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Step-by-Step Solution
Step 1: Understand the problem
We have three harmonic waves with:
Equal frequency, denoted by $ \nu $.
Same intensity, $ I_0 $.
Phase angles: $0$, $ \frac{\pi}{4} $, and $ -\frac{\pi}{4} $.
We need to find the resultant intensity when these three waves are superimposed.
Step 2: Relate intensity and amplitude
For a wave of intensity $ I_0 $, the corresponding amplitude (let us denote it by $ A_0 $) is:
$$ A_0 = \sqrt{I_0}. $$
Thus, each of the three waves has the same amplitude $ \sqrt{I_0} $.
Step 3: Represent the waves as phasors
Let us denote the amplitudes in complex form:
Wave 1 (phase $0$):
$$ A_1 = \sqrt{I_0} \, e^{i \cdot 0} = \sqrt{I_0}. $$
Wave 2 (phase $\frac{\pi}{4}$):
$$ A_2 = \sqrt{I_0} \, e^{i \frac{\pi}{4}} = \sqrt{I_0} \left(\cos \frac{\pi}{4} + i \,\sin \frac{\pi}{4}\right)
= \sqrt{I_0} \left(\frac{1}{\sqrt{2}} + i \,\frac{1}{\sqrt{2}}\right). $$
Wave 3 (phase $-\frac{\pi}{4}$):
$$ A_3 = \sqrt{I_0} \, e^{-\,i \frac{\pi}{4}} = \sqrt{I_0} \left(\cos \left(-\frac{\pi}{4}\right) + i \,\sin \left(-\frac{\pi}{4}\right)\right)
= \sqrt{I_0} \left(\frac{1}{\sqrt{2}} - i \,\frac{1}{\sqrt{2}}\right). $$
Step 4: Add the amplitudes
The total amplitude $ A_{\text{total}} $ is the vector (phasor) sum of $ A_1 $, $ A_2 $, and $ A_3 $:
$$
A_{\text{total}} = A_1 + A_2 + A_3.
$$
Substitute the values:
$$
A_{\text{total}} = \sqrt{I_0} \; + \;
\sqrt{I_0} \left(\frac{1}{\sqrt{2}} + i \,\frac{1}{\sqrt{2}}\right)
\;+\;
\sqrt{I_0} \left(\frac{1}{\sqrt{2}} - i \,\frac{1}{\sqrt{2}}\right).
$$
Combine like terms:
Real part:
$$
\sqrt{I_0} \Big( 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \Big)
= \sqrt{I_0} \Big( 1 + \sqrt{2} \Big).
$$
Imaginary part:
$$
\sqrt{I_0} \left( i \,\frac{1}{\sqrt{2}} - i \,\frac{1}{\sqrt{2}} \right) = 0.
$$
Therefore,
$$
A_{\text{total}} = \sqrt{I_0} \big( 1 + \sqrt{2} \big).
$$
Step 5: Calculate the resultant intensity
Intensity is proportional to the square of the amplitude. Hence the resultant intensity $ I_{\text{resultant}} $ is:
$$
I_{\text{resultant}} = \big| A_{\text{total}} \big|^2
= \Big[ \sqrt{I_0} \big( 1 + \sqrt{2} \big) \Big]^2
= I_0 \, \big( 1 + \sqrt{2} \big)^2.
$$
Now expand $ (1 + \sqrt{2})^2 $:
$$
(1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}.
$$
Therefore,
$$
I_{\text{resultant}} = I_0 \, (3 + 2\sqrt{2}).
$$
Step 6: Numerical approximation
The approximate value of $ 2\sqrt{2} $ is about $ 2 \times 1.414 = 2.828 $. Hence:
$$
3 + 2\sqrt{2} \approx 3 + 2.828 = 5.828.
$$
This is approximately $ 5.8 $. Therefore,
$$
I_{\text{resultant}} \approx 5.8 \, I_0.
$$
Final Answer
The resultant intensity is approximately $ 5.8 \, I_0 $, which matches the given correct option.
