Two coherent waves are represented by `y_(1)=a_(1)cos_(omega)` t and `y_(2)=a_(2)sin_(omega)`t. The resultant intensity due to interference will be

Options

A. `(a_(1)^(2)-a_(2)^(2))`
B. `(a_(1)^(2)+a_(2)^(2))`
C. `(a_(1)-a_(2))`
D. `(a_(1)+a_(2))`

Answer

To find the resultant intensity due to the interference of two coherent waves represented by \( y_1 = A_1 \cos(\omega t) \) and \( y_2 = A_2 \sin(\omega t) \), we can follow these steps:

### Step 1: Identify the Intensities of Individual Waves
The intensity of a wave is proportional to the square of its amplitude. Therefore, the intensity of the first wave \( y_1 \) is given by:
\[ I_1 = k A_1^2 \]
where \( k \) is a constant of proportionality.

Similarly, the intensity of the second wave \( y_2 \) is:
\[ I_2 = k A_2^2 \]

### Step 2: Calculate the Resultant Intensity
When two coherent waves interfere, the resultant intensity \( I \) can be calculated using the formula:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
where \( \phi \) is the phase difference between the two waves.

In this case, since \( y_1 \) is a cosine wave and \( y_2 \) is a sine wave, they are \( 90^\circ \) out of phase. Thus, \( \cos(\phi) = \cos(90^\circ) = 0 \).

So, the formula simplifies to:
\[ I = I_1 + I_2 \]

### Step 3: Substitute the Intensities
Substituting the expressions for \( I_1 \) and \( I_2 \):
\[ I = k A_1^2 + k A_2^2 \]

### Step 4: Factor Out the Constant
Factoring out the constant \( k \):
\[ I = k (A_1^2 + A_2^2) \]

### Step 5: Conclusion
The resultant intensity due to the interference of the two coherent waves is:
\[ I = k (A_1^2 + A_2^2) \]

### Final Result
Thus, the resultant intensity is proportional to the sum of the squares of the amplitudes of the two waves.