Two light waves are represented by `y_(1)=a sin_(omega)t and y_(2)= a sin(omega t+delta)`. The phase of the resultant wave is

Options

A. `2delta`
B. `(delta)/(2)`
C. `(delta)/(3)`
D. `(delta)/(4)`

Answer

To find the phase of the resultant wave formed by the superposition of two light waves represented by the equations \( y_1 = a \sin(\omega t) \) and \( y_2 = a \sin(\omega t + \delta) \), we can follow these steps:

### Step 1: Write the equations of the two waves
The two waves are given as:
\[ y_1 = a \sin(\omega t) \]
\[ y_2 = a \sin(\omega t + \delta) \]

### Step 2: Use the sine addition formula
Using the sine addition formula, we can express \( y_2 \):
\[ y_2 = a \sin(\omega t + \delta) = a \left( \sin(\omega t) \cos(\delta) + \cos(\omega t) \sin(\delta) \right) \]

### Step 3: Combine the two wave equations
Now, we can add \( y_1 \) and \( y_2 \):
\[ y = y_1 + y_2 = a \sin(\omega t) + a \left( \sin(\omega t) \cos(\delta) + \cos(\omega t) \sin(\delta) \right) \]
This simplifies to:
\[ y = a \sin(\omega t) (1 + \cos(\delta)) + a \cos(\omega t) \sin(\delta) \]

### Step 4: Factor out the amplitude
We can factor out the amplitude \( a \):
\[ y = a \left( \sin(\omega t) (1 + \cos(\delta)) + \cos(\omega t) \sin(\delta) \right) \]

### Step 5: Identify the resultant phase
The resultant wave can be expressed in the form \( R \sin(\omega t + \phi) \), where \( R \) is the resultant amplitude and \( \phi \) is the phase. To find \( \phi \), we can use the coefficients of \( \sin(\omega t) \) and \( \cos(\omega t) \):
- Coefficient of \( \sin(\omega t) \): \( a(1 + \cos(\delta)) \)
- Coefficient of \( \cos(\omega t) \): \( a \sin(\delta) \)

Using the tangent function:
\[ \tan(\phi) = \frac{\text{Coefficient of } \cos(\omega t)}{\text{Coefficient of } \sin(\omega t)} = \frac{a \sin(\delta)}{a(1 + \cos(\delta))} = \frac{\sin(\delta)}{1 + \cos(\delta)} \]

### Step 6: Final expression for phase
Thus, the phase of the resultant wave is given by:
\[ \phi = \tan^{-1}\left(\frac{\sin(\delta)}{1 + \cos(\delta)}\right) \]