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A slit 0.360 \(\mathrm{mm}\) wide is illuminated by parallel rays of light that have a wavelength of 540 \(\mathrm{nm}\) . The diffraction pattern is observed on a screen that is 1.20 \(\mathrm{m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\) . (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?(\text { See Problem } 36.53 \text { , part (a), for a hint about how to }\) solve for the phase angle \(\beta . )\)

Short Answer

Expert verified
(a) 1.8 mm; (b) 2.5 mm.

Step by step solution

01

Determine the Condition for the First Minimum

The first minimum occurs in a single-slit diffraction pattern where the equation \( a \sin \theta = m\lambda \) applies, with \( m \) being an integer representing the order of the minimum. For the first minimum, \( m = 1 \). Here, \( a = 0.360 \times 10^{-3} \ \text{m} \), \( \lambda = 540 \times 10^{-9} \ \text{m} \). So, \( 0.360 \times 10^{-3} \sin \theta = 540 \times 10^{-9} \).
02

Calculate the Angle for the First Minimum

Solve for \( \theta \) using the equation from Step 1:\[ \sin \theta = \frac{540 \times 10^{-9}}{0.360 \times 10^{-3}} \]Calculate \( \sin \theta = \frac{540}{360,000} = 0.0015 \). Thus, \( \theta \approx \arcsin(0.0015) \).
03

Find the Distance to the First Minimum on the Screen

The distance \( y \) from the central maximum to the first minimum is given by \( y = L \tan \theta \), where \( L = 1.20 \ \text{m} \) is the distance from the slit to the screen. Since \( \theta \) is small, \( \tan \theta \approx \sin \theta \). Thus, \( y = 1.20 \times 0.0015 = 0.0018 \ \text{m} = 1.8 \ \text{mm} \).
04

Calculate the Phase Angle for Half Intensity

For half intensity, the equation \( I = I_0 \left(\frac{\sin \beta}{\beta}\right)^2 = \frac{I_0}{2} \) applies. Solving for \( \beta \), we have:\[ \frac{\sin \beta}{\beta} = \frac{1}{\sqrt{2}} \].
05

Solve for \( \beta \) that Satisfies Half Intensity

Using numerical methods or approximation, find \( \beta \) such that \( \frac{\sin \beta}{\beta} \approx 0.707 \). This comes to approximately \( \beta \approx 1.39 \ \, \text{radians} \).
06

Determine the Angle \(\theta\) for Half Intensity

From the relation \( \beta = \frac{\pi a \sin \theta}{\lambda} \), use \( \beta = 1.39 \), \( a = 0.360 \times 10^{-3} \), and \( \lambda = 540 \times 10^{-9} \). Solve \[ 1.39 = \frac{\pi \times 0.360 \times 10^{-3} \sin \theta}{540 \times 10^{-9}} \] giving \[ \sin \theta \] approximately \( \theta = \arcsin( \frac{1.39 \times 540 \times 10^{-9}}{\pi \times 0.360 \times 10^{-3}} ) \).
07

Find Distance to Half Intensity on the Screen

Calculate distance using \( y = L \sin \theta \) (\( \theta \) being small allows for this approximation). Substituting the calculated \( \sin \theta \), we find the distance approximately \( y \approx \frac{1.20 \times 1.39 \times 540 \times 10^{-9}}{\pi \times 0.360 \times 10^{-3}} \approx 0.0025 \text{m} = 2.5 \ \text{mm} \).

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Key Concepts

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

Diffraction Pattern
The diffraction pattern in a single-slit experiment is a result of light waves bending around the edges of the slit. When light passes through a small opening, it doesn’t just continue in a straight line. Instead, it spreads out and creates a pattern of dark and light bands on a screen placed behind the slit.
This pattern is due to interference, where waves that are in phase amplify each other (bright spots) and waves that are out of phase cancel each other out (dark spots). The central maximum is the brightest and widest part of the diffraction pattern, flanked by alternating dark and less bright regions known as minima and secondary maxima.
These patterns can be predicted using constructive and destructive interference concepts and mathematical equations like the single-slit diffraction formula. In our specific exercise, the width of the slit and the wavelength of the light are key inputs for calculating where these minima occur. This setup produces a recognizable pattern that provides valuable information about the light and its behavior.
Central Maximum Intensity
In single-slit diffraction, the central maximum is the most significant part of the diffraction pattern. It is the wide, bright band located directly opposite the slit on the screen. The intensity at the center of this maximum is considered the highest, denoted as \( I_0 \).
The width and brightness of this central band are due to constructive interference of light waves as they travel similar paths through the slit. However, as you move away from the center, the waves slightly differ in phase, reducing the intensity.
Understanding how this intensity changes helps in determining key points in the pattern, like where half intensity, \( I_0/2 \), is observed. This requires precise calculations involving phase differences. Calculating these variations in intensity is crucial for practical applications, such as resolving images in optical equipment or analyzing light properties.
Phase Angle Calculation
The phase angle \( \beta \) represents the relative phase difference between light waves fronting each other after passing through the slit.
For practical calculations, especially in determining points of half-intensity, the relationship \( I = I_0 \left(\frac{\sin \beta}{\beta}\right)^2 = \frac{I_0}{2} \) is employed. This equation helps in understanding how the intensity diminishes when moving from bright peaks towards the minima.
Solving for \( \beta \) using this relationship provides quantitative measurements critical for predicting where light intensity hits specific values on the screen, such as half of the central maximum intensity. Estimating \( \beta \) often involves iterative or numerical methods, particularly as analytic solutions become complex. This knowledge is not only theoretical but also assists with designing and interpreting experiments involving light and diffraction.

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Most popular questions from this chapter

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 \(\mathrm{mm}\) wide. At the point in the pattern \(3.25^{\circ}\) from the center of the central maximum, the total phase difference between wavelets from the tup and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maximum is \(I_{0} ?\)

A diffraction grating has 650 slits \(/ \mathrm{mm}\) . What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately \(400-700 \mathrm{nm} . )\)

Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interference pattern is at an angle of \(8.94^{\circ} .\) What is the angular position of the fourth-order maximum?

Parallel rays of green mercury light with a wavelength of 546 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of \(60.0 \mathrm{cm} .\) In the focal plane of the lens the distance from the central maximum to the first minimum is 10.2 \(\mathrm{mm}\) . What is the width of the slit?

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute). (a) Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 550 -nm light. (b) Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10- diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope ( 5 -m diameter) on Palomar Mountain in California? In what ways are they not superior? Explain.

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