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Chapter 25: Problem 8

Coherent light from a laser is split into two beams with intensities \(I_{0}\) and \(4 I_{0},\) respectively. What is the intensity of the light when the beams are recombined? If there is more than one possibility, give the range of possibilities. (tutorial: two waves).

Short Answer

Expert verified
Answer: The range of intensities when the beams are recombined is from \(I_0\) to \(9I_0\).

Step by step solution

01

Find the Amplitudes of the Waves

The intensity of a wave is proportional to the square of its amplitude. Let's suppose that the amplitudes of the waves are \(A_1\) and \(A_2\), respectively. From the given intensities, we have: \(I_0 \propto A_1^2\) \(4I_0 \propto A_2^2\) By using proportionality, we can find the amplitudes of both waves: \(A_1 = k\sqrt{I_0}\) \(A_2 = k\sqrt{4I_0}\)
02

Constructive Interference

For the waves to interfere constructively, they must have a phase difference of \(2n\pi\) where \(n\) is an integer. In this case, the total amplitude of the recombined wave will be the sum of the individual amplitudes: \(A_{tot} = A_1 + A_2 = k\sqrt{I_0} + k\sqrt{4I_0} = k\sqrt{I_0} + 2k\sqrt{I_0}\) The intensity of the recombined wave is the square of the total amplitude: \(I_{max} \propto (A_{tot})^2 \propto (k\sqrt{I_0} + 2k\sqrt{I_0})^2\)
03

Destructive Interference

For the waves to interfere destructively, they must have a phase difference of \((2n+1)\pi\) where \(n\) is an integer. In this case, the total amplitude of the recombined wave will be the difference between the individual amplitudes: \(A_{tot} = |A_1 - A_2| = |k\sqrt{I_0} - 2k\sqrt{I_0}|\) The intensity of the recombined wave is the square of the total amplitude: \(I_{min} \propto (A_{tot})^2 \propto (k\sqrt{I_0} - 2k\sqrt{I_0})^2\)
04

Range of Intensities

Now, we can find the range of intensities for the recombined wave by finding the maximum and minimum intensities: \(I_{max} \propto (k\sqrt{I_0} + 2k\sqrt{I_0})^2 = (3\sqrt{I_0})^2 = 9I_0\) \(I_{min} \propto (k\sqrt{I_0} - 2k\sqrt{I_0})^2 = (-\sqrt{I_0})^2 = I_0\) So, the range of intensities when the beams are recombined is from \(I_0\) to \(9I_0\).

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Most popular questions from this chapter

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.
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Two radio towers are a distance \(d\) apart as shown in the overhead view. Each antenna by itself would radiate equally in all directions in a horizontal plane. The radio waves have the same frequency and start out in phase. A detector is moved in a circle around the towers at a distance of $100 \mathrm{km}.$ The waves have frequency \(3.0 \mathrm{MHz}\) and the distance between antennas is \(d=0.30 \mathrm{km} .\) (a) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=0^{\circ} ?\) (b) What is the difference in the path lengths traveled by the waves that arrive at the detector at \(\theta=90^{\circ} ?\) (c) At how many angles $\left(0 \leq \theta<360^{\circ}\right)$ would you expect to detect a maximum intensity? Explain. (d) Find the angles \((\theta)\) of the maxima in the first quadrant \(\left(0 \leq \theta \leq 90^{\circ}\right) .\) (e) Which (if any) of your answers to parts (a) to (d) would change if the detector were instead only $1 \mathrm{km}$ from the towers? Explain. (Don't calculate new values for the answers.)
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