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

The wavelength range of the visible spectrum is approximately 380-750 nm. White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (\(Note\): An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example 36.4.)

Short Answer

Expert verified
In the first order, the angular width is \( \Delta \theta_1 \), and in the third order, it is \( \Delta \theta_3 \). These are calculated using the diffraction equation.

Step by step solution

01

Understanding the Problem

The problem involves calculating the angular width of the visible spectrum when white light passes through a diffraction grating. We are given the range of the visible spectrum (380 nm to 750 nm) and the number of slits per millimeter on the grating (350 slits/mm). The task is to find the angular width for the first and third order diffraction.
02

Calculate Grating Spacing

The first step in solving this problem is to convert the number of slits per millimeter to grating spacing. This is done using the formula \[ d = \frac{1}{\text{slits/mm}} \], where \( d \) is the spacing in meters. Thus, \( d = \frac{1}{350,000} \) meters because there are 1,000 millimeters in a meter.
03

Implement the Diffraction Equation

For diffraction, the equation \( d \sin(\theta) = m \lambda \) is used, where \( \theta \) is the angle of diffraction, \( m \) is the order number, and \( \lambda \) is the wavelength. We need to calculate \( \theta \) for both the shortest wavelength (380 nm) and the longest wavelength (750 nm) separately in both the first (\( m = 1 \)) and third order (\( m = 3 \)).
04

Calculate First Order Angles

For \( m = 1 \):Using \( \lambda = 380 \) nm, \[ \theta_1 = \arcsin\left( \frac{1 \times 380 \times 10^{-9}}{\frac{1}{350,000}} \right) \].Similarly, for \( \lambda = 750 \) nm, \[ \theta_2 = \arcsin\left( \frac{1 \times 750 \times 10^{-9}}{\frac{1}{350,000}} \right) \].Calculate these angles to find the angular spread \( \Delta \theta_1 = \theta_2 - \theta_1 \) for the first order.
05

Calculate Third Order Angles

For \( m = 3 \):Using \( \lambda = 380 \) nm, \[ \theta_1 = \arcsin\left( \frac{3 \times 380 \times 10^{-9}}{\frac{1}{350,000}} \right) \].Similarly, for \( \lambda = 750 \) nm, \[ \theta_2 = \arcsin\left( \frac{3 \times 750 \times 10^{-9}}{\frac{1}{350,000}} \right) \].Calculate these angles to find the angular spread \( \Delta \theta_3 = \theta_2 - \theta_1 \) for the third order.
06

Conclude with Angular Widths

Using the calculations from Steps 4 and 5, we find the angular width of the visible spectrum for the first order (\( m = 1 \)) as \( \Delta \theta_1 \) and for the third order (\( m = 3 \)) as \( \Delta \theta_3 \). These provide the desired angular spreads of light at first and third order.

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.

Angular Width
Angular width is an essential concept when discussing diffraction gratings and light dispersion. It refers to the spread of angles over which the light of different wavelengths is diffracted. In this exercise, we explore how white light's different wavelengths spread out as they pass through a diffraction grating. The angular width is defined as the difference between the angles of diffraction for the light's shortest and longest wavelengths within the specified order.

For example, in the first order diffraction, we calculate the diffraction angles for the wavelengths at 380 nm and 750 nm. The difference between these two angles is the angular width. This helps us understand how diverse the spread of colors will be in the spectrum that the grating projects. By comparing the angular width of different diffraction orders, we can appreciate why higher-order diffractions, like third order, can provide greater resolution despite the complications of overlapping spectra.
Visible Spectrum
The visible spectrum is a portion of the electromagnetic spectrum that is visible to the human eye. It typically ranges from approximately 380 nm to 750 nm in wavelength. When white light, which contains all the colors of the visible spectrum, is passed through a diffraction grating, it spreads out into its individual colors due to the variation in wavelengths.
  • This separation occurs because each wavelength is diffracted at a different angle. Shorter wavelengths, like violet, diffract at smaller angles than longer wavelengths, like red.
  • The visible spectrum’s spread through a diffraction grating provides a practical demonstration of the dispersion of light.
The visible spectrum is crucial for various applications, including optical instruments and astronomy, where distinguishing between different light wavelengths can offer insights into materials and physical conditions.
First Order Diffraction
First order diffraction refers to the first set of angles where constructive interference occurs as white light passes through a diffraction grating. In this order, each wavelength of the visible spectrum is diffracted once, resulting in the first occurrence of the full color spectrum being displayed.

The angle of diffraction for any wavelength is found through the formula \(d \sin(\theta) = m \lambda\), where \(m = 1\) for the first order. This equation shows the relationship between the spacing of the grating \(d\), the diffraction angle \(\theta\), and the wavelength \(\lambda\).
  • The angular width in the first order is calculated by determining the angles for the wavelengths at both ends of the spectrum—violet (380 nm) and red (750 nm).
  • The difference between the angles gives the angular spread or width.
Observing the first order diffraction is crucial for basic spectrum analysis, as it often provides a clear distinct view of the light’s color components.
Third Order Diffraction
Third order diffraction occurs when light interferes constructively for the third time, crossing beyond the first and second orders. It involves higher angles of diffraction compared to the first order. This results in potentially greater angular width, offering improved resolution of the visible spectrum. However, third order diffraction can overlap with higher wavelength light from earlier orders, which might complicate spectrum analysis.

By applying the diffraction equation \(d \sin(\theta) = m \lambda\) with \(m = 3\), we can calculate the angles of diffraction for the visible spectrum.
  • The increased angular spread in third order diffraction is advantageous for applications requiring detailed spectral analysis, like in spectroscopy where fine wavelength distinction is essential.
  • Despite its benefits, caution is advised due to potential overlapping of spectra from different orders.
Understanding third order diffraction helps in choosing appropriate settings for fine-resolution optical instruments, maximizing the clarity and detail of data provided by the visible spectrum.

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

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 wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture \(f/\)4.00 to photograph a bear that is 11.5 m away. Assume the wavelength is 550 nm. (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f/\)22.0, what would be the width of the smallest resolvable feature on the bear?

Coherent monochromatic light of wavelength l passes through a narrow slit of width \(a\), and a diffraction pattern is observed on a screen that is a distance \(x\) from the slit. On the screen, the width \(w\) of the central diffraction maximum is twice the distance \(x\). What is the ratio \(a/ \lambda\) of the width of the slit to the wavelength of the light?

Monochromatic electromagnetic radiation with wavelength \(\lambda\) from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width \(a\) if the wavelength is (a) 500 nm (visible light); (b) 50.0 \(\mu\)m (infrared radiation); (c) 0.500 nm (x rays)?

(a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\). Show that at a time \(t\), the electric field at a distant point \(P\) is $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ where \(E_0\) is the amplitude at \(P\) of the electric field due to an individual slit, \(\phi = (2\pi d\) sin \(\theta)/\lambda\), \(\theta\) is the angle of the rays reaching \(P\) (as measured from the perpendicular bisector of the slit arrangement), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d\). (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship \(e^{iz} = cos z + i \space sin \space z\), where \(i = \sqrt{-1}\). In this expression, cos \(z\) is the \(real\) \(part\) of the complex number \(e^{iz}\), and sin \(z\) is its \(imaginary\) \(part\). Show that the electric field \(E_P(t)\) is equal to the real part of the complex quantity $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that \(e^Ae^B = e^{(A+B)}\) and \((e^A)^n = e^{nA}\), show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship \(e^{iz}\) = cos \(z\) + \(i\) sin \(z\), show that the (real) electric field at point \(P\) is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ The quantity in the first square brackets in this expression is the amplitude of the electric field at \(P\). (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle \(\theta\) is $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ where \(I_0\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N\) = 2. It will help to recall that sin 2\(A\) = 2 sin \(A\) cos \(A\). Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. (\(Hint\): Is I0 defined in the same way in both expressions?)

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