In YDSE, `D = 2 mm, D = 2 m,` and `lambda = 500 nm`. If intensities of two slits are `I_(0)` and `9I_(0)`, then find intensity at `y = (1)/(6) mm`.

Options

A. `7 I_(0)`
B. `10 I_(0)`
C. `16 I_(0)`
D. `4 I_(0)`

Explanation

To solve the problem step by step, we will follow the concepts of Young's Double Slit Experiment (YDSE) and the formula for intensity at a point on the screen.

### Step 1: Identify the Given Values
- Distance between the slits, \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 2 \, \text{m} \)
- Wavelength, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- Intensities of the two slits: \( I_1 = I_0 \) and \( I_2 = 9I_0 \)
- Distance from the center to the point of interest on the screen, \( y = \frac{1}{6} \, \text{mm} = \frac{1}{6} \times 10^{-3} \, \text{m} \)

### Step 2: Calculate the Path Difference
The path difference \( \Delta x \) at a point on the screen can be calculated using the formula:
\[ \Delta x = \frac{d \cdot y}{D} \]
Substituting the values:
\[ \Delta x = \frac{(2 \times 10^{-3}) \cdot \left(\frac{1}{6} \times 10^{-3}\right)}{2} \]
\[ \Delta x = \frac{2 \times 10^{-3} \cdot \frac{1}{6} \times 10^{-3}}{2} = \frac{1 \times 10^{-6}}{2} = 0.5 \times 10^{-6} \, \text{m} \]

### Step 3: Calculate the Phase Difference
The phase difference \( \phi \) can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting the values:
\[ \phi = \frac{2\pi}{500 \times 10^{-9}} \cdot (0.5 \times 10^{-6}) \]
\[ \phi = \frac{2\pi \cdot 0.5 \times 10^{-6}}{500 \times 10^{-9}} = \frac{\pi \times 10^{-6}}{250 \times 10^{-9}} = \frac{\pi}{0.25} = 4\pi \, \text{radians} \]

### Step 4: Calculate the Intensity at the Point
The intensity \( I \) at the point on the screen can be calculated using the formula:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
Substituting the values:
\[ I = I_0 + 9I_0 + 2\sqrt{I_0 \cdot 9I_0} \cos(4\pi) \]
Since \( \cos(4\pi) = 1 \):
\[ I = 10I_0 + 2\sqrt{9I_0^2} \cdot 1 \]
\[ I = 10I_0 + 6I_0 = 16I_0 \]

### Final Answer
The intensity at \( y = \frac{1}{6} \, \text{mm} \) is:
\[ I = 16I_0 \]