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Chapter 9: Problem 54

Two waves having the intensities in the ratio of \(9: 1\) produce interference. The ratio of maximum to minimum intensity is equal to (A) \(10: 8\) (B) \(9: 1\) (C) \(4: 1\) (D) \(2: 1\)

Short Answer

Expert verified
The ratio of maximum to minimum intensity when two waves with intensities in the ratio of \(9:1\) produce interference is (C) \(4:1\).

Step by step solution

01

Understanding constructive and destructive interference

Constructive interference occurs when two waves meet and combine to create a wave with greater amplitude. On the other hand, destructive interference occurs when two waves meet and cancel each other out due to their opposite phases, resulting in a wave with lower amplitude.
02

Finding the amplitudes of the two waves

Given the intensity ratio, we are asked to find the ratio of maximum to minimum intensity following interference. We first need to find the amplitudes of the two waves. Since the intensity is proportional to the square of amplitude, we can write it as: \[\frac{I_1}{I_2} = \frac{A_1^2}{A_2^2}\] Substituting the given ratio of intensities, \(9:1\): \[\frac{9}{1} = \frac{A_1^2}{A_2^2}\] Take the square root of both sides to obtain: \[3 = \frac{A_1}{A_2}\] So, the ratio of the amplitudes is \(3:1\). Let's call the amplitudes \(3A\) and \(A\) for waves 1 and 2, respectively.
03

Calculating maximum and minimum intensity

For constructive interference, the amplitudes combine, resulting in maximum combined amplitude (\(A_{max}\)): \[A_{max} = A_1 + A_2 = 3A + A = 4A\] For destructive interference, the amplitudes cancel each other, resulting in minimum combined amplitude (\(A_{min}\)): \[A_{min} = \lvert A_1 - A_2 \rvert = \lvert 3A - A \rvert = 2A\] Now, we need to find the maximum and minimum intensities using these amplitudes: \[I_{max} = (A_{max})^2 = (4A)^2 = 16A^2\] \[I_{min} = (A_{min})^2 = (2A)^2 = 4A^2\]
04

Finding the ratio of maximum to minimum intensity

Now that we have the maximum and minimum intensities, we can find the ratio: \[\frac{I_{max}}{I_{min}} = \frac{16A^2}{4A^2} = \frac{16}{4} = 4\] Thus, the ratio of maximum to minimum intensity is \(4:1\), which matches option (C). So the correct answer is (C) \(4:1\).

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Two monochromatic coherent point sources \(S_{I}\) and \(S_{2}\) are separated by a distance \(L .\) Each source emits light of wavelength \(\lambda\), where \(L \gg \lambda\). The line \(S_{1} S_{2}\) when extended meets a screen perpendicular to it at a point \(A\). (A) The interference fringes on the screen are circular in shape. (B) The interference fringes on the screen are straight lines perpendicular to the line \(S_{1} S_{2} A\). (C) The point \(A\) is an intensity maxima if \(L=n \lambda\). (D) The point \(A\) is always an intensity maxima for any separation \(L\).
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