Incident wave `y= A sin (ax + bt+ pi/2)` is reflected by an obstacle at x = 0 which reduces intensity of reflected wave by 36%. Due to superposition, the resulting wave consists of a standing wave and a travelling wave given by
`y= -1.6 sin ax sin bt + cA cos (bt + ax)`
where A, a, b and c are positive constants.
1. Amplitude of reflected wave is

Options

A. 0.6A
B. 0.8A
C. 0.4A
D. 0.2A

Explanation

To find the amplitude of the reflected wave, we can follow these steps:

### Step 1: Understand the relationship between intensity and amplitude
The intensity \( I \) of a wave is proportional to the square of its amplitude \( A \):
\[ I \propto A^2 \]

### Step 2: Define the intensities of the incident and reflected waves
Let \( I_0 \) be the intensity of the incident wave and \( I_R \) be the intensity of the reflected wave. According to the problem, the intensity of the reflected wave is reduced to 64% of the incident wave:
\[ I_R = 0.64 I_0 \]

### Step 3: Relate the intensities to the amplitudes
Using the relationship between intensity and amplitude, we can express the intensities in terms of their respective amplitudes:
\[ I_0 \propto A_0^2 \quad \text{and} \quad I_R \propto A_R^2 \]
Thus, we can write:
\[ I_R = k A_R^2 \quad \text{and} \quad I_0 = k A_0^2 \]
where \( k \) is a proportionality constant.

### Step 4: Substitute the intensity relationship
From the relationship \( I_R = 0.64 I_0 \), we can substitute:
\[ k A_R^2 = 0.64 (k A_0^2) \]
This simplifies to:
\[ A_R^2 = 0.64 A_0^2 \]

### Step 5: Solve for the amplitude of the reflected wave
Taking the square root of both sides gives:
\[ A_R = \sqrt{0.64} A_0 \]
Calculating the square root:
\[ A_R = 0.8 A_0 \]

### Conclusion
Thus, the amplitude of the reflected wave is:
\[ A_R = 0.8 A \]