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Chapter 4: Problem 24

(a) Find the angle between the first minima for the two sodium vapor lines, which have wavelengths of 589.1 and \(589.6 \mathrm{nm},\) when they fall upon a single slit of width \(2.00 \mu \mathrm{m} .\) (b) What is the distance between these minima if the diffraction pattern falls on a screen \(1.00 \mathrm{m}\) from the slit? (c) Discuss the ease or difficulty of measuring such a distance.

Short Answer

Expert verified
(a) The angles between the first minima for the two sodium vapor lines are approximately \(17.11^\circ\) and \(17.13^\circ\). (b) The distance between the first minima for the two sodium vapor lines on the screen is approximately \(0.25\ mm\). (c) Measuring this small distance of \(0.25\ mm\) can be challenging using standard tools but can be feasible with specialized instruments or experimental techniques.

Step by step solution

01

Calculate the angles between the first minima for both sodium vapor lines

We will use the single-slit diffraction formula \(\sin \theta = \frac{m\lambda}{a}\) to find the angles for both sodium vapor lines' first minima. For the first sodium vapor line (\(589.1\ nm\)): \(\sin \theta_1 = \frac{m\lambda_1}{a}\) \[\sin \theta_1 = \frac{(1)(589.1 \times 10^{-9}\ \mathrm{m})}{(2.00 \times 10^{-6}\ \mathrm{m})}\] \(\sin \theta_1 \approx 0.29455\) Now, find \(\theta_1\): \[\theta_1 = \arcsin(0.29455)\] \[\theta_1 \approx 17.11^\circ\] For the second sodium vapor line (\(589.6\ nm\)): \(\sin \theta_2 = \frac{m\lambda_2}{a}\) \[\sin \theta_2 = \frac{(1)(589.6 \times 10^{-9}\ \mathrm{m})}{(2.00 \times 10^{-6}\ \mathrm{m})}\] \(\sin \theta_2 \approx 0.29480\) Now, find \(\theta_2\): \[\theta_2 = \arcsin(0.29480)\] \[\theta_2 \approx 17.13^\circ\] The angles between the first minima for the two sodium vapor lines are approximately \(17.11^\circ\) and \(17.13^\circ\).
02

Calculate the distance between the minima on the screen

Now that we have the angles between the first minima for both sodium vapor lines, we can calculate the distance between these minima on the screen using the small angle approximation (\(\tan \theta \approx \sin \theta \approx \frac{y}{L}\), where \(y\) is the distance between the minimum and the central maximum on the screen, and \(L\) is the distance between the slit and the screen): For the first sodium vapor line: \(y_1 = L\sin \theta_1\) \[y_1 = (1.00\ \mathrm{m})(0.29455)\] \[y_1 \approx 0.29455\ \mathrm{m}\] For the second sodium vapor line: \(y_2 = L\sin \theta_2\) \[y_2 = (1.00\ \mathrm{m})(0.29480)\] \[y_2 \approx 0.29480\ \mathrm{m}\] Now, find the distance between the minima on the screen (\(\Delta y\)): \[\Delta y = |y_2 - y_1|\] \[\Delta y \approx 0.00025\ \mathrm{m}\] The distance between the first minima for the two sodium vapor lines on the screen is approximately \(0.25\ mm\).
03

Discuss the ease or difficulty of measuring the distance

The distance between the first minima for the two sodium vapor lines on the screen is \(0.25\ mm\), which is a small distance. Measuring such a small distance can be challenging using standard tools like a ruler or calipers. However, using specialized instruments (e.g., high-resolution cameras, microscopes) or experimental techniques (e.g., interference patterns, spectroscopy), it may be possible to accurately measure this distance. Overall, measuring this distance would likely be somewhat difficult but feasible with the right tools and techniques.

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Key Concepts

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

Wavelength
The core of understanding light's interaction with slits is the concept of wavelength, symbolized as \(\lambda\). Wavelength is the distance between successive crests of a wave, typically measured in meters for light waves. In our scenario, the two sodium vapor lines have wavelengths of 589.1 nm and 589.6 nm. The different wavelengths cause each color of light to diffract differently when passing through the same slit.

Since light behaves as both a particle and a wave, its wavelength determines how it will spread out after passing through a narrow opening, like the single slit in our exercise. When the wavelength is similar to the size of the slit, significant diffraction occurs, creating a pronounced diffraction pattern. As the wavelength changes, even slightly, this will alter the location of the minima and maxima in the pattern.
Diffraction Pattern
A diffraction pattern is a series of light and dark bands resulting from the wave-like characteristics of light. When light passes through a single slit and encounters edges, it disperses and interferes with itself, resulting in areas of constructive and destructive interference. These are seen as the maxima (bright bands) and minima (dark bands) on a screen.

The appearance of the diffraction pattern depends on the slit's width and the light's wavelength. In our case, the slit width is \(2.00\ \mu\mathrm{m}\), which affects how sharply light spreads out into the pattern. Our exercise calculates the angle to the first minimum of the pattern for two closely related wavelengths, illustrating the sensitivity of the pattern to small changes in wavelength.
Angle of Diffraction
The angle of diffraction, denoted as \(\theta\), is the angle at which light spreads out from the original direction of the wave after passing through the slit. It is measured between the direction of the incident light and the line drawn from the slit to a particular minimum or maximum in the diffraction pattern.

In the solution steps, the angle of diffraction was calculated using the formula \(\sin \theta = \frac{m\lambda}{a}\), where \(m\) is the order of the minimum, \(\lambda\) is the wavelength, and \(a\) is the slit width. The angles were then used to find the position of the first minima on the screen for two different wavelengths, which allows us to understand and predict the location and spacing of features within a diffraction pattern.
Small Angle Approximation
In physics and astronomy, the small angle approximation is a useful tool for simplifying calculations. When the angle is small, measured in radians, the sine and tangent of the angle are approximately equal to the angle itself when the angle is expressed in radians. This is written as \(\tan(\theta) \approx \sin(\theta) \approx \theta\).

The approximation greatly simplifies the math in many cases, often without significantly compromising accuracy. We applied this approximation in our exercise to convert the calculated angles of diffraction into the linear distance between minima on the screen. When \(\theta\) is small, we can use \(y = L\sin(\theta)\approx L\theta\) to find the linear distance \(y\) on the screen, with \(L\) being the distance between the slit and screen. This makes it easier to calculate distances when we know the angle and vice versa.

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Most popular questions from this chapter

Blue light of wavelength 450 nm falls on a slit of width \(0.25 \mathrm{mm}\). A converging lens of focal length \(20 \mathrm{cm}\) is placed behind the slit and focuses the diffraction pattern on a screen. (a) How far is the screen from the lens? (b) What is the distance between the first and the third minima of the diffraction pattern?

(a) What is the minimum angular spread of a 633 -nm wavelength He-Ne laser beam that is originally \(1.00 \mathrm{mm}\) in diameter? (b) If this laser is aimed at a mountain cliff 15.0 km away, how big will the illuminated spot be? (c) How big a spot would be illuminated on the moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round- trip time and, hence, distance.)

A source of light having two wavelengths \(550 \mathrm{nm}\) and \(600 \mathrm{nm}\) of equal intensity is incident on a slit of width \(1.8 \mu \mathrm{m} .\) Find the separation of the \(m=1\) bright spots of the two wavelengths on a screen \(30.0 \mathrm{cm}\) away.

Determine the intensities of two interference peaks other than the central peak in the central maximum of the diffraction, if possible, when a light of wavelength \(628 \mathrm{nm}\) is incident on a double slit of width \(500 \mathrm{nm}\) and separation \(1500 \mathrm{nm}\). Use the intensity of the central spot to be \(1 \mathrm{mW} / \mathrm{cm}^{2}\)

Suppose that the central peak of a single-slit diffraction pattem is so wide that the first minima can be assumed to occur at angular positions of \(\pm 90^{\circ} .\) For this case, what is the ratio of the slit width to the wavelength of the light?
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