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Chapter 27: Problem 27

In a single-slit diffraction pattern, the central fringe is 450 times as wide as the slit. The screen is 18000 times farther from the slit than the slit is wide. What is the ratio \(\lambda / W\) where \(\lambda\) is the wavelength of the light shining through the slit and \(W\) is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that \(\sin \theta \approx \tan \theta\)

Short Answer

Expert verified
\(\frac{\lambda}{W} = \frac{1}{40}\)

Step by step solution

01

Identify the Given and Required

We are given that the central fringe is 450 times as wide as the slit. The screen is 18,000 times farther than the slit width. We need to find the ratio \( \lambda / W \), where \( \lambda \) is the wavelength and \( W \) the width of the slit.
02

Relate Fringe Width to Angular Displacement

In single-slit diffraction, the angular width of the central fringe from minima to minima is given by \( \theta = \frac{\lambda}{W} \). For small angles, we assume \( \tan \theta \approx \sin \theta \approx \theta \).
03

Use the Distance to Screen

The width of the central maximum \( x \) on the screen is given by \( x = L \cdot \theta \), where \( L \) is the distance to the screen. We know \( x = 450W \) and \( L = 18000W \).
04

Substitute and Solve for \( \frac{\lambda}{W} \)

Substitute the expressions into \( x = L \cdot \theta \): \[ 450W = 18000W \cdot \frac{\lambda}{W} \]This simplifies to:\[ 450 = 18000 \cdot \frac{\lambda}{W} \]Solve for \( \frac{\lambda}{W} \): \[ \frac{\lambda}{W} = \frac{450}{18000} \]
05

Simplify the Fraction

Simplify \( \frac{450}{18000} \) by dividing both the numerator and the denominator by 450.\[ \frac{450}{18000} = \frac{1}{40} \] Thus, \( \frac{\lambda}{W} = \frac{1}{40} \).

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Key Concepts

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

Central Fringe Width
The central fringe in a single-slit diffraction pattern refers to the bright band that appears on a screen as light passes through a narrow slit. This band is the widest compared to other fringes and is formed due to constructive interference of light rays entering through the slit. The width of the central fringe is an important characteristic because it helps us understand how light behaves when it is diffracted.
For the given exercise, the central fringe is stated to be 450 times the width of the slit, indicating how pronounced the diffraction effect is. The width of the central fringe, often denoted as \( x \), is directly related to both the properties of the light and the slit geometry.
In mathematical terms, if \( W \) is the slit width and \( x = 450W \), this fracture illustrates the central fringe's dependence on the slit dimensions, providing a solid representation of diffraction's scale.
Angular Width Approximation
In the context of diffraction, the angle that describes how far light spreads from the direction of the slit straight through to the screen is significant. Under small angle approximations, the trigonometric functions \( \sin \theta \) and \( \tan \theta \) can be approximated by the angle \( \theta \) itself. This allows a mathematical simplification that makes understanding diffraction patterns more manageable.
This exercise utilizes this concept, as it deals with the approximation that \( \sin \theta \approx \tan \theta \approx \theta \). Such approximations are valid when the angles involved are very small, which is typical in diffraction scenarios. In particular, this means that the calculations can be done using linear equations without losing much accuracy. This approximation is key to deriving the relationship between the fringe width, the slit width, and the wavelength.
Distance to Screen in Diffraction
The distance from the slit to the screen, denoted here as \( L \), plays a crucial role in determining the diffraction pattern's characteristics. In the exercise, \( L \) is given as 18000 times the width of the slit \( W \).
This large factor underscores how even though the slit is narrow, the screen is positioned far enough away to enable the formation of clear and measurable interference patterns. This considerable distance ensures that the light waves have enough space to spread out and interfere, creating visible diffraction fringes on the screen.
The formula \( x = L \cdot \theta \) used in the solution describes how the width of the central fringe (\( x \)) is determined by both the distance to the screen (\( L \)) and the angular width (\( \theta \)). This interconnectedness between distance and fringe width is essential to predict and analyze diffraction phenomena.
Wavelength to Slit Width Ratio
The ratio \( \frac{\lambda}{W} \), where \( \lambda \) is the wavelength of the light, and \( W \) is the width of the slit, is critical in characterizing diffraction patterns. It represents the proportionate comparison of the light's wave nature to the geometric constraint of the slit.
In the given exercise, determining this ratio involves understanding how the wavelength affects the diffraction pattern. The derived ratio \( \frac{1}{40} \) indicates that the wavelength is much smaller relative to the slit width. This suggests that the diffraction is relatively subtle compared to scenarios where the wavelength is closer in size to the slit width. This calculation and the understanding of its significance help underline how light's wave properties are influenced by the environment they traverse.
This ratio is crucial because it gives insights into both the optical properties of the setup and the mechanics behind wave diffracting through the slit itself.

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Most popular questions from this chapter

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.

A diffraction grating is \(1.50 \mathrm{~cm}\) wide and contains 2400 lines. When used with light of a certain wavelength, a third-order maximum is formed at an angle of \(18.0^{\circ} .\) What is the wavelength (in \(\mathrm{nm}\) )?

Concept Questions An inkjet color printer uses tiny dots of red, green, and blue ink to produce an image. At normal viewing distances, the eye does not resolve the individual dots, so that the image has a normal look. The angle \(\theta_{\text {min }}\) is the minimum angle that two dots can subtend at the eye and still be resolved separately. (a) For which color does \(\theta_{\text {min }}\) have the largest value? (b) For which color does \(\theta_{\text {min }}\) have the smallest value? (c) Corresponding to each value of \(\theta_{\text {trin, }}\) there is a value for the separation distance \(s\) between the dots. How is \(\theta_{\text {rnin }}\) related to \(s\) and the viewing distance \(L ?\) (d) Assume that the dot separation on the printed page is the same for all colors and is chosen so that none of the colored dots can be seen as separate objects. Should the maximum allowable dot separation be less than \(s_{\text {red }}, s_{\text {green }},\) or \(s_{\text {blue }}\) ? For each answer, give your reasoning. Problem 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 value for the dot separation identified in Concept Question (d)?

Two gratings \(A\) and \(B\) have slit separations \(d_{\mathrm{A}}\) and \(d_{\mathrm{B}},\) respectively. They are used with the same light and the same observation screen. When grating A is replaced with grating \(\mathrm{B},\) it is observed that the first-order maximum of \(\mathrm{A}\) is exactly replaced by the secondorder maximum of B. (a) Determine the ratio \(d_{\mathrm{B}} / d_{\mathrm{A}}\) of the spacings between the slits of the gratings. (b) Find the next two principal maxima of grating A and the principal maxima of \(\mathrm{B}\) that exactly replace them when the gratings are switched. Identify these maxima by their order numbers.

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 \(498 \mathrm{~nm}\).
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