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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 \(660 \mathrm{~nm}\). What is (are) the wavelength(s) of the incident light that could have produced this maximum?

Short Answer

Expert verified
Wavelengths are found using orders that satisfy 410 nm ≤ λ ≤ 660 nm.

Step by step solution

01

Understand the Diffraction Grating Formula

The diffraction grating formula is given by \( d \sin \theta = m \lambda \), where \( d \) is the distance between the slits (grating spacing), \( \theta \) is the angle of diffraction, \( m \) is the order of the maximum, and \( \lambda \) is the wavelength of light. We are asked to find the wavelengths that result in a principal maximum at \( \theta = 30.0^{\circ} \).
02

Calculate the Grating Spacing

Given that the grating has 2604 lines per centimeter, the grating spacing \( d \) in centimeters is the reciprocal of the number of lines, i.e., \( d = \frac{1}{2604} \) cm. To convert to meters, use: \( d = \frac{1}{2604 \times 100} \) m.
03

Rearrange the Diffraction Equation for Wavelength

Rearrange the diffraction grating formula to solve for \( \lambda \):\[ \lambda = \frac{d \sin \theta}{m} \]Substitute \( \sin(30.0^{\circ}) = 0.5 \) into the equation.
04

Identify the Possible Orders of Maximum

Determine which orders \( m \) result in wavelengths within the range 410 nm to 660 nm. Since \( \lambda = \frac{d \cdot 0.5}{m} \), you need to find values of \( m \) that keep \( \lambda \) in this range.
05

Calculate Wavelengths for Different Orders

Use the values determined from the previous step for \( m \) to calculate the corresponding wavelength using \( \lambda = \frac{d \cdot 0.5}{m} \). Ensure \( \lambda \) falls between 410 nm and 660 nm for an order to be valid.
06

Verify the Valid Wavelengths

After calculating the possible wavelengths, verify that each calculated wavelength fits within the range 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.

Wavelength Calculation
The first step in understanding how light behaves through a diffraction grating is to calculate the wavelength of light that produces a specific maximum. Using the fundamental equation for diffraction gratings, \( d \sin \theta = m \lambda \), we can find the wavelength \( \lambda \). Here, \( d \) is the grating spacing, \( \theta \) is the angle of diffraction, and \( m \) is the order of the maximum.
First, rearrange the equation to solve for \( \lambda \):
  • \( \lambda = \frac{d \sin \theta}{m} \)
By substituting the known values of \( d \), \( \sin \theta \), and \( m \), we can determine the specific wavelength that results in the observed diffraction pattern.
This process connects the geometric arrangement of the grating to the physical properties of the light. It allows us to pinpoint how different wavelengths are spread out by the grating's interference effects.
Grating Spacing
Grating spacing is a crucial part of understanding how diffraction gratings work. It refers to the distance between adjacent slits or lines in the grating. In this exercise, the grating has 2604 lines per centimeter, which is essential for defining the grating spacing \( d \).
To find \( d \), you take the reciprocal of the number of lines per centimeter:
  • \( d = \frac{1}{2604} \) cm
Convert this to meters for standardized calculations:
  • \( d = \frac{1}{2604 \times 100} \) m
Grating spacing determines how the light waves will interfere with each other when passing through the grating.
A smaller spacing means the light waves will spread out more, creating more distinct interference patterns. This is a foundational concept that helps explain why different wavelengths of light will appear at different angles.
Angle of Diffraction
The angle of diffraction, \( \theta \), is an angle where a specific wavelength of light constructively interferes, leading to a noticeable fringe or principal maximum. For our exercise, \( \theta \) is given as \( 30.0^{\circ} \).
The sine of \( \theta \) is a component of the diffraction grating formula:
  • Using \( \sin(30.0^{\circ}) = 0.5 \)
This value represents the extent to which the light is deviated from its original path due to the grating.
Understanding this helps to determine which specific wavelengths are causing the bright spots at particular angles. This angle dependence is what makes diffraction gratings a powerful tool for analyzing light.
Order of Maximum
In diffraction gratings, the term 'order of maximum', denoted as \( m \), is an integer representing the sequence of bright spots produced by constructive interference. Each order corresponds to a different set of conditions where light waves reinforce each other.
For the exercise, you need to find which orders \( m \) produce wavelengths falling within our defined range of 410 nm to 660 nm. Start with the equation:
  • \( \lambda = \frac{d \cdot 0.5}{m} \)
Figure out which values of \( m \) will keep \( \lambda \) within this range.
This allows us to identify multiple wavelengths resulting in maxima at the given angle. Each order reflects how different wavelengths are resolved through the grating, further illustrating the separation capability of diffraction gratings.

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Most popular questions from this chapter

(a) In a single-slit diffraction pattern the width of the central bright fringe is defined by the location of the first dark fringe that lies on either side of it. For a given slit width, does the width of the central bright fringe increase, decrease, or remain the same as the wavelength of the light increases? (b) For a given wavelength, does the width of the central bright fringe increase, decrease, or remain the same as the slit width increases? (c) When both the wavelength and the slit width change, it is possible for the width of the central bright fringe to remain the same. What condition must be satisfied for this to happen? In each case, give your reasoning. A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{~m}\). When light with a wavelength of \(\lambda_{1}=510 \mathrm{~nm}\) passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2}\) ) and a wavelength of \(\lambda_{2}=740 \mathrm{~nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2}\).

A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light (wavelength \(=640.0 \mathrm{~nm}\) in vacuum). Assuming that the visible spectrum extends from 380 to \(750 \mathrm{~nm}\), for which visible wavelength(s) (in vacuum) will the film appear bright due to constructive interference?

A hunter who is a bit of a braggart claims that, from a distance of \(1.6 \mathrm{~km}\), he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What's more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of \(498 \mathrm{~nm}\) (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to \(8 \mathrm{~mm}\), the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of \(498 \mathrm{~nm}\).

Light shines on a diffraction grating, and a diffraction pattern is produced on a viewing screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) From trigonometry, how is the distance \(y\) from the central bright fringe to the second-order bright fringe related to the diffraction angle \(\theta\) and the distance \(L\) between the grating and the screen? (b) From physics, how is \(\theta\) related to the order \(m\) of the bright fringe, the wavelength \(\lambda\) of the light, and the separation \(d\) between the slits? (c) In this problem, the angle \(\theta\) is small (less than a few degrees). When the angle is small, \(\tan \theta\) is approximately equal to \(\sin \theta\), or \(\tan \theta \approx \sin \theta\). Using this approximation, obtain an expression for \(y\) in terms of \(L, m, \lambda\), and \(d\). (d) If the entire apparatus in the drawing is submerged in water, would you expect the distance \(y\) to increase, decrease, or remain unchanged? Why? Light of wavelength \(480 \mathrm{~nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(5.0 \times 10^{-7} \mathrm{~m}\). The distance between the grating and the viewing screen is \(0.15 \mathrm{~m}\). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right)\), what is the distance \(y ?\) Be sure your answer is consistent with part (d) of the Concept Questions.

For a wavelength of \(420 \mathrm{~nm}\), a diffraction grating produces a bright fringe at an angle of \(26^{\circ}\). For an unknown wavelength, the same grating produces a bright fringe at an angle of \(41^{\circ} .\) In both cases the bright fringes are of the same order \(m\). What is the unknown wavelength?

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