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Chapter 36: Problem 9

A slit \(1.00 \mathrm{~mm}\) wide is illuminated by light of wavelength \(589 \mathrm{nm}\). We see a diffraction pattern on a screen \(3.00 \mathrm{~m}\) away. What is the distance between the first two diffraction minima on the same side of the central diffraction maximum?

Short Answer

Expert verified
The distance between the first two diffraction minima is \(1.767 \text{ mm}\).

Step by step solution

01

Understanding the Problem

We are dealing with a single-slit diffraction pattern where we need to calculate the distance between the first two diffraction minima. The width of the slit is given as \(1.00 \mathrm{~mm}\), the wavelength of the light is \(589 \mathrm{nm}\), and the screen is \(3.00 \mathrm{~m}\) away from the slit.
02

Determine the Position of the First Minimum

The position of the first diffraction minimum is given by the formula \( a \sin \theta_1 = m \lambda \), where \(a\) is the slit width, \(\lambda\) is the wavelength, and \(m\) is the order of the minimum (\(m = 1\) for the first minimum). Since \(a = 1.00 \times 10^{-3} \mathrm{~m}\) and \(\lambda = 589 \times 10^{-9} \mathrm{~m}\), we use \(m = 1\) and find \(\sin \theta_1 = \frac{1 \times 589 \times 10^{-9}}{1.00 \times 10^{-3}} = 5.89 \times 10^{-4}\).
03

Calculate the Angle to the First Minimum

Since \(\sin \theta_1\) is very small, we use the approximation \(\theta_1 \approx \sin \theta_1\), therefore, \(\theta_1 \approx 5.89 \times 10^{-4} \text{ radians}\).
04

Determine the Position of the Second Minimum

Similarly, for the second minimum, we use \(a \sin \theta_2 = 2 \lambda\), giving us \(\sin \theta_2 = \frac{2 \times 589 \times 10^{-9}}{1.00 \times 10^{-3}} = 1.178 \times 10^{-3}\).
05

Calculate the Angle for the Second Minimum

Using the small angle approximation again, we find \(\theta_2 \approx 1.178 \times 10^{-3} \text{ radians}\).
06

Calculate the Tangential Distances on the Screen

The linear distance \(y\) from the central maximum on the screen to a minimum is given by \(y_m = L \tan \theta_m\). With \(L = 3.00 \text{ m}\):For the first minimum: \(y_1 \approx 3.00 \times 5.89 \times 10^{-4} = 1.767 \times 10^{-3} \text{ m}\).For the second minimum: \(y_2 \approx 3.00 \times 1.178 \times 10^{-3} = 3.534 \times 10^{-3} \text{ m}\).
07

Find the Distance Between the Minima

The distance between the first and second minima is calculated as the difference \(y_2 - y_1 = 3.534 \times 10^{-3} - 1.767 \times 10^{-3} = 1.767 \times 10^{-3} \text{ m}\) or \(1.767 \text{ mm}\).

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Key Concepts

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

Diffraction Pattern
When light passes through a narrow opening or single slit, it doesn't just go straight through. Instead, it spreads out and creates a diffraction pattern. This spreading of light waves is a result of diffraction, a phenomenon that occurs when waves encounter obstacles or openings. A diffraction pattern consists of a series of light and dark bands observed on a screen. These are formed because light waves interfere with each other, creating areas of constructive (light) and destructive (dark) interference. - The central bright band is called the central diffraction maximum. - Flanking this central maximum, you will find dark areas known as minima and additional, smaller maxima. Understanding this pattern is crucial because it allows us to measure various properties like the wavelength of light and slit dimensions.
Wavelength
Wavelength is one of the fundamental parameters that define a wave. It is the distance between two consecutive points that are in phase, such as crest to crest or trough to trough. In the context of diffraction:- Wavelength, denoted by \( \lambda \), plays a significant role in determining the diffraction pattern you observe.- Longer wavelengths generally result in broader diffraction patterns, whereas shorter wavelengths produce tighter patterns.In the given exercise, the wavelength of the light illuminating the slit is specified as 589 nm (nanometers). This specific wavelength will influence the size and positions of those light and dark bands on the screen behind the slit, using the relationship \( a \sin \theta = m \lambda \) to calculate exact positions for minima.
Diffraction Minima
Diffraction minima are the dark spots in a diffraction pattern where the intensity of light falls to zero. These occur because the waves interfere destructively at these points.The positions of these minima can be calculated using the formula \( a \sin \theta = m \lambda \), where - \( a \) is the slit width,- \( \lambda \) is the wavelength,- \( m \) is the order of the minimum.For the first minimum, \( m = 1 \), for the second, \( m = 2 \), and so on.To find where these minima appear on the screen, we often use the small-angle approximation \( \sin \theta \approx \theta \), especially when \( \theta \) is small. This allows us to relate angles to positions on the screen behind the slit easily.By precisely understanding the spacing between these minima, one can determine critical optical properties and verify theoretical models.

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Most popular questions from this chapter

Light at wavelength \(589 \mathrm{nm}\) from a sodium lamp is incident perpendicularly on a grating with 40000 rulings over width \(76 \mathrm{~mm}\). What are the first-order (a) dispersion \(D\) and (b) resolving power \(R,\) the second-order (c) \(D\) and (d) \(R\), and the third-order (e) \(D\) and (f) \(R ?\)

Two emission lines have wavelengths \(\lambda\) and \(\lambda+\Delta \lambda,\) respectively, where \(\Delta \lambda<\lambda\). Show that their angular separation \(\Delta \theta\) in a grating spectrometer is given approximately by $$\Delta \theta=\frac{\Delta \lambda}{\sqrt{(d / m)^{2}-\lambda^{2}}}$$ where \(d\) is the slit separation and \(m\) is the order at which the lines are observed. Note that the angular separation is greater in the higher orders than the lower orders.

Millimeter-wave radar generates a narrower beam than conventional microwave radar, making it less vulnerable to antiradar missiles than conventional radar. (a) Calculate the angular width \(2 \theta\) of the central maximum, from first minimum to first minimum, produced by a 220 GHz radar beam emitted by a \(55.0-\mathrm{cm}-\) diameter circular antenna. (The frequency is chosen to coincide with a low-absorption atmospheric "window.") (b) What is \(2 \theta\) for a more conventional circular antenna that has a diameter of \(2.3 \mathrm{~m}\) and emits at wavelength \(1.6 \mathrm{~cm} ?\)

A beam of light consisting of wavelengths from \(460.0 \mathrm{nm}\) to \(640.0 \mathrm{nm}\) is directed perpendicularly onto a diffraction grating with 160 lines/mm. (a) What is the lowest order that is overlapped by another order? (b) What is the highest order for which the complete wavelength range of the beam is present? In that highest order, at what angle does the light at wavelength (c) \(460.0 \mathrm{nm}\) and (d) \(640.0 \mathrm{nm}\) appear? (e) What is the greatest angle at which the light at wavelength \(460.0 \mathrm{nm}\) appears?

A grating has 350 rulings/mm and is illuminated at normal incidence by white light. A spectrum is formed on a screen \(30.0 \mathrm{~cm}\) from the grating. If a hole \(10.0 \mathrm{~mm}\) square is cut in the screen, its inner edge being \(50.0 \mathrm{~mm}\) from the central maximum and parallel to it, what are the (a) shortest and (b) longest wavelengths of the light that passes through the hole?
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