The superposing waves are represented by the following equations :
`y_(1) 5 sin 2 pi (10 t-0.1 x ), y_(2)=10 sin 2 pi (20t -0.2 x)` Ratio of intensities `(I_("max"))/(I_("min")) ` will be

Options

A. 1
B. 9
C. 4
D. 16

Answer

To solve the problem of finding the ratio of maximum intensity to minimum intensity for the given superimposing waves, we will follow these steps:

### Step 1: Identify the Amplitudes
The equations of the two waves are given as:
- \( y_1 = 5 \sin(2\pi(10t - 0.1x)) \)
- \( y_2 = 10 \sin(2\pi(20t - 0.2x)) \)

From these equations, we can identify the amplitudes:
- Amplitude of \( y_1 \) (denoted as \( A_1 \)) = 5
- Amplitude of \( y_2 \) (denoted as \( A_2 \)) = 10

### Step 2: Calculate Maximum Amplitude
The maximum amplitude \( A_{\text{max}} \) when two waves interfere constructively is given by:
\[ A_{\text{max}} = A_1 + A_2 \]
Substituting the values:
\[ A_{\text{max}} = 5 + 10 = 15 \]

### Step 3: Calculate Minimum Amplitude
The minimum amplitude \( A_{\text{min}} \) when two waves interfere destructively is given by:
\[ A_{\text{min}} = |A_1 - A_2| \]
Substituting the values:
\[ A_{\text{min}} = |5 - 10| = 5 \]

### Step 4: Calculate Intensities
The intensity \( I \) of a wave is proportional to the square of its amplitude:
\[ I \propto A^2 \]
Thus, we can express the maximum and minimum intensities as:
- Maximum Intensity \( I_{\text{max}} \):
\[ I_{\text{max}} \propto A_{\text{max}}^2 = (15)^2 = 225 \]
- Minimum Intensity \( I_{\text{min}} \):
\[ I_{\text{min}} \propto A_{\text{min}}^2 = (5)^2 = 25 \]

### Step 5: Calculate the Ratio of Intensities
Now, we can find the ratio of maximum intensity to minimum intensity:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = \frac{225}{25} = 9 \]

### Final Answer
The ratio of maximum intensity to minimum intensity is:
\[ \frac{I_{\text{max}}}{I_{\text{min}}} = 9 \]