = 0.15 \times 10^{-3} \) m, and \( x = 0.005 \) m.

Step 5: Perform the substitution and simplify the expression for \( L \). This will give the distance to the screen. Ensure units are consistent throughout the calculation.

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Key Concepts

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

Single Slit Diffraction

Single slit diffraction occurs when light passes through a narrow opening, causing the light waves to spread out and create an interference pattern. The width of the slit and the wavelength of the light determine the pattern's characteristics, including the position and intensity of the bright and dark fringes observed on a screen.

Wavelength and Frequency

Wavelength is the distance between successive peaks of a wave, while frequency is the number of peaks that pass a point in a given time. In this context, the wavelength of 630 nm (nanometers) is crucial for calculating the diffraction pattern, as it influences the angle and spacing of the intensity maxima and minima on the screen.

Intensity Pattern

The intensity pattern observed on the screen is a result of the constructive and destructive interference of light waves emanating from different parts of the slit. The central maximum is the brightest point, with subsequent maxima and minima decreasing in intensity, which can be quantitatively analyzed to determine the distance to the screen based on the slit width and wavelength.

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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

The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 500 lines/mm, and the light is observed on a screen 1.50 m behind the grating. What is the distance between the first-order red and blue fringes?

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Textbook Question

In a single-slit experiment, the slit width is 200 times the wavelength of the light. What is the width (in mm) of the central maximum on a screen 2.0 m behind the slit?

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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

FIGURE EX33.17 shows the interference pattern on a screen 1.0 m behind an 800 lines/mm diffraction grating. What is the wavelength (in nm) of the light?

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