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Chapter 36: Problem 75

Visible light is incident perpendicularly on a diffraction grating of 200 rulings/mm. What are the (a) longest, (b) second longest, and (c) third longest wavelengths that can be associated with an intensity maximum at \(\theta=30.0^{\circ}\) ?

Short Answer

Expert verified
(a) 2500 nm, (b) 1250 nm, (c) 833 nm.

Step by step solution

01

Understand the Diffraction Grating Equation

The diffraction grating equation that relates the angle of diffraction (\(\theta\)) with the wavelength (\(\lambda\)) and the order of the maximum (\(m\)) is given by \(d \sin \theta = m \lambda\), where \(d\) is the distance between adjacent rulings on the grating.
02

Calculate the Grating Spacing (d)

Given that the grating has 200 rulings per millimeter, we find the grating spacing \(d\) as follows: \[ d = \frac{1\text{ mm}}{200} = 5 \times 10^{-6} \text{ m} \].
03

Calculate the Wavelength for the First Order (m=1)

To find the longest wavelength, set \(m = 1\). Substitute into the equation \(d \sin \theta = m \lambda\): \[ \lambda_{1} = \frac{d \sin \theta}{1} = \frac{5 \times 10^{-6} \times \sin(30^{\circ})}{1} \]. Calculate \(\sin(30^{\circ}) = 0.5\), hence \[ \lambda_{1} = 2.5 \times 10^{-6} \text{ m} = 2500 \text{ nm}\].
04

Calculate the Wavelength for the Second Order (m=2)

For the second longest wavelength, let \(m = 2\). The wavelength \(\lambda_2\) is given by: \[ \lambda_{2} = \frac{d \sin \theta}{2} = \frac{5 \times 10^{-6} \times 0.5}{2} = 1.25 \times 10^{-6} \text{ m} = 1250 \text{ nm}\].
05

Calculate the Wavelength for the Third Order (m=3)

For the third longest wavelength, let \(m = 3\). The wavelength \(\lambda_3\) is given by: \[ \lambda_{3} = \frac{d \sin \theta}{3} = \frac{5 \times 10^{-6} \times 0.5}{3} \approx 0.833 \times 10^{-6} \text{ m} = 833 \text{ nm}\].

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Key Concepts

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

Wavelength Calculation
Wavelength calculation in a diffraction grating setup is a fundamental aspect of understanding how light interacts with a periodic structure. When light strikes a grating, it is diffracted, and the resulting pattern can be analyzed to determine the specific wavelengths present in the light source. To find these wavelengths, we use the formula derived from Young's double-slit experiment, where the condition for maximum intensity is defined at specific angles. The wavelengths correspond to different orders of these maxima. The diffraction equation that relates wavelength (\( \lambda \)) to grating spacing and other factors cleans up the path for calculation. By knowing the angle (\( \theta \)) at which these maxima occur, and the order of maximum (\( m \)) we're interested in, we can back-calculate the wavelength. For a given problem, this provides a powerful way to decompose a light wave into its constituent colors or wavelengths.
Grating Spacing
Grating spacing (\( d \)) is a critical parameter when dealing with diffraction gratings. Gratings consist of a large number of equally spaced parallel lines or rulings. The distance between these lines, called 'grating spacing,' radically affects how light is diffracted. To find this spacing, you need to know the number of lines per millimeter specified in the grating.For instance, if a grating has 200 rulings per millimeter, then the grating spacing is the reciprocal of this number in meters: \[ d = \frac{1\text{ mm}}{200} = 5 \times 10^{-6} \text{ m} \] The significance of this calculation is key. Smaller spacing means light can be diffracted at larger angles, allowing for a finer resolution in identifying wavelengths. Once you know the spacing, you can proceed with using the diffraction grating equation to find specific wavelengths of light at given angles.
Diffraction Equation
The diffraction equation is central to understanding how light behaves when it encounters a diffraction grating. This equation is typically written as:\[ d \sin \theta = m \lambda \] Here, \( d \) represents the grating spacing, \( \theta \) the diffraction angle, \( \lambda \) the wavelength, and \( m \) the order of the maximum.This formula essentially links the physical dimensions of the grating (via \( d \)) to the wavelength and its visible diffraction pattern. It allows us to calculate the wavelengths that correspond to different orders of intensity maxima visible at specific angles. By understanding this relationship, students can deepen their comprehension of light interference, enabling practical applications in spectroscopy, for instance. Working through the calculations helps tie theory directly to observable phenomena.
Order of Maximum
The order of maximum (\( m \)) refers to the multiple points where light intensity is maximized due to constructive interference in a diffraction pattern. Each order corresponds to an integer (\( m = 0, 1, 2, \ldots \)) that signifies repeated or higher-order diffraction peaks occurring due to the same wavelength.To discover these orders, solve the diffraction equation for different values of \( m \). The first order \( m=1 \) gives the longest wavelength, as \( m \) increases the wavelength associated with maxima decreases, hence higher orders correspond to shorter wavelengths. These orders play a pivotal role in experiments, especially in identifying unknown wavelengths. In scientific examinations like spectroscopy, they provide a meticulous way to separate and analyze light for understanding the constituents and properties of a light source.

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Most popular questions from this chapter

An x-ray beam of wavelength \(A\) undergoes first-order reflection (Bragg law diffraction) from a crystal when its angle of incidence to a crystal face is \(23^{\circ}\), and an \(\mathrm{x}\) -ray beam of wavelength \(97 \mathrm{pm}\) undergoes third-order reflection when its angle of incidence to that face is \(60^{\circ}\). Assuming that the two beams reflect from the same family of reflecting planes, find (a) the interplanar spacing and (b) the wavelength \(A\).

In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui, Hawaii, and reflected back from the shuttle Discovery as it sped by \(354 \mathrm{~km}\) overhead. The diameter of the central maximum of the beam at the shuttle position was said to be \(9.1 \mathrm{~m}\). and the beam wavelength was \(500 \mathrm{~nm}\). What is the effective diameter of the laser aperture at the Maui ground station? (Hint: A laser beam spreads only because of diffraction; assume a circular exit aperture.)

(a) A circular diaphragm \(60 \mathrm{~cm}\) in diameter oscillates at a frequency of \(25 \mathrm{kHz}\) as an underwater source of sound used for submarine detection. Far from the source, the sound intensity is distributed as the diffraction pattern of a circular hole whose diameter equals that of the diaphragm. Take the speed of sound in water to be \(1450 \mathrm{~m} / \mathrm{s}\) and find the angle between the normal to the diaphragm and a line from the diaphragm to the first minimum. (b) Is there such a minimum for a source having an (audible) frequency of \(1.0 \mathrm{kHz}\) ?

Babinet's principle. A monochromatic beam of parallel light is incident on a "collimating" hole of diameter \(x>\lambda\). Point \(P\) lies in the geometrical shadow region on a distant screen (Fig. \(36-39 a\) ). Two diffracting objects, shown in Fig. \(36-39 b\), are placed in turn over the collimating hole. Object \(A\) is an opaque circle with a hole in it, and \(B\) is the "photographic negative" of \(A\). Using superposition concepts, show that the intensity at \(P\) is identical for the two diffracting objects \(A\) and \(B\).

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?
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