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Chapter 27: Problem 49

A diffraction grating has 2604 lines per centimeter, and it produces a principal maximum at \(\theta=30.0^{\circ} .\) The grating is used with light that contains all wavelengths between 410 and 660 nm. What is (are) the wavelength(s) of the incident light that could have produced this maximum?

Short Answer

Expert verified
The wavelengths are 426 nm and 642 nm.

Step by step solution

01

Understand the Diagram Elements

A diffraction grating consists of many closely spaced lines. When light hits it, it bends or diffracts and produces principal maxima in certain directions (angles). Here, the diffraction angle (\(\theta\)) is \(30.0^{\circ}\).
02

Apply Grating Equation

The formula for the principal maxima in a diffraction grating is given by the grating equation: \(d \sin \theta = m \lambda\), where \(d\) is the distance between adjacent slits/grating spacing, \(\theta\) is the angle of the maximum, \(m\) is the order of the maximum, and \(\lambda\) is the wavelength of light.
03

Calculate Grating Spacing \(d\)

Given that there are 2604 lines per centimeter, we can calculate the spacing \(d\) by taking the reciprocal: \(d = \frac{1}{2604} \text{ cm} = \frac{1}{260400} \text{ m}\).
04

Solve for Wavelength \(\lambda\)

Substitute the known values into the grating equation: \[\frac{1}{260400} \sin(30.0^{\circ}) = m\lambda\].Since \(\sin(30.0^{\circ}) = 0.5\), we have \[\frac{0.5}{260400} = m\lambda\].Solving for \(\lambda\) gives us \[\lambda = \frac{0.5}{260400m}\].
05

Evaluate Possible \(m\) Values

Given that \(\lambda\) must be between 410 nm and 660 nm, calculate possible \(m\) values: - For \(\lambda = 410 \text{ nm} = 410 \times 10^{-9} \text{ m}\), rearranging gives \[m = \frac{0.5}{410 \times 10^{-9} \times 260400}\].- Similarly, calculate for \(\lambda = 660 \text{ nm}\).
06

Confirm Valid Wavelengths

Calculations show:- If \(m = 1\), \(\lambda \approx 642 \text{ nm}\)- If \(m = 2\), \(\lambda \approx 426 \text{ nm}\), both within 410 nm to 660 nm.

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Key Concepts

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

Grating Equation
The grating equation is fundamental when studying diffraction patterns in optics. This equation is expressed as \( d \sin \theta = m \lambda \), where:
  • \(d\) represents the distance between adjacent slits, or the grating spacing,
  • \(\theta\) is the angle at which a particular order of maximum intensity occurs,
  • \(m\) is the order of the maxima (an integer value),
  • \(\lambda\) is the wavelength of the light.

The role of this equation is to relate the geometrical arrangement of a diffraction grating to the physical properties of the light passing through it. By solving this equation, we can predict where the light will produce bright spots, known as principal maxima, on a screen. The angle and spacing dictate which wavelengths of light will constructively interfere, leading to observable maxima.
Wavelength Calculation
Calculating the wavelength using the grating equation involves knowing the grating spacing \(d\) and the angle \(\theta\) of diffraction. For instance, if a grating has 2604 lines per centimeter, the spacing \(d\) can be calculated by taking the reciprocal, which converts the lines to meters 鈥嬧�媆( d = \frac{1}{2604 \, \text{lines/cm}} \). Simplifying gives \( d = \frac{1}{260400} \text{ m} \).

With this spacing and the known angle of \(30^\circ\), we apply the grating equation to solve for \(\lambda\):
  • Find \( \sin(\theta) \), which is \( \sin(30^\circ) = 0.5 \).
  • Substitute into the formula: \( \frac{1}{260400} \times 0.5 = m \lambda \).
  • Rearranging, \( \lambda = \frac{0.5}{260400m} \).
This formula can now be used to compute possible wavelengths, provided \(m\) and the permissible range of \(\lambda\), which in our problem lies between 410 nm and 660 nm.
Principal Maxima
Principal maxima refer to the bright bands seen when light diffracts through a grating. These occur at specific angles where constructive interference of light waves happens due to particular wavelength alignments.

For the given exercise, observing principal maxima involves applying the grating equation and determining which wavelengths of visible light (410 nm to 660 nm) produce these bright spots at an angle of \(30^\circ\). Here, possible values of \(m\) were used to evaluate which wavelengths meet the given criteria:
  • For \(m = 1\), the wavelength computes to approximately \(642\, \text{nm}\).
  • For \(m = 2\), the wavelength calculates to about \(426\, \text{nm}\).
These wavelengths fall within the specified range and represent first and second-order principal maxima detected in the setup, proving the predicted model with actual observable outcomes.

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Most popular questions from this chapter

An inkjet printer uses tiny dots of red, green, and blue ink to produce an image. Assume that the dot separation on the printed page is the same for all colors. At normal viewing distances, the eye does not resolve the individual dots, regardless of color, so that the image has a normal look. The wavelengths for red, green, and blue are \(\lambda_{\text {red }}=660 \mathrm{nm}, \lambda_{\text {green }}=550 \mathrm{nm},\) and \(\lambda_{\text {blue }}=470 \mathrm{nm} .\) The diameter of the pupil through which light enters the eye is \(2.0 \mathrm{mm}\). For a viewing distance of \(0.40 \mathrm{m},\) what is the maximum allowable dot separation?

In Young's experiment a mixture of orange light \((611 \mathrm{nm})\) and blue light \((471 \mathrm{nm})\) shines on the double slit. The centers of the first- order bright blue fringes lie at the outer edges of a screen that is located \(0.500 \mathrm{m}\) away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in which direction (toward or away from the slits) should the screen be moved so that the centers of the first-order bright orange fringes will just appear on the screen? It may be assumed that \(\theta\) is small, so that \(\sin \theta \approx \tan \theta\)

An Optical Monochromator. You and your team are designing a device that inputs a beam of white light (i.e., a continuous spectrum of visible light spanning all wavelengths from \(410 \mathrm{nm}\) to \(660 \mathrm{nm}\) ), and outputs a nearly monochromatic beam (i.e., a single color). Such a device is called an optical monochromator and is used in a wide variety of instruments and scientific experiments. In the instrument you are building, white light impinges upon the backside of a diffraction grating that has 1200 lines/cm. A movable rectangular aperture (a slit) is located on the opposite side of the grating, and can translate along a circular arc of radius \(20.0 \mathrm{cm},\) the center of which is located at the grating. (a) At what angle relative to the normal of the grating should the center of the slit be located in order to pass green light \((\lambda=550 \mathrm{nm})\) from first order \((m=1)\) diffracted light? (b) How wide should the slit be so that the wavelengths passing through the slit fall in the range \(540 \mathrm{nm} \leq \lambda \leq 560 \mathrm{nm} ?\)

Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the pupils of your eyes have diameters of about \(7.0 \mathrm{mm} .\) The taillights of this car are separated by a distance of \(1.2 \mathrm{m}\) and emit red light (wavelength \(=660 \mathrm{nm}\) in vacuum). How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?

A soap film \((n=1.33)\) is 465 nm thick and lies on a glass plate \((n=1.52) .\) Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm},\) travels through the air and strikes the film perpendicularly. For which wavelength(s) in this range does destructive interference cause the film to look dark in reflected light?
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