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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 grating \(1.600 \mathrm{cm}\) wide has exactly 12000 slits. The grating is used to resolve two nearly equal wavelengths in a light source: \(\lambda_{\mathrm{a}}=440.000 \mathrm{nm}\) and $\lambda_{\mathrm{b}}=440.936 \mathrm{nm}$ (a) How many orders of the lines can be seen with the grating? (b) What is the angular separation \(\theta_{b}-\theta_{a}\) between the lines in each order? (c) Which order best resolves the two lines? Explain.
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