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 principles of Young's Double Slit Experiment (YDSE) and use the given parameters to find the intensity at a specific point.

### Step 1: Calculate the path difference (Δx) at y = (1/6) mm

The path difference in YDSE is given by:
\[ \Delta x = d \sin \theta \]
For small angles, we can use the approximation \(\sin \theta \approx \frac{y}{D}\), where \(y\) is the distance from the central maximum and \(D\) is the distance from the slits to the screen.

Given:
- \(d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}\)
- \(D = 2 \, \text{m}\)
- \(y = \frac{1}{6} \, \text{mm} = \frac{1}{6} \times 10^{-3} \, \text{m}\)

Substituting the values:
\[ \Delta x = d \cdot \frac{y}{D} = (2 \times 10^{-3}) \cdot \left(\frac{1/6 \times 10^{-3}}{2}\right) \]
\[ \Delta x = 2 \times 10^{-3} \cdot \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{6} \times 10^{-6} \, \text{m} \]

### Step 2: Calculate the phase difference (Δφ)

The phase difference is given by:
\[ \Delta \phi = \frac{2\pi \Delta x}{\lambda} \]
Given \(\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}\), we substitute:
\[ \Delta \phi = \frac{2\pi \left(\frac{1}{6} \times 10^{-6}\right)}{500 \times 10^{-9}} \]
\[ \Delta \phi = \frac{2\pi}{6} \cdot \frac{10^{-6}}{500 \times 10^{-9}} = \frac{2\pi}{6} \cdot \frac{10^{-6}}{5 \times 10^{-7}} = \frac{2\pi}{6} \cdot \frac{2}{1} = \frac{2\pi}{3} \]

### Step 3: Calculate the resultant intensity (I)

The resultant intensity \(I_R\) is given by:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) \]
Where:
- \(I_1 = 9I_0\)
- \(I_2 = I_0\)

Substituting these values:
\[ I_R = 9I_0 + I_0 + 2\sqrt{9I_0 \cdot I_0} \cos\left(\frac{2\pi}{3}\right) \]
\[ I_R = 10I_0 + 2\sqrt{9I_0^2} \cdot \left(-\frac{1}{2}\right) \]
\[ I_R = 10I_0 - 3I_0 = 7I_0 \]

### Final Answer

Thus, the intensity at \(y = \frac{1}{6} \, \text{mm}\) is:
\[ \boxed{7I_0} \]