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A diffraction grating with 345 lines/mm is \(1.00 \mathrm{m}\) in front of a screen. What is the wavelength of light whose first-order maxima will be \(16.4 \mathrm{cm}\) from the central maximum on the screen?

Short Answer

Expert verified
The wavelength is approximately \(475 \text{ nm}\) (nanometers).

Step by step solution

01

Identify the Grating Equation

The diffraction grating equation is given by \(d\sin\theta = m\lambda\), where \(d\) is the distance between adjacent grating lines, \(\theta\) is the angle of diffraction, \(m\) is the order of the maxima, and \(\lambda\) is the wavelength of the light. As we know the number of lines per millimeter, we can find \(d\).
02

Calculate the Distance 'd' Between Grating Lines

Given that the grating has 345 lines/mm, convert this to meters to find \(d\). The distance between lines \(d\) is \(\frac{1}{345 \, \text{lines/mm}} = \frac{1}{345000} \, \text{m}\).
03

Determine the Angle θ

Use trigonometry to find \(\theta\). The first-order maximum is \(16.4\,\text{cm} = 0.164\,\text{m}\) from the central maximum on a screen 1.00 m away. The tangent of this angle is given by \(\tan\theta = \frac{0.164}{1.00}\), which helps us find \(\theta\).
04

Use Small Angle Approximation (if applicable)

Since \(\theta\) is likely small, we approximate \(\sin\theta \approx \tan\theta = 0.164\). This simplification allows us to use the smaller angle to find the wavelength more directly.
05

Plug Values into the Grating Equation

Using the simplified equation, \(d\sin\theta = m\lambda\) becomes \(\frac{1}{345000} \cdot 0.164 = 1 \cdot \lambda\). Solve for \(\lambda\) to find the wavelength of light.
06

Solve for λ to Find the Wavelength

Rearrange and solve: \(\lambda = \frac{0.164}{345000}\), resulting in \(\lambda \approx 4.75 \times 10^{-7} \text{m}\).

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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 the context of diffraction is a fascinating process where we determine the distance between successive peaks of a wave. In scenarios involving diffraction gratings, the wavelength is a crucial variable that represents a specific characteristic of light. When solving for the wavelength, we start with the grating equation: \[ d\sin\theta = m\lambda \]Here, \( \lambda \) is the wavelength, \( d \) is the distance between grating lines, and \( \theta \) is the angle of diffraction for a certain order \( m \). By calculating the wavelength, students can better understand the light's nature and behavior as it interacts with different optical elements.
  • Determine the distance \( d \) from the grating lines.
  • Find the angle \( \theta \) using the setup's geometry.
  • Apply these values in the grating equation to solve for \( \lambda \).
This intuitive approach helps make the abstract concept of wavelength more tangible.
Grating Equation
The grating equation forms the backbone of diffraction calculations. By defining the relationship between the grating's physical properties and the resulting wave patterns, it provides a clear pathway to understanding how light behaves. The general form is:\[ d\sin\theta = m\lambda \]In this equation:- \( d \) represents the distance between lines on the grating.- \( \theta \) is the angle at which light of order \( m \) diffracts.- \( m \) indicates the diffraction order, which can be any integer.- \( \lambda \) is the wavelength of the light.This relationship is powerful in that it allows us to calculate any of these variables if the others are known. The grating equation is essential for experiments dealing with spectrometry and understanding light properties in practical situations.
  • Grating equation links optical properties and observed phenomena.
  • Useful in spectral analysis and determining light wavelengths.
Angle of Diffraction
The angle of diffraction, denoted by \( \theta \), is critical in analyzing wave behavior as they pass through a diffraction grating. It is essentially the angle at which a certain order of maxima is observed.To find \( \theta \), we use the formula:\[ \tan\theta = \frac{\text{distance from central maximum}}{\text{distance from grating to screen}} \]For the given exercise:- The distance from the central maximum is 16.4 cm (0.164 m).- The distance from the grating to the screen is 1 meter.Using trigonometry, we solve for \( \theta \), which plays a pivotal role in the further analysis of wavelength and diffraction order. This step is vital in ensuring accurate computation, as it directly influences the wavelength derived through the grating equation.
Small Angle Approximation
The small angle approximation is a mathematical simplification used when dealing with tiny angles of diffraction. This approximation assumes that for small angles, the sine and tangent of the angle are nearly equal which can simplify calculations significantly.Mathematically, it is expressed as:\[ \sin\theta \approx \tan\theta \approx \theta \] provided \( \theta \) is expressed in radians. In the context of diffraction, this is especially useful when the angle \( \theta \) is small enough that the values of \( \sin\theta \) and \( \tan\theta \) are approximately the same. This allows for the easier determination of \( \lambda \) in the grating equation, making the equation:\[ d \cdot \theta = m \lambda \]The small angle approximation simplifies the analysis and computation without losing significant accuracy, especially in educational settings where the focus is on basic principles rather than computational complexity.

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Most popular questions from this chapter

A grating has 787 lines per centimeter. Find the angles of the first three principal maxima above the central fringe when this grating is illuminated with \(655-\mathrm{nm}\) light.

The yellow light of sodium, with wavelengths of 588.99 nm and \(589.59 \mathrm{nm},\) is normally incident on a grating with 494 lines/cm. Find the linear distance between the first-order maxima for these two wavelengths on a screen \(2.55 \mathrm{m}\) from the grating.

A lens that is "optically perfect" is still limited by diffraction effects. Suppose a lens has a diameter of \(120 \mathrm{mm}\) and a focal length of \(640 \mathrm{mm}\). (a) Find the angular width (that is, the angle from the bottom to the top) of the central maximum in the diffraction pattern formed by this lens when illuminated with \(540-\mathrm{nm}\) light. (b) What is the linear width (diameter) of the central maximum at the focal distance of the lens?

A thin film of oil \((n=1.30)\) floats on water \((n=1.33) .\) When sunlight is incident at right angles to this film, the only colors that are enhanced by reflection are blue \((458 \mathrm{nm})\) and red \((687 \mathrm{nm})\). Estimate the thickness of the oil film.

A two-slit experiment with slits separated by \(48.0 \times 10^{-5} \mathrm{m}\) produces a second-order maximum at an angle of \(0.0990^{\circ}\) (a) Find the wavelength of the light used in this experiment. (b) If the slit separation is increased but the secondorder maximum stays at the same angle, does the wavelength increase, decrease, or stay the same? Explain. (c) Calculate the wavelength for a slit separation of \(68.0 \times 10^{-5} \mathrm{m}\).

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