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In a large vacuum chamber, monochromatic laser light passes through a narrow slit in a thin aluminum plate and forms a diffraction pattern on a screen that is 0.620 \(\mathrm{m}\) from the slit. When the aluminum plate has a temperature of \(20.0^{\circ} \mathrm{C},\) the width of the central maximum in the diffraction pattern is 2.75 \(\mathrm{mm}\) . What is the change in the width of the central maximum when the temperature of the plate is raised to \(520.0^{\circ} \mathrm{C}\) ? Does the width of the central diffraction maximum increase or decrease when the temperature is increased?

Short Answer

Expert verified
The width of the central maximum increases.

Step by step solution

01

Identify the Formula for Diffraction

In a single-slit diffraction pattern, the width of the central maximum is given by the formula:\[ W = \frac{2 \lambda D}{a} \]where \( \lambda \) is the wavelength of the light, \( D \) is the distance from the slit to the screen, and \( a \) is the width of the slit.
02

Understand the Effect of Temperature on the Slit Width

As the temperature increases, the width of the slit \( a \) also changes due to thermal expansion. The change in slit width is given by:\[ \Delta a = a_0 \alpha \Delta T \]where \( a_0 \) is the initial width of the slit, \( \alpha \) is the coefficient of linear expansion for aluminum, and \( \Delta T \) is the change in temperature.
03

Calculate the Change in Slit Width

Calculate \( \Delta a \) using the formula for thermal expansion. Note that the coefficient of linear expansion for aluminum is approximately \(23 \times 10^{-6} \ \mathrm{^{\circ}C}^{-1}\), and the temperature change \( \Delta T \) is \(520.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 500.0^{\circ} \mathrm{C}\). Substitute \( a_0 \) and \( \alpha \) in the formula to find \( \Delta a \).
04

Find the New Width of the Slit

The new width of the slit is \( a = a_0 + \Delta a \). Use the initial slit width \( a_0 \) and the change in slit width \( \Delta a \) to find \( a \).
05

Calculate the New Width of the Central Maximum

Substitute the new slit width into the diffraction formula to find the new width \( W' \) of the central maximum: \[ W' = \frac{2 \lambda D}{a} \].
06

Determine the Change in Width of the Central Maximum

The change in the width of the central maximum is \( \Delta W = W' - W_0 \), where \( W_0 \) is the initial width of 2.75 mm.

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Key Concepts

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

Laser Light
Laser light is a remarkable form of light with distinct properties that make it different from regular light sources. One of its main characteristics is its coherence, meaning the light waves are in phase and travel in a precise direction. This syncronization allows for laser beams to remain narrow over long distances, making them ideal for experiments like diffraction.
Additionally, laser light is monochromatic, meaning it is comprised of a single color or wavelength. This uniformity is crucial in experiments involving diffraction, as it ensures consistent results. In the exercise, a monochromatic laser shines through a slit, allowing us to analyze effects such as the central maximum more precisely.
Another defining feature is the high intensity of laser light. This helps create clear and visible patterns on screens, enabling more detailed studies of phenomena like single-slit diffraction.
Thermal Expansion
Thermal expansion is the tendency of matter to change its shape, area, and volume in response to a change in temperature. In this context, it affects the slit in the aluminum plate. When the temperature increases, the atoms in the material vibrate more vigorously and take up more space.
This process leads to an increase in the dimensional space the object occupies. The formula to calculate how much the width of the slit changes due to temperature is: \[ \Delta a = a_0 \alpha \Delta T \]where \( a_0 \) is the original width, \( \alpha \) is the thermal expansion coefficient, and \( \Delta T \) is the change in temperature. For aluminum, \( \alpha \) is approximately \( 23 \times 10^{-6} \, \mathrm{^{\circ}C}^{-1} \). The slit widens slightly when the temperature is increased, affecting the diffraction pattern.
Single-Slit Diffraction
Single-slit diffraction occurs when light waves pass through a narrow opening and spread out. This spreading causes the light to interact, resulting in a pattern of bright and dark bands. The light waves interfere with each other - where crests and troughs align, bright spots appear, while destructive interference forms dark areas.
The central hallmark of this pattern is the central maximum - the brightest band, which appears directly opposite the slit. The width of this central bright band is determined by the relationship: \[ W = \frac{2 \lambda D}{a} \]where \( \lambda \) is the wavelength, \( D \) is the distance to the screen, and \( a \) is the slit width. When the width of the slit changes due to thermal expansion, it directly affects this central maximum's size.
Central Maximum
The central maximum is the most prominent feature in a diffraction pattern. It is the wide bright band directly in front of the slit where the laser light first encounters. This maximum is centered around the undiffracted light ray passing straight through the center of the slit.
Central maxima result from constructive interference where light waves overlap, reinforcing each other. It's often the brightest and widest part of the diffraction pattern.
In the exercise, the central maximum is initially 2.75 mm wide. However, when the temperature of the aluminum plate increases, the slit width increases slightly as a result, leading to a narrower central maximum. The calculation shows that as the slit width widens, the central maximum width decreases. This occurs because the formula \( W = \frac{2 \lambda D}{a} \) indicates that as \( a \) increases, \( W \) decreases.

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Most popular questions from this chapter

Monochromatic light of wavelength \(\lambda=620 \mathrm{nm}\) from a distant source passes through a slit 0.450 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slit. In terms of the intensity \(I_{0}\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) \(1.00 \mathrm{mm} ;\) (b) 3.00 \(\mathrm{mm}\) ; (c) 5.00 \(\mathrm{mm}\) ?

The light from an iron arc includes many different wavelengths. Two of these are at \(\lambda=587.9782 \mathrm{nm}\) and \(\lambda=\) 587.8002 \(\mathrm{nm} .\) You wish to resolve these spectral lines in first order using a grating 1.20 \(\mathrm{cm}\) in length. What minimum number of slits per centimeter must the grating have?

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 \(\mathrm{cm} / \mathrm{s}\) on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 \(\mathrm{m}\) away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at \(\pm 61.3 \mathrm{cm}\) from the point directly opposite the hole, but waves do reach the shore every- where within this distance. (a) How wide is the hole in the barrier? (b) At what other angles do you find no waves hitting the shore?

A converging lens 7.20 \(\mathrm{cm}\) in diameter has a focal length of 300 \(\mathrm{mm} .\) If the resolution is diffraction limited, how far away can an object be if points on it 4.00 \(\mathrm{mm}\) apart are to be resolved (according to Rayleigh's criterion)? Use \(\lambda=550 \mathrm{nm} .\)

A diffraction experiment involving two thin parallel slits yields the pattern of closely spaced bright and dark fringes shown in Fig. E36.26. Only the central portion of the pattern is shown in the figure. The bright spots are equally spaced at 1.53 \(\mathrm{mm}\) center to center (except for the missing spots) on a screen 2.50 \(\mathrm{m}\) from the slits. The light source was a He-Ne laser producing a wavelength of 632.8 nm. (a) How far apart are the two slits? (b) How wide is each one?

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