At two point P and Q on screen in Young's double slit experiment, waves from slits `S_(1)` and `S_(2)` have a path difference of 0 and `(lamda)/(4)` respectively. The ratio of intensities at P and Q will be:

Options

A. `2 : 1`
B. `sqrt2 : 1`
C. `4 : 1`
D. `3 : 2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given conditions
We have two points, P and Q, on a screen in Young's double slit experiment. The path difference for point P is 0, and for point Q, it is \( \frac{\lambda}{4} \).

### Step 2: Define the intensities from the slits
Let the intensity of light from slit \( S_1 \) be \( I_1 = I_0 \) and from slit \( S_2 \) be \( I_2 = I_0 \) (assuming both slits emit light of equal intensity).

### Step 3: Write the formula for resultant intensity
The resultant intensity \( I_R \) at any point on the screen can be expressed as:
\[ I_R = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( \phi \) is the phase difference between the two waves arriving at that point.

### Step 4: Calculate the phase difference at point P
For point P, the path difference is 0:
\[ \phi_P = \frac{2\pi}{\lambda} \times 0 = 0 \]
Substituting this into the intensity formula:
\[ I_P = 4 I_0 \cos^2(0) = 4 I_0 \cdot 1 = 4 I_0 \]

### Step 5: Calculate the phase difference at point Q
For point Q, the path difference is \( \frac{\lambda}{4} \):
\[ \phi_Q = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \]
Now substituting this into the intensity formula:
\[ I_Q = 4 I_0 \cos^2\left(\frac{\pi}{4}\right) = 4 I_0 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = 4 I_0 \cdot \frac{1}{2} = 2 I_0 \]

### Step 6: Find the ratio of intensities at points P and Q
Now we can find the ratio of intensities:
\[ \frac{I_P}{I_Q} = \frac{4 I_0}{2 I_0} = 2 \]
Thus, the ratio of intensities at points P and Q is:
\[ \text{Ratio } I_P : I_Q = 2 : 1 \]

### Final Answer
The ratio of intensities at points P and Q is \( 2 : 1 \).

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