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A diffraction grating illuminated by monochromatic light normal to the grating produces a certain line at angle \(\theta .(\mathrm{a}) \mathrm{What}\) is the product of that line's half-width and the grating's resolving power? (b) Evaluate that product for the first order of a grating of slit separation \(900 \mathrm{~nm}\) in light of wavelength \(600 \mathrm{~nm}\).

Short Answer

Expert verified
The product is \(\frac{2}{3}\).

Step by step solution

01

Understand the Problem

We need to find the product of the line's half-width and the grating's resolving power. We're given a grating illuminated by light, with a line forming at angle \( \theta \). For part (b), specific values for slit separation and wavelength are provided.
02

Define the Resolving Power

The resolving power of a diffraction grating, \( R \), is defined as \( R = \frac{\lambda_0}{\Delta \lambda} = mN \), where \( \lambda_0 \) is the wavelength of light, \( \Delta \lambda \) is the smallest difference in wavelengths that can be resolved, \( m \) is the order of diffraction, and \( N \) is the number of slits.
03

Calculate the Line's Half-width

The angular half-width of a diffraction line can be approximated as \( \Delta \theta = \frac{\lambda_0}{Nd \cos \theta} \), where \( d \) is the slit separation, and \( N \) is the number of slits in the grating.
04

Calculate the Product

The product of the line's half-width \( \Delta \theta \) and the resolving power \( R \) is then \( (\Delta \theta)R = \frac{\lambda_0}{Nd \cos \theta} \times mN = \frac{m\lambda_0}{d \cos \theta} \).
05

Simplify for First Order in Part (b)

For part (b), assume first-order diffraction \( m = 1 \), \( \lambda_0 = 600 \text{ nm} \), and \( d = 900 \text{ nm} \). Since \( \theta \) is not specified, assume \( \theta = 0 \) for simplification (normal incidence), hence \( \cos \theta = 1 \). Evaluate the expression: \( \frac{600 \text{ nm}}{900 \text{ nm} \cdot 1} = \frac{2}{3} \).
06

Evaluate the Expression

Multiply the value obtained for first-order: \( \frac{2}{3} \) is the product of the line's half-width and the grating's resolving power for the given conditions.

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Key Concepts

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

Resolving Power
The resolving power of a diffraction grating is a crucial concept in optics. It is essentially the ability of the grating to separate or distinguish between two closely spaced wavelengths of light. This capability is mathematically defined by the formula \( R = \frac{\lambda_0}{\Delta \lambda} = mN \), where:
  • \( \lambda_0 \) is the central wavelength of the light.
  • \( \Delta \lambda \) is the minimum wavelength difference that can be resolved.
  • \( m \) is the order of diffraction, which indicates the number of wavelengths by which adjacent slits are out of phase.
  • \( N \) is the total number of slits in the grating.
Greater resolving power means better ability to distinguish between two wavelengths. Thus, a grating with a higher resolving power can detect finer details in the spectral lines which are closely spaced, making it invaluable in spectroscopy and other optical applications.
Angular Half-width
Angular half-width is an important parameter related to the width of the diffracted lines. It tells us how broad the spectral line appears and is approximated by \( \Delta \theta = \frac{\lambda_0}{Nd \cos \theta} \), where:
  • \( \lambda_0 \) is the wavelength of light.
  • \( N \) is the number of slits in the grating.
  • \( d \) is the separation between adjacent slits.
  • \( \theta \) is the angle at which the line is observed.
This formula implies that as the number of slits \( N \) increases or the slit separation \( d \) decreases, the angular width becomes narrower. Consequently, the spectral lines become sharper, which is advantageous for precise measurements. In practical applications, optimizing the angular half-width can improve the clarity and resolution of the diffraction pattern produced.
Monochromatic Light
Monochromatic light refers to light that has a single wavelength or color. In the context of diffraction gratings, using monochromatic light ensures that the resulting diffraction pattern is clear and consists of discrete bright lines called diffraction orders. This property is central to many optical experiments and applications such as spectroscopy.
When a diffraction grating is illuminated by monochromatic light, each order of the pattern provides information related to the wavelength and the angle of incidence. The degree to which these orders can be separated and accurately analyzed is largely dependent on the grating's properties, such as its resolving power.
Monochromatic light sources, like lasers, are often preferred in optical setups because they produce minimal overlapping of spectral features. This helps in accurate determination and analysis of the light's properties, which is essential in scientific research and technology.

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Most popular questions from this chapter

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?

If Superman really had \(\mathrm{x}\) -ray vision at \(0.10 \mathrm{~nm}\) wavelength and a \(4.0 \mathrm{~mm}\) pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by \(5.0 \mathrm{~cm}\) to do this?

A diffraction grating \(20.0 \mathrm{~mm}\) wide has 6000 rulings. Light of wavelength \(589 \mathrm{~nm}\) is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of \(\theta\) at which maxima appear on a distant viewing screen?

The radar system of a navy cruiser transmits at a wavelength of \(1.6 \mathrm{~cm}\), from a circular antenna with a diameter of \(2.3 \mathrm{~m}\). At a range of \(6.2 \mathrm{~km}\), what is the smallest distance that two speedboats can be from each other and still be resolved as two separate objects by the radar system?

Light of wavelength \(633 \mathrm{~nm}\) is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is \(1.20^{\circ} .\) What is the width of the slit?

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