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Chapter 35: Problem 19

35.19. In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_{0}\) . (a) At a point in the pattern where the phase difference between the waves from the two slits is \(60.0^{\circ}\) , what is the intensity? (b) What is the path difference for 480 -nm light from the two slits at a point where the phase angle is \(60.0^{\circ} ?\)

Short Answer

Expert verified
(a) The intensity is \(\frac{3}{4} I_{0}\); (b) the path difference is 80 nm.

Step by step solution

01

Understanding Intensity Relation

The intensity of light in a two-slit interference pattern can be calculated using the formula for interference: \[ I = I_{0} \cos^2\left(\frac{\phi}{2}\right) \]where \(I_0\) is the intensity at the central maximum and \(\phi\) is the phase difference between the waves from the two slits.
02

Calculate Intensity with Phase Difference

We are given a phase difference \(\phi = 60.0^{\circ}\). Convert the degree to radians since trigonometric functions typically use radians:\[\phi = 60.0^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{\pi}{3} \text{ radians}\].Substitute \(\phi\) into the intensity formula:\[ I = I_{0} \cos^2\left(\frac{\phi}{2}\right) = I_{0} \cos^2\left(\frac{\pi}{6}\right) = I_{0} \times \left(\frac{\sqrt{3}}{2}\right)^2 = I_{0} \times \frac{3}{4} \].Thus, the intensity is \(\frac{3}{4} I_{0}\).
03

Understanding Path Difference Relation

The phase difference \(\phi\) is related to the path difference \(\Delta r\) by the equation:\[ \phi = \frac{2\pi}{\lambda} \Delta r \]where \(\lambda\) is the wavelength of light.
04

Calculate Path Difference

We know \(\phi = \frac{\pi}{3} \) and \(\lambda = 480 \) nm (convert nm to meters if necessary: \(1\, \text{nm} = 10^{-9}\) m). Substitute into the formula:\[ \frac{\pi}{3} = \frac{2\pi}{480 \times 10^{-9}} \Delta r \]Solving for \(\Delta r\), we get:\[ \Delta r = \frac{480 \times 10^{-9}}{2} \times \frac{1}{3} = 80 \times 10^{-9} \text{ m (or }80 \text{ nm)} \].

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Key Concepts

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

Phase Difference
When light waves pass through two slits, the difference in phase between the waves greatly affects the interference pattern produced on a screen. The phase difference, denoted as \( \phi \), occurs when the waves from the two slits start at different positions in their wave cycles. This often results in either constructive or destructive interference. - **Constructive interference** happens when the waves arrive in phase, peaking at the central maximum with maximum brightness.- **Destructive interference** occurs when they are out of phase, reducing intensity at certain points.The phase difference \( \phi \) can be given in degrees or converted into radians, which is often more useful in calculations involving trigonometric functions. For example, \( 60.0^{\circ} \) converts to \( \frac{\pi}{3} \) radians. Understanding this relationship is crucial for determining the resulting intensity at various points on the interference pattern.
Intensity Calculation
To find the intensity of light at different points in the two-slit interference pattern, you use the formula:\[I = I_{0} \cos^2\left(\frac{\phi}{2}\right)\]Here's what each part represents:- \( I \): Intensity at a point in the interference pattern- \( I_{0} \): Maximum intensity at the central maximum- \( \phi \): Phase difference in radiansWith a phase difference of \( 60.0^{\circ} \) (or \( \frac{\pi}{3} \) radians), plug it into the formula to get:\[I = I_{0} \cos^2\left(\frac{\pi}{6}\right) = I_{0} \times \left(\frac{\sqrt{3}}{2}\right)^2 = I_{0} \times \frac{3}{4}\]So, the intensity is \( \frac{3}{4} I_{0} \). This means that at the given point where the phase difference is \( 60.0^{\circ} \), the light intensity is 75% of the maximum intensity.
Path Difference
The concept of path difference plays a key role in understanding how the interference pattern is formed in a two-slit experiment. Path difference \( \Delta r \) is the physical difference in the distance traveled by two waves from the slits to a particular point on a screen.The relationship between phase difference \( \phi \) and path difference \( \Delta r \) is described by the equation:\[\phi = \frac{2\pi}{\lambda} \Delta r\]where \( \lambda \) represents the wavelength of the light used. For example, with \( \phi = \frac{\pi}{3} \) and \( \lambda = 480 \) nm, you can calculate \( \Delta r \) as:\[\Delta r = \frac{480 \times 10^{-9}}{2} \times \frac{1}{3} = 80 \times 10^{-9} \text{ m}\]This shows that the path difference is 80 nm when the phase difference is \( 60.0^{\circ} \). Comprehending path difference helps predict where in the pattern on the screen constructive or destructive interference will occur.

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Most popular questions from this chapter

35.3. A radio transmitting station operating at a frequency of 120 \(\mathrm{MHz}\) has two identical antennas that radiate in phase. Antenna \(B\) is 9.00 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(P\) between the antennas and along the line connecting them, a horizontal distance \(x\) to the right of antenna \(A\) . For what values of \(x\) will constructive interference occur at point \(P ?\)

35.2. Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A\) . Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q\) ?

35.20. Coherent sources \(A\) and \(B\) emit electromagnetic waves with wavelength 2.00 \(\mathrm{cm}\) . Point \(P\) is 4.86 \(\mathrm{m}\) from \(A\) and 5.24 \(\mathrm{m}\) from \(B\) . What is the phase difference at \(P\) between these two waves?

35.15. Coherent light with wavelength 600 \(\mathrm{nm}\) passes through two very narrow slits and the interference pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slits. The first-order bright fringe is at 4.84 \(\mathrm{mm}\) from the center of the central bright fringe. For what wavelength of light will the first order dark fringe be observed at this same point on the screen?

35.51. A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index 1.50 . At room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) , this film is just thick enough for light with wavelength 582.4 \(\mathrm{nm}\) reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to \(170^{\circ} \mathrm{C},\) you find that the film cancels reflected light with wavelength 588.5 \(\mathrm{nm}\) . What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)
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