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How narrow is a slit that produces a diffraction pattem on a screen \(1.8 \mathrm{m}\) away whose central peak is \(1.0 \mathrm{m}\) wide? Assume \(\lambda=589 \mathrm{nm}\)

Short Answer

Expert verified
The width of the slit that produces a diffraction pattern on a screen 1.8 m away, whose central peak is 1.0 m wide, and given a wavelength of \(λ=589 \mathrm{nm}\), is approximately 3.974 micrometers.

Step by step solution

01

Recall the formula for the angular width of the central peak

The angular width of the central peak of the diffraction pattern can be calculated using the formula: \[θ = \frac{2λ}{w}\] where \(θ\) is the angular width, \(λ\) is the wavelength of the light, and \(w\) is the width of the slit.
02

Calculate the angular width of the central peak

We know the width of the central peak on the screen, which is 1.0 m, and the distance from the slit to the screen, which is 1.8 m. To find the angular width, we can calculate the angle by using the formula: \[tan(θ) = \frac{1.0}{2 \cdot 1.8}\]
03

Find the actual angular width

Now let's calculate the actual angular width \(θ\): \[θ = arctan\left(\frac{1}{3.6}\right)\]
04

Use the angular width to find the width of the slit

Now that we have the angular width, we can use the formula from step 1 to calculate the width of the slit: \[w = \frac{2λ}{θ}\] Substitute the given values, where \(λ = 589 \times 10^{-9} m\) and \(θ\) is the value calculated in step 3: \[w = \frac{2(589 \times 10^{-9})}{arctan\left(\frac{1.0}{3.6}\right)}\]
05

Calculate the width of the slit

Finally, we can calculate the width of the slit: \[w = \frac{2(589 \times 10^{-9})}{arctan\left(\frac{1}{3.6}\right)} = 3.974 \times 10^{-6} m\]
06

Convert to the appropriate units

Now, we can convert the width of the slit to more convenient units, such as micrometers: \[w = 3.974 \times 10^{-6} m \times \frac{10^{6} \mu m}{1 m} = 3.974\, \mu m\] So, the width of the slit is approximately 3.974 micrometers.

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Key Concepts

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

Angular Width of Diffraction Peak
Understanding the angular width of a diffraction peak is fundamental to interpreting diffraction patterns. In simple terms, diffraction occurs when waves, such as light, encounter an obstacle or pass through a narrow opening, known as a slit. The waves spread out, or diffract, creating a pattern of bright and dark regions on a screen. The central bright region, or the central peak, has a certain width that can be measured in angular terms.

The formula to calculate the angular width (\theta) of the central peak is given by:
\[\theta = \frac{2 \lambda}{w}\]
where \(\theta\) represents the angular width of the central peak, \(\theta\) the wavelength of the light, and \(w\) the slit width. To measure this angle accurately, we can use trigonometric functions based on the geometry of the setup, considering the distance from the slit to the screen and the physical width of the peak observed. After finding the angle, you can rearrange this formula to solve for unknown variables such as the slit width, which is exactly what you'd need in exercises like determining the narrowness of a slit based on the diffraction pattern produced.
Wavelength of Light
The wavelength of light, denoted as \(\theta\), plays a critical role in diffraction and optics. It is the distance over which the light wave's shape repeats and is a key parameter in understanding how light interacts with objects. For instance, in the case of diffraction through a slit, different wavelengths will produce diffraction patterns with varying angular widths, which affects the visibility and clarity of the fringes on the screen.

Light's wavelength is often measured in nanometers (nm), where 1 nm equals 10-9 meters. Visible light—the light we can see—varies in wavelength from around 400 nm (violet) to 700 nm (red). In many diffraction problems, including the textbook example, a specific wavelength is given, which corresponds to yellow light in this case. This information is essential for calculating the diffraction pattern because as the wavelength changes, so does the pattern.
Slit Width Calculation
Calculating the width of a slit, given the diffraction pattern, is a typical application of the principles of wave optics. In the scenario provided, the process requires initially determining the angular width of the central diffraction peak. This angle can be found using:
\[\theta = \frac{1.0}{2 \cdot 1.8}\]
, where the width of the peak on the screen and the distance between the screen and the slit are known. From that angle, we calculate the slit width (w) with the formula:
\[w = \frac{2\theta}{\tan^{-1}(\theta)}\]

By substituting the angle and known wavelength into this equation, we can solve for the slit's width. It's important to convert this final measurement into practical units like micrometers (\textmu m) for easier comprehension. The ability to calculate slit width from diffraction patterns is essential for understanding optical phenomena and designing experiments involving light diffraction.

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Most popular questions from this chapter

(a) How wide is a single slit that produces its first minimum for 633 -nm light at an angle of \(28.0^{\circ}\) ? (b) At what angle will the second minimum be?

(a) The dwarf planet Pluto and its moon, Charon, are separated by \(19,600 \mathrm{km}\). Neglecting atmospheric effects, should the 5.08 -m-diameter Palomar Mountain telescope be able to resolve these bodies when they are \(4.50 \times 10^{9} \mathrm{km} \quad\) from Earth? Assume an average wavelength of \(550 \mathrm{nm}\). (b) In actuality, it is just barely possible to discern that Pluto and Charon are separate bodies using a ground-based telescope. What are the reasons for this?

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The width of the central peak in a single-slit diffraction pattern is \(5.0 \mathrm{mm}\). The wavelength of the light is \(600 \mathrm{nm}\), and the screen is \(2.0 \mathrm{m}\) from the slit. (a) What is the width of the slit? (b) Determine the ratio of the intensity at \(4.5 \mathrm{mm}\) from the center of the pattern to the intensity at the center.

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