/*! 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 102 Monochromatic light (wavelength ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Monochromatic light (wavelength \(=450 \mathrm{~nm}\) ) is incident perpendicularly on a single slit (width \(=0.40 \mathrm{~mm}\) ). A screen is placed parallel to the slit plane, and on it the distance between the two minima on either side of the central maximum is \(1.8 \mathrm{~mm}\). (a) What is the distance from the slit to the screen? (Hint: The angle to either minimum is small enough that \(\sin \theta \approx \tan \theta .\) ) (b) What is the distance on the screen between the first minimum and the third minimum on the same side of the central maximum?

Short Answer

Expert verified
(a) 0.8 m; (b) 1.8 mm

Step by step solution

01

Understand the Problem

We have monochromatic light with wavelength \( \lambda = 450 \mathrm{~nm} \) passing through a single slit with a width \( a = 0.40 \mathrm{~mm} \). The screen is placed parallel to the slit. We need to calculate (a) the distance from the slit to the screen, given that the distance between the first minima on either side of the central maximum is \( 1.8 \mathrm{~mm} \), and (b) the distance between the first and third minima on the same side of the central maximum.
02

Apply Formula for Minima

For single slit diffraction, the condition for minima is given by \( a \sin \theta = m \lambda \), where \( m \) is the order of the minimum (an integer \( m = \pm 1, \pm 2, \pm 3, \ldots\)).
03

Determine Small Angle Approximation

Since the angle \( \theta \) is very small, we use the approximation \( \sin \theta \approx \tan \theta \). This simplifies the calculation, as \( \tan \theta = \frac{x}{L} \), where \( x \) is the distance on the screen from the central maximum to the first minimum.
04

Calculate Distance from Slit to Screen (a)

First, substitute \( a \sin \theta = m \lambda \) with \( a \tan \theta = m \lambda \), simplifying to \( \frac{am \lambda}{L} = x \), where \( x = 0.9 \mathrm{~mm} \) (half the distance for \( m = \pm 1 \)).\a. \( m = \pm 1 \)\\( x = \frac{\lambda L}{a} \ L = \frac{a \cdot x}{\lambda} = \frac{0.40 \times 10^{-3} \times 0.9 \times 10^{-3}}{450 \times 10^{-9}} \approx 0.8 \mathrm{~m}\).
05

Calculate Distance for First to Third Minima (b)

For the third minimum on the same side, \( m = 3 \). The distance on the screen from the central maximum to this point is calculated as \( x_3 = \frac{3\lambda L}{a} \). \b. \( x_3 = \frac{3 \times 450 \times 10^{-9} \times 0.8}{0.40 \times 10^{-3}} \approx 2.7 \mathrm{~mm}\) \The distance between the first (\( x_1 = 0.9 \mathrm{~mm} \)) and third minimum (\( x_3 = 2.7 \mathrm{~mm} \)) on the same side of the central maximum is \( x_3 - x_1 = 2.7 - 0.9 = 1.8 \mathrm{~mm} \).

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.

Monochromatic Light
Monochromatic light refers to light composed of a single wavelength or color. In the context of diffraction, using monochromatic light simplifies observations of interference patterns, as there are no overlapping waves of different wavelengths to complicate the pattern. This results in clear and distinct diffraction patterns that are easier to analyze and predict. For this exercise, the wavelength of the light used is specified as 450 nm, ensuring that the light is purely one color within the visible spectrum.range. When passed through a slit, this monochromatic light produces a distinct pattern of maxima and minima on a screen, which is ideal for studying the effects of diffraction.
Wavelength
The wavelength of light, denoted by the symbol \( \lambda \), represents the distance between two consecutive peaks or troughs of a wave. It is a critical factor in diffraction experiments as it directly influences the pattern of light observed on a screen. In this exercise, the wavelength is given as 450 nm, which is a typical wavelength within the blue part of the visible light spectrum. Light's wavelength affects the positions of minima and maxima in diffraction patterns due to constructive and destructive interference: as wavelength increases, the separation of these features on the screen also increases. Therefore, knowing the wavelength is essential for predicting the positions of these patterns when light passes through a single slit.
Screen Distance
The distance from the slit to the screen, labeled as \( L \), plays a crucial role in determining where the diffraction pattern will appear. It influences the angle \( \theta \), which is related to how far apart the minima and maxima will spread on the screen. The greater the distance \( L \), the more spread out the diffraction pattern becomes. Using the small angle approximation, where \( \sin \theta \approx \tan \theta \approx \frac{x}{L} \) (with \( x \) as the distance on the screen from the central maximum to a minimum), allows for simpler calculations, especially when \( \theta \) is small. In the exercise, by observing the positions of the first minima, the distance \( L \) is calculated to be approximately 0.8 m, signifying how essential this measure is for accurately mapping the diffraction pattern.
Minima Calculation
Calculating the position of minima in a diffraction pattern is a fundamental aspect of understanding single slit diffraction phenomena. Minima occur at certain angles \( \theta \), which can be calculated using the equation \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \lambda \) is the light's wavelength, and \( m \) is the order of the minimum. Here, \( m \) is typically an integer, and solutions are found for all \( m = \pm 1, \pm 2, \pm 3, \ldots \). However, for small angles, the approximation \( \sin \theta \approx \frac{x}{L} \) allows us to write \( x = \frac{m \lambda L}{a} \), simplifying calculations. This lets us find the position \( x \) on the screen, where each minimum order is observed. For example, the distance between the first and third minima on the same side is determined using this approach, showcasing the utility of this formula in practical applications.

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

Perhaps to confuse a predator, some tropical gyrinid beetles (whirligig beetles) are colored by optical interference that is due to scales whose alignment forms a diffraction grating (which scatters light instead of transmitting it). When the incident light rays are perpendicular to the grating, the angle between the firstorder maxima (on opposite sides of the zeroth-order maximum) is about \(26^{\circ}\) in light with a wavelength of \(550 \mathrm{~nm}\). What is the grating spacing of the beetle?

The wall of a large room is covered with acoustic tile in which small holes are drilled \(5.0 \mathrm{~mm}\) from center to center. How far can a person be from such a tile and still distinguish the individual holes, assuming ideal conditions, the pupil diameter of the observer's eye to be \(4.0 \mathrm{~mm}\), and the wavelength of the room light to be \(550 \mathrm{~nm}\) ?

In the single-slit diffraction experiment of Fig. \(36-4\), let the wavelength of the light be \(500 \mathrm{~nm}\), the slit width be \(6.00 \mu \mathrm{m}\), and the viewing screen be at distance \(D=3.00 \mathrm{~m}\). Let a \(y\) axis extend upward along the viewing screen, with its origin at the center of the diffraction pattern. Also let \(I_{P}\) represent the intensity of the diffracted light at point \(P\) at \(y=15.0 \mathrm{~cm} .\) (a) What is the ratio of \(I_{P}\) to the intensity \(I_{m}\) at the center of the pattern? (b) Determine where point \(P\) is in the diffraction pattern by giving the maximum and minimum between which it lies, or the two minima between which it lies.

Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to \(2000 \mathrm{~km}\). One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of \(1.40 \mathrm{~nm}\). The element that emits light is the end of a wire with diameter \(0.200 \mathrm{~mm}\). (a) Calculate the diameter of the central beam at a target \(2000 \mathrm{~km}\) away from the beam source. (b) What is the ratio of the beam intensity at the target to that at the end of the wire? (The laser is fired from space, so neglect any atmospheric absorption.)

A plane wave of wavelength \(590 \mathrm{~nm}\) is incident on a slit with a width of \(a=0.40 \mathrm{~mm}\). A thin converging lens of focal length \(+70 \mathrm{~cm}\) is placed between the slit and a viewing screen and focuses the light on the screen. (a) How far is the screen from the lens? (b) What is the distance on the screen from the center of the diffraction pattern to the first minimum?

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.