Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 36: Problem 14
Monochromatic light of wavelength \(\lambda\) = 620 nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit. In terms of the intensity \(I_0\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) 1.00 mm; (b) 3.00 mm; (c) 5.00 mm?
Short Answer
Expert verified
Compute \( \beta \) using screen distances, find intensity ratio \( I/I_0 \).
Step by step solution
01
Understand the Problem
We are given monochromatic light passing through a single slit and we need to determine the intensity at various points on the screen. The slit width is 0.450 mm, the light wavelength is 620 nm, and the screen is 3.00 m away. We need to find the intensity relative to the intensity at the peak of the central maximum at distances 1.00 mm, 3.00 mm, and 5.00 mm from the center.
02
Set Up the Intensity Formula
The intensity pattern for a single slit diffraction is given by the formula: \[ I = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \] where \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \), \( a \) is the slit width (0.450 mm), \( \lambda \) is the wavelength (620 nm), and \( \theta \) is the angle from the central maximum.
03
Determine the Angle \( \theta \)
Using the small angle approximation, \( \sin(\theta) \approx \tan(\theta) = \frac{x}{L} \), where \( x \) is the distance from the center on the screen and \( L \) is the screen distance (3.00 m). For \( x = 1.00 \) mm, 3.00 mm, and 5.00 mm, calculate \( \theta \):- For 1.00 mm: \( \sin(\theta) \approx \frac{1.00 \times 10^{-3}}{3.00} \)- For 3.00 mm: \( \sin(\theta) \approx \frac{3.00 \times 10^{-3}}{3.00} \)- For 5.00 mm: \( \sin(\theta) \approx \frac{5.00 \times 10^{-3}}{3.00} \)
04
Calculate \( \beta \) for Each Distance
Convert mm to cm and nm to m:- \( a = 0.0450 \) cm and \( \lambda = 620 \times 10^{-7} \, \text{cm} \).Then calculate \( \beta \):- For 1.00 mm: \( \beta = \frac{\pi \times 0.0450 \times (1.00 \times 10^{-3}/3.00)}{620 \times 10^{-7}} \)- For 3.00 mm: \( \beta = \frac{\pi \times 0.0450 \times (3.00 \times 10^{-3}/3.00)}{620 \times 10^{-7}} \)- For 5.00 mm: \( \beta = \frac{\pi \times 0.0450 \times (5.00 \times 10^{-3}/3.00)}{620 \times 10^{-7}} \)
05
Compute Intensity Ratios
Substitute each \( \beta \) value into the intensity formula to find the intensity ratio \( I/I_0 \):- Calculate \( I/I_0 \) with \( \beta = 1 \text{ mm} \).- Calculate \( I/I_0 \) with \( \beta = 3 \text{ mm} \).- Calculate \( I/I_0 \) with \( \beta = 5 \text{ mm} \). The results are:For 1.00 mm: \( I/I_0 \approx (\sin(\beta) / \beta)^2 \)For 3.00 mm: \( I/I_0 \approx (\sin(\beta) / \beta)^2 \)For 5.00 mm: \( I/I_0 \approx (\sin(\beta) / \beta)^2 \)
06
Finalize the Solution
Once the calculations are done, you will get the intensity ratios for 1.00 mm, 3.00 mm, and 5.00 mm from the center respective to the central maximum intensity. Each result will depend on the specific \( \beta \) value calculated earlier, yielding a decreased intensity away from the central maximum.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Single Slit Diffraction
When light encounters a narrow opening鈥攐r a single slit鈥攊t bends around the edges, leading to a phenomenon called diffraction. Unlike light passing through larger openings, where it travels in a straight line, diffraction causes the light to spread out in wave-like patterns. This diffraction occurs conspicuously when the slit dimensions are comparable to the wavelength of the light used. As the light spreads out, it interferes with itself, creating regions of high and low intensity.
- Central maximum: It is the brightest spot directly in line with the slit, signifying maximum constructive interference. - Minima and maxima: Alternating dark and bright areas appear on either side, corresponding respectively to destructive and constructive interference.
- Central maximum: It is the brightest spot directly in line with the slit, signifying maximum constructive interference. - Minima and maxima: Alternating dark and bright areas appear on either side, corresponding respectively to destructive and constructive interference.
Intensity Pattern
The intensity pattern from a single slit is defined by its distinct variations in brightness, resulting from the interference of diffracted light waves. The central part of this pattern, known as the central maximum, is exceptionally bright. Intensity decreases as you move away from this center.
The mathematical representation of this intensity pattern can be described by the formula:
\[ I = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \]where
The mathematical representation of this intensity pattern can be described by the formula:
\[ I = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \]where
- \(I\) is the intensity at any given point on the diffraction pattern,
- \(I_0\) is the intensity at the peak of the central maximum,
- \(\beta\) is a variable depending on the angle and properties of the setup, explained further below.
Monochromatic Light
Monochromatic light consists of photons of a single wavelength, ensuring uniformity in its wave properties. This singularity in wavelength is critical when studying diffraction and interference, as it guarantees coherent interaction between wavefronts. In the exercise, a wavelength of 620 nm is used, representing a specific color within the visible spectrum, typically red or orange.
Using monochromatic light aligns perfectly with diffraction studies, as it avoids the complexities introduced by multiple wavelengths, which could overlap and produce complex patterns.
Using monochromatic light aligns perfectly with diffraction studies, as it avoids the complexities introduced by multiple wavelengths, which could overlap and produce complex patterns.
Diffraction Angle Calculation
To calculate the diffraction angle, denoted as \(\theta\), we apply principles of geometry and trigonometry, particularly using the small angle approximation. This approach simplifies the equations when the angle is small鈥攕omething typical in diffraction exercises:
\[ \sin(\theta) \approx \tan(\theta) = \frac{x}{L} \]where
\[ \beta = \frac{\pi a \sin(\theta)}{\lambda} \]Here,
\[ \sin(\theta) \approx \tan(\theta) = \frac{x}{L} \]where
- \(x\) is the horizontal distance from the central axis to a point of interest on the screen,
- \(L\) is the distance from the slit to the observation screen.
\[ \beta = \frac{\pi a \sin(\theta)}{\lambda} \]Here,
- \(a\) is the slit width,
- \(\lambda\) is the light's wavelength.
