/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Monochromatic light of wavelengt... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Monochromatic light of wavelength \(\lambda=620 \mathrm{nm}\) from a distant source passes through a slit 0.450 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) 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) \(1.00 \mathrm{mm} ;\) (b) 3.00 \(\mathrm{mm}\) ; (c) 5.00 \(\mathrm{mm}\) ?

Short Answer

Expert verified
Calculate \( \beta \) for each distance, then use \( I = I_0 (\sin(\beta)/\beta)^2 \) to find intensities at 1 mm, 3 mm, and 5 mm.

Step by step solution

01

Understand the Diffraction Formula

For light passing through a single slit, the intensity distribution can be described by the formula:\[ I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \]where \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \), \( a \) is the slit width, \( \lambda \) is the wavelength of the light, and \( \theta \) is the angle of diffraction.
02

Relate Angle and Distance for Small Angles

For small angles, the angle \( \theta \) can be related to distance \( y \) from the central maximum by the approximation:\[ \sin(\theta) \approx \tan(\theta) \approx \frac{y}{L} \]where \( L = 3.00 \ \mathrm{m} \) is the distance to the screen.
03

Calculate Intensity at 1.00 mm

Substitute \( y = 1.00 \ \mathrm{mm} = 1.00 \times 10^{-3} \ \mathrm{m} \) in the formula to find \( \theta \):\[ \sin(\theta) \approx \frac{1.00 \times 10^{-3}}{3.00} \]Now calculate \( \beta \):\[ \beta = \frac{\pi \times 0.450 \times 10^{-3} \times 1.00 \times 10^{-3}}{620 \times 10^{-9} \times 3.00} \]Compute \( \left( \frac{\sin(\beta)}{\beta} \right)^2 \) to get the intensity \( I \) relative to \( I_0 \).
04

Calculate Intensity at 3.00 mm

Substitute \( y = 3.00 \ \mathrm{mm} = 3.00 \times 10^{-3} \ \mathrm{m} \):\[ \sin(\theta) \approx \frac{3.00 \times 10^{-3}}{3.00} \]Now calculate \( \beta \) similarly:\[ \beta = \frac{\pi \times 0.450 \times 10^{-3} \times 3.00 \times 10^{-3}}{620 \times 10^{-9} \times 3.00} \]Compute the intensity \( I \) relative to \( I_0 \) using \( \left( \frac{\sin(\beta)}{\beta} \right)^2 \).
05

Calculate Intensity at 5.00 mm

Substitute \( y = 5.00 \ \mathrm{mm} = 5.00 \times 10^{-3} \ \mathrm{m} \):\[ \sin(\theta) \approx \frac{5.00 \times 10^{-3}}{3.00} \]Now calculate \( \beta \) again:\[ \beta = \frac{\pi \times 0.450 \times 10^{-3} \times 5.00 \times 10^{-3}}{620 \times 10^{-9} \times 3.00} \]Compute \( \left( \frac{\sin(\beta)}{\beta} \right)^2 \) to determine the intensity \( I \) as a fraction of \( I_0 \).

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 91Ó°ÊÓ!

Key Concepts

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

Intensity Distribution
When light waves pass through a narrow opening like a single slit, they spread out and form patterns of light and darkness on a screen—a phenomenon called diffraction. The intensity of light, or how bright it appears, at various points on the screen is not uniform but varies according to a specific mathematical formula. This distribution of light intensity can be described using the equation:\[ I(\theta) = I_0 \left( \frac{ \sin(\beta) }{ \beta } \right)^2 \]where:
  • \( I(\theta) \) is the light intensity at an angle \( \theta \).
  • \( I_0 \) is the intensity at the central maximum (the brightest point directly in line with the slit).
  • \( \beta \) is a parameter that relates to the slit width \( a \) and the wavelength \( \lambda \) of the light.
  • \( \beta = \frac{ \pi a \sin(\theta) }{ \lambda } \).
This equation shows how light intensity decreases with distance from the center of the central maximum. The central point is the brightest, and the intensity becomes less as you move towards the edges.
Diffraction Pattern
A beautiful outcome of passing light through a slit is the creation of a diffraction pattern. When you observe a diffraction pattern, you'll usually see a bright central band, known as the central maximum, surrounded by alternating dark and bright bands. These bands occur due to the way light waves overlap and interfere with each other upon emerging from the slit. As these waves spread out after passing through the slit, they can either constructively or destructively interfere with one another.
  • In constructive interference, light waves add up to enhance brightness, leading to the bright bands.
  • In destructive interference, waves cancel each other out, creating the dark areas in between.
This interference is why you'll often see distinct dark and bright areas on the screen, which collectively forms the diffraction pattern.
Small Angle Approximation
When studying single slit diffraction, something called the small angle approximation is often used to simplify calculations. In this context, when the angle \( \theta \) is relatively small (usually in situations where the slit-to-screen distance is large compared to the slit width), it can be assumed that:\[ \sin(\theta) \approx \tan(\theta) \approx \frac{y}{L} \]where:
  • \( y \) is the distance from the central maximum to a point on the screen.
  • \( L \) is the distance from the slit to the screen.
This approximation simplifies calculations because it allows us to use simple linear relations instead of trigonometric functions. For example, if you want to find the intensity at a specific point on the screen, the small angle approximation lets you relate \( \theta \) directly to \( y \) and \( L \), making the math much more manageable and the concepts easier to grasp.

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

A slit 0.360 \(\mathrm{mm}\) wide is illuminated by parallel rays of light that have a wavelength of 540 \(\mathrm{nm}\) . The diffraction pattern is observed on a screen that is 1.20 \(\mathrm{m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) 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 \(I_{0} / 2 ?\)

Identifying Isotones hy Snectra Different isotones of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is 656.45 nm; for deuterium, the corresponding wavelength is 656.27 \(\mathrm{nm} .\) (a) What minimum number of slits is required to resolve these two wavelengths in second order? (b) If the grating has 500.00 slits/mm, find the angles and angular separation of these two wavelengths in the second order.

If a diffraction grating produces its third-order bright band at an angle of \(78.4^{\circ}\) for light of wavelength \(681 \mathrm{nm},\) find (a) the number of slits per centimeter for the grating and (b) the angular location of the first-order and second-order bright bands. (c) Will there be a fourth-order bright band? Explain.

An interference pattern is produced by light of wave- length 580 \(\mathrm{nm}\) from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 \(\mathrm{mm}\) . (a) If the slits are very narrow, what would be the angular positions of the first-order and second- order, two-slit, interference maxima? (b) Let the slits have width 0.320 \(\mathrm{mm}\) . In terms of the intensity \(I_{0}\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

Monochromatic light from a distant source is incident on a slit 0.750 \(\mathrm{mm}\) wide. On a screen 2.00 \(\mathrm{m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 \(\mathrm{mm}\) . Calculate the wavelength of the light.

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.