/*! 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 maxi- mum: (a) \(1.00 \mathrm{~mm}\) (b) \(3.00 \mathrm{~mm}\) (c) \(5.00 \mathrm{~mm} ?\)

Short Answer

Expert verified
The respective intensities at 1.00mm, 3.00mm and 5.00mm from the centre of the central maximum can be found by substituting these values one by one in the formula mentioned in step 2. Note that the angle \(\Theta\) and hence the intensity will change for each of these points.

Step by step solution

01

Calculate Angle: Step 1

First, calculate the angle \(\Theta\) which the light makes at the point of observation using the formula: \(\Theta = \tan^{-1}(Y/L)\). Here, Y is the distance from the centre of the central maximum (1.00mm, 3.00mm and 5.00mm respectively) and L is the distance of the screen from the slit which is 3.00m.
02

Find Intensity of Light: Step 2

Next, use the following formula derived from Babinet's principle and the approximation method of stationary phase to obtain the intensity of the light: \(I = I_{0} [\frac{\sin(\pi a \sin \Theta / \lambda)}{\pi a \sin \Theta / \lambda}]^{2}\). In here, \(I_{0}\) is the intensity at the peak of the central maximum, a is the width of the slit, \(\Theta\) is the angle calculated from the previous step, and \(\lambda\) is the wavelength of light.
03

Repeat for all Views: Step 3

After calculating the intensity for one point, repeat the steps for the other distances from the centre of 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 91Ó°ÊÓ!

Key Concepts

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

Monochromatic Light
When we talk about monochromatic light, we are referring to light that consists of a single wavelength. This type of light is crucial in experiments and calculations involving diffraction patterns because it ensures that the interference is coherent and predictable. Monochromatic light can be produced by lasers or by filtering light to isolate a single color. In the exercise, the monochromatic light has a wavelength of 620 nm, which falls within the visible red part of the spectrum.

Having a uniform wavelength guarantees that all photons of the light wave are in phase, which means that both the peaks and troughs of the light waves align. This phase alignment is necessary to produce distinct and measurable diffraction patterns when the monochromatic light passes through a slit in experiments like the one described in the textbook exercise.
Slit Diffraction
The phenomenon of slit diffraction is an illustration of the wave nature of light. It occurs when a wavefront of monochromatic light encounters an obstacle, such as a slit, that is comparable in size to its wavelength. When passing through this slit, the light waves spread out and interfere with one another, creating a pattern of bright and dark regions on a screen placed behind the slit. This pattern is called a diffraction pattern.

Factors influencing the sharpness and spacing of the diffraction fringes include the slit width and the wavelength of the incident light. In the exercise, a slit 0.450 mm wide is used, and the resulting diffraction pattern will vary in intensity across the screen, which is the basis for the experiment and subsequent calculations.
Babinet's Principle
Babinet's principle is a useful concept in the field of wave optics. It states that the diffraction pattern from an opaque body is identical to that from a hole of the same shape and size in an opaque screen, provided that the screen is illuminated from behind by a coherent light source. This principle allows us to predict diffraction patterns using complementary screens.

Even though Babinet's principle does not directly appear in the formula used in the exercise for intensity calculations, it underpins the concepts of diffraction patterns used to derive more complex formulas. Understanding this principle is key to predicting the behavior of light in diffraction-related phenomena.
Wavelength
Wavelength is a critical concept in the study of optics and is defined as the distance between consecutive crests or troughs in a wave. In the context of light, the wavelength determines the color we perceive and is typically measured in nanometers (nm). The wavelength is inversely proportional to the frequency of the light wave, meaning that shorter wavelengths correspond to higher frequencies and vice versa.

For the purposes of the exercise, the wavelength of the monochromatic light is given as 620 nm. The wavelength directly affects the diffraction pattern produced, with longer wavelengths resulting in wider spacing of the fringes in the pattern.
Central Maximum Intensity
The central maximum intensity, often denoted as is the brightest point in the diffraction pattern and corresponds to the perpendicular line from the slit to the observation screen or, in other words, the point of zero angle deviation. At the central maximum, all the diffracted waves constructively interfere, resulting in the maximum intensity of light.

In the context of the exercise, represents the maximum intensity at this central point. Calculating the intensity at different points away from the central maximum involves using the central maximum intensity as a reference point, as other points on the diffraction pattern will have a lower intensity due to the varying degrees of constructive and destructive interference.
Stationary Phase Approximation
The stationary phase approximation is a method used in physics for approximating the integral of oscillatory functions. It is particularly useful in wave optics for estimating the intensity of light in diffraction problems. It is based on the principle that the most significant contributions to the integral come from regions where the phase of the oscillations is stationary - that is, not varying rapidly.

While the step-by-step solution refers to using the approximation to find the light intensity, it is indirectly referring to the simplification of more complex wave integrals that underlie the formulation of the diffraction pattern. This approximation makes it possible to calculate intensity without cumbersome integral calculus, which is particularly advantageous for educational purposes.
Intensity Calculation
The calculation of intensity in the context of a diffraction pattern involves determining the distribution of light's power per unit area on a screen after it passes through a slit. Linked directly to the squared amplitude of the wave, intensity can be visualized as the brightness in the diffraction pattern.

The formula used in the exercise, derived from principles including Babinet's and the stationary phase approximation, allows for the computation of intensity at varying distances from the central maximum. The intensity calculation involves sinusoidal functions that account for the wave properties of light and yield the square of the ratio between the sine of an angle related to the observation point and the angle itself, all of which is squared and multiplied by the central maximum intensity to find the resulting intensity at various points on the screen.

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

Coherent electromagnetic waves with wavelength \(\lambda=500 \mathrm{nm}\) pass through two identical slits. The width of each slit is \(a,\) and the distance between the centers of the slits is \(d=9.00 \mathrm{~mm}\). (a) What is the smallest possible width \(a\) of the slits if the \(m=3 \max -\) imum in the interference pattern is not present? (b) What is the next larger value of the slit width for which the \(m=3\) maximum is absent?

Your physics study partner tells you that the width of the central bright band in a single-slit diffraction pattern is inversely proportional to the width of the slit. This means that the width of the central maximum increases when the width of the slit decreases. The claim seems counterintuitive to you, so you make measurements to test it. You shine monochromatic laser light with wavelength \(\lambda\) onto a very narrow slit of width \(a\) and measure the width \(w\) of the central maximum in the diffraction pattern that is produced on a screen \(1.50 \mathrm{~m}\) from the slit. (By "width," you mean the distance on the screen between the two minima on either side of the central maximum.) Your measurements are given in the table. $$ \begin{array}{l|llllllllr} a(\mu \mathrm{m}) & 0.78 & 0.91 & 1.04 & 1.82 & 3.12 & 5.20 & 7.80 & 10.40 & 15.60 \\ \hline w(\mathrm{~m}) & 2.68 & 2.09 & 1.73 & 0.89 & 0.51 & 0.30 & 0.20 & 0.15 & 0.10 \end{array} $$ (a) If \(w\) is inversely proportional to \(a\), then the product \(a w\) is constant, independent of \(a\). For the data in the table, graph \(a w\) versus \(a\). Explain why \(a w\) is not constant for smaller values of \(a\). (b) Use your graph in part (a) to calculate the wavelength \(\lambda\) of the laser light. (c) What is the angular position of the first minimum in the diffraction pattern for (i) \(a=0.78 \mu \mathrm{m}\) and (ii) \(a=15.60 \mu \mathrm{m} ?\)

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longer-wavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

Laser light of wavelength \(632.8 \mathrm{nm}\) falls normally on a slit that is \(0.0250 \mathrm{~mm}\) wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is \(8.50 \mathrm{~W} / \mathrm{m}^{2}\). (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

An interference pattern is produced by light of wavelength \(580 \mathrm{nm}\) 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)?

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.