Skip to main content
Ch 36: Diffraction
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Young & Freedman Calc 14th EditionCh 36: Diffraction문제 17b
36장, 문제 17b

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 mm wide. At the point in the pattern 3.25° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. What is the intensity at this point, if the intensity at the center of the central maximum is I0?

검증된 단계별 안내
1
Understand the problem: We are dealing with a single-slit diffraction pattern. The intensity at a given angle θ from the central maximum is determined by the interference of wavelets originating from different parts of the slit. The phase difference between the wavelets is given, and we need to calculate the intensity at this angle relative to the central maximum intensity I₀.
Recall the formula for the intensity in a single-slit diffraction pattern: \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \), where \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \). Here, \( a \) is the slit width, \( \lambda \) is the wavelength of the light, and \( \theta \) is the angle from the central maximum.
Relate the given phase difference \( \Delta \phi \) to the parameter \( \beta \). The phase difference \( \Delta \phi \) is related to \( \beta \) by \( \Delta \phi = 2\beta \). Using the given \( \Delta \phi = 56.0 \) rad, calculate \( \beta \) as \( \beta = \frac{\Delta \phi}{2} \).
Substitute \( \beta \) into the intensity formula. Compute \( \sin(\beta) \) and divide it by \( \beta \). Then square the result to find the relative intensity \( \frac{I(\theta)}{I_0} = \left( \frac{\sin(\beta)}{\beta} \right)^2 \).
Interpret the result: The value of \( \frac{I(\theta)}{I_0} \) represents the fraction of the central maximum intensity at the given angle. Multiply this fraction by \( I_0 \) to find the intensity \( I(\theta) \) at the specified angle.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
5m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Single-Slit Diffraction

Single-slit diffraction occurs when waves pass through a narrow opening, causing them to spread out and create a pattern of light and dark fringes on a screen. The width of the slit and the wavelength of the light determine the characteristics of the diffraction pattern, including the angular position of the minima and maxima.
추천 영상:
가이드 코스
12:33
Single Slit Diffraciton

Phase Difference

Phase difference refers to the difference in the phase of two wavefronts at a given point in space. In the context of diffraction, it is crucial for determining how wavelets from different parts of the slit interfere with each other, affecting the intensity of the resulting pattern. A phase difference of 56.0 rad indicates how much the waves are out of sync at a specific angle.
추천 영상:
가이드 코스
08:59
Phase Constant of a Wave Function

Intensity of Light

The intensity of light is a measure of the power per unit area carried by a wave, often represented as I. In diffraction patterns, the intensity at any point can be calculated using the central maximum intensity (I0) and the phase difference, which influences constructive and destructive interference. The intensity decreases as one moves away from the central maximum due to the varying contributions of the wavelets.
추천 영상:
가이드 코스
02:32
Wave Intensity
관련 실천
교과서 질문

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 mm wide. At the point in the pattern 3.25° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. What is the wavelength of the radiation?

1856
views
교과서 질문

Laser light of wavelength 500.0 nm illuminates two identical slits, producing an interference pattern on a screen 90.0 cm from the slits. The bright bands are 1.00 cm apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

2071
views
1
rank
교과서 질문

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ±17.8° from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

383
views
교과서 질문

Parallel rays of monochromatic light with wavelength 568 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 5.00 x 10-4 W/m2, what is the intensity at a point on the screen that is 0.900 mm from the center of the central maximum?

2026
views
교과서 질문

Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum (θ = 0°) is 4.00x10-5 W/m2. What is the intensity at a point on the screen that corresponds to θ = 1.20°?

2251
views
교과서 질문

A slit 0.240 mm wide is illuminated by parallel light rays of wavelength 540 nm. The diffraction pattern is observed on a screen that is 3.00 m from the slit. The intensity at the center of the central maximum (θ = 0°) is 6.00 x 10-6 W/m2. What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

1770
views