Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \((\theta = 0^\circ)\) is 4.00 \(\times\) 10\(^{-5}\) W/m\(^2\). What is the intensity at a point on the screen that corresponds to \(\theta\) = 1.20\(^\circ\)?

Short Answer

Expert verified
The intensity at \( \theta = 1.20^\circ \) is approximately \( 1.33 \times 10^{-8} \text{ W/m}^2 \).

Step by step solution

01

Understand the Diffraction Equation

The intensity of light in a single-slit diffraction pattern is given by the equation: \[ I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \]where \( I_0 \) is the intensity at the center, \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \), and \( a \) is the slit width. We are tasked with finding \( I(1.20^\circ) \).
02

Convert Units and Calculate \( \beta \)

Convert wavelength \( \lambda = 592 \text{ nm} = 592 \times 10^{-9} \text{ m} \) and slit width \( a = 0.029 \text{ mm} = 0.029 \times 10^{-3} \text{ m} \). For \( \theta = 1.20^\circ \), \( \sin(\theta) \approx \sin(1.20^\circ) = 0.0209 \). Calculate:\[ \beta = \frac{\pi \times 0.029 \times 10^{-3} \times 0.0209}{592 \times 10^{-9}} \]
03

Simplify and Insert Values

Calculate \( \beta \) using the given values:\[ \beta = \frac{\pi \times 0.029 \times 10^{-3} \times 0.0209}{592 \times 10^{-9}} \approx 3.20 \]This will be used to find the new intensity at \( \theta = 1.20^\circ \).
04

Compute the Intensity Ratio

Using the intensity equation:\[ \frac{I(\theta)}{I_0} = \left( \frac{\sin(3.20)}{3.20} \right)^2 \]Calculate \( \sin(3.20) \) and divide by \( 3.20 \). Then square the result to find the intensity ratio.
05

Determine the Intensity at \( 1.20^\circ \)

Calculate \( \sin(3.20) \approx -0.0584 \) (note the negative value, but the square is taken):\[ \frac{I(1.20^\circ)}{4.00 \times 10^{-5}} = \left( \frac{-0.0584}{3.20} \right)^2 \approx \left( -0.01825 \right)^2 \approx 0.0003335 \]Therefore, \( I(1.20^\circ) = 4.00 \times 10^{-5} \times 0.0003335 \approx 1.33 \times 10^{-8} \text{ W/m}^2 \).

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diffraction Pattern
When light encounters an obstacle, such as a slit, it tends to bend and spread out. This is known as diffraction. In a single-slit diffraction setup, when monochromatic light passes through a narrow slit, it produces a characteristic pattern of alternating bright and dark bands on a screen, known as a diffraction pattern.
This pattern results from the constructive and destructive interference of light waves. The central bright band, called the central maximum, is the brightest part of the pattern. Flanking it are alternating bands of light (maxima) and darkness (minima), whose intensity gradually decreases with distance from the center.
Intensity Calculation
The intensity of light in a diffraction pattern is not uniform. It varies based on the angle \(\theta\)\, which measures the position on the screen relative to the central maximum. The intensity at any point in the pattern can be calculated using:
  • \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2\)
Where \(\beta = \frac{\pi a \sin(\theta)}{\lambda}\)\, \(I_0\) is the intensity at the central maximum, and \(\lambda\) is the wavelength. This equation accounts for the wave nature of light and demonstrates how intensity diminishes as one moves away from the central axis. Calculating \(\beta\) provides insight into intensity changes due to different angles and slit dimensions.
Monochromatic Light
Monochromatic light is light consisting of a single wavelength and frequency. This type of light is crucial for diffraction experiments because it ensures clear and distinguishable patterns.
In the given exercise, the light used has a wavelength of 592 nm. As all photons have the same energy and wavelength, the pattern produced is coherent and precise, aiding in accurate determination of diffraction characteristics such as maxima and minima positions.
Wavelength
The wavelength \(\lambda\) of light is the distance between consecutive peaks (or troughs) in a wave. It’s a vital parameter in understanding many optical phenomena, including diffraction. The wavelength determines the scale of the diffraction pattern.
  • Longer wavelengths create more spread-out patterns,
  • Shorter wavelengths yield narrower bands.
In our context, the light's wavelength is 592 nm, translating to 592 \(\times\) 10^{-9}\ m. This small scale emphasizes the microscopic nature of diffraction patterns.
Slit Width
Slit width is the width (denoted as \(a\)) of the opening through which light passes. It significantly influences the diffraction pattern's appearance. As the width changes, so does the interference pattern's distribution:
  • Narrower slits cause more pronounced diffraction, leading to wider separation between maxima.
  • Wider slits result in less diffraction and narrower spacing of the bands.
In this exercise, the slit width is 0.0290 mm, which defines the initial condition for the diffraction phenomena observed.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Monochromatic light with wavelength 620 nm passes through a circular aperture with diameter 7.4 \(\mu\)m. The resulting diffraction pattern is observed on a screen that is 4.5 m from the aperture. What is the diameter of the Airy disk on the screen?

When the light is passed through the bottom of the sample container, the interference maximum is observed to be at 41\(^\circ\); when it is passed through the top, the corresponding maximum is at 37\(^\circ\). What is the best explanation for this observation? (a) The microspheres are more tightly packed at the bottom, because they tend to settle in the suspension. (b) The microspheres aremore tightly packed at the top, because they tend to float to the top of the suspension. (c) The increased pressure at the bottom makes the microspheres smaller there. (d) The maximum at the bottom corresponds to \(m\) = 2, whereas the maximum at the top corresponds to \(m\) = 1.

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.350 mm wide. The diffraction pattern is observed on a screen 3.00 m away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

A slit 0.360 mm wide is illuminated by parallel rays of light that have a wavelength of 540 nm. The diffraction pattern is observed on a screen that is 1.20 m from the slit. The intensity at the center of the central maximum \((\theta = 0^\circ)\) is \(I_0\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_0\)/2?

The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0 \(\mu\)m in diameter) limits the size of an object at the near point (25 cm) of the eye to a height of about 50 \(\mu\)m. (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50-\(\mu\)m- tall object at 25 cm from the eye with light of wavelength 550 nm? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the 25-cm near point with light of wavelength 550 nm? (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes (60 min = 1\(^\circ\)), and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free