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Chapter 25: Problem 28

Show that the interference fringes in a double-slit experiment are equally spaced on a distant screen near the center of the interference pattern. [Hint: Use the small angle approximation for \(\boldsymbol{\theta} .]\)

Short Answer

Expert verified
Question: Demonstrate that the interference fringes in a double-slit experiment are equally spaced near the center of the interference pattern on a distant screen. Answer: The interference fringes are equally spaced near the center of the interference pattern because the fringe spacing, 螖y, is constant and does not depend on the fringe order, m. This can be shown using the small-angle approximation and the equation \(y = mL\frac{\lambda}{d}\), where y is the fringe position, m is the fringe order, 位 is the wavelength of light, and d is the distance between the slits. The fringe spacing is calculated as 螖y = L\frac{\lambda}{d}, which does not depend on m, indicating that the fringes are equally spaced.

Step by step solution

01

Understand the double-slit experiment setup

In a double-slit experiment, a light source produces parallel beams that pass through two narrow slits, separated by a distance, d. The lightwaves interfere as they pass through the slits, and a series of alternating bright and dark fringes are observed on a screen, placed at a distance L away from the slits. The position of bright fringes (maxima) is determined by constructive interference of the lightwaves, and the position of dark fringes (minima) is determined by destructive interference.
02

Write the equation for fringe position

The general equation for the position of bright fringes (maxima) on the screen is given by: \(\sin \theta = m\frac{\lambda}{d}\) where \(\theta\) is the angle between the center of the screen and the fringe, m is an integer called the fringe order, 位 is the wavelength of the light, and d is the distance between the slits. To find the y-coordinate of the fringe, we can use the small-angle approximation: \(\sin\theta \approx \tan\theta \approx \frac{y}{L}\), where y is the fringe position on the screen and L is the distance between the screen and the slits.
03

Replace sin(theta) with the small-angle approximation

Substituting the small-angle approximation into the fringe position equation, we get: \(\frac{y}{L} = m\frac{\lambda}{d}\) Next, we need to solve the equation for y.
04

Solve for y, the fringe position on the screen

Rearrange the equation to solve for y: \(y = mL\frac{\lambda}{d}\) This formula gives the position of the bright fringes (maxima) on the screen. The positions of the dark fringes (minima) can be found by replacing m with \((m + \frac{1}{2})\) in the above formula.
05

Find the fringe spacing

To determine the spacing between the fringes, let's take the difference between the positions of two consecutive bright fringes (maxima). Let the position of the m-th and (m+1)-th fringe be \(y_m\) and \(y_{m+1}\), respectively: \(y_m = mL\frac{\lambda}{d}\) \(y_{m+1} = (m + 1)L\frac{\lambda}{d}\) Now, subtract \(y_m\) from \(y_{m+1}\) to obtain the fringe spacing, 螖y: \(\Delta y = y_{m+1} - y_m = L\frac{\lambda}{d}\)
06

Analyze the result

The result obtained in step 5 shows that the fringe spacing, 螖y, is constant and does not depend on the fringe order, m. This means that the interference fringes are equally spaced near the center of the interference pattern. Note that the small-angle approximation is valid as long as \(\theta\) is small, which is the case near the center of the pattern.

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Most popular questions from this chapter

A steep cliff west of Lydia's home reflects a \(1020-\mathrm{kHz}\) radio signal from a station that is \(74 \mathrm{km}\) due east of her home. If there is destructive interference, what is the minimum distance of the cliff from her home? Assume there is a \(180^{\circ}\) phase shift when the wave reflects from the cliff.
The diffraction pattern from a single slit is viewed on a screen. Using blue light, the width of the central maximum is \(2.0 \mathrm{cm} .\) (a) Would the central maximum be narrower or wider if red light is used instead? (b) If the blue light has wavelength \(0.43 \mu \mathrm{m}\) and the red light has wavelength \(0.70 \mu \mathrm{m},\) what is the width of the central maximum when red light is used?
Red light of 650 nm can be seen in three orders in a particular grating. About how many slits per centimeter does this grating have?
A spectrometer is used to analyze a light source. The screen-to-grating distance is \(50.0 \mathrm{cm}\) and the grating has 5000.0 slits/cm. Spectral lines are observed at the following angles: $12.98^{\circ}, 19.0^{\circ}, 26.7^{\circ}, 40.6^{\circ}, 42.4^{\circ}\( \)63.9^{\circ},\( and \)77.6^{\circ} .$ (a) How many different wavelengths are present in the spectrum of this light source? Find each of the wavelengths. (b) If a different grating with 2000.0 slits/cm were used, how many spectral lines would be seen on the screen on one side of the central maximum? Explain.
A grating spectrometer is used to resolve wavelengths \(660.0 \mathrm{nm}\) and \(661.4 \mathrm{nm}\) in second order. (a) How many slits per centimeter must the grating have to produce both wavelengths in second order? (The answer is either a maximum or a minimum number of slits per centimeter.) (b) The minimum number of slits required to resolve two closely spaced lines is $N=\lambda /(m \Delta \lambda),\( where \)\lambda$ is the average of the two wavelengths, \(\Delta \lambda\) is the difference between the two wavelengths, and \(m\) is the order. What minimum number of slits must this grating have to resolve the lines in second order?
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