In Young's double - slit experiment, the distance between slits is d = 0.25 cm and the distance of the screen D = 120 cm from the slits. If the wavelength of light used is `lambda=6000Å` and `I_(0)` is the intensity of central maximum, then the minimum distance of the point from the centre, where the intensity is `(I_(0))/(2)` is `kxx10^(-5)m`. What is the value of k?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between intensity and phase difference
In Young's double-slit experiment, the intensity \( I \) at a point on the screen can be expressed in terms of the central maximum intensity \( I_0 \) and the phase difference \( \phi \):
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
For the intensity to be \( \frac{I_0}{2} \), we set up the equation:
\[ \frac{I_0}{2} = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Dividing both sides by \( I_0 \):
\[ \frac{1}{2} = \cos^2\left(\frac{\phi}{2}\right) \]
Taking the square root:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \]
This implies:
\[ \frac{\phi}{2} = \frac{\pi}{4} \quad \Rightarrow \quad \phi = \frac{\pi}{2} \]

### Step 2: Relate phase difference to path difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \phi = \frac{\pi}{2} \):
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x \]
Solving for \( \Delta x \):
\[ \Delta x = \frac{\lambda}{4} \]

### Step 3: Substitute the values for \( \lambda \)
Given that the wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \):
\[ \Delta x = \frac{6 \times 10^{-7}}{4} = 1.5 \times 10^{-7} \, \text{m} \]

### Step 4: Find the position on the screen
Using the small angle approximation, the position \( x_n \) of the point on the screen can be calculated using:
\[ x_n = \frac{\Delta x \cdot D}{d} \]
Where:
- \( D = 120 \, \text{cm} = 1.2 \, \text{m} \)
- \( d = 0.25 \, \text{cm} = 0.0025 \, \text{m} \)

Substituting the values:
\[ x_n = \frac{(1.5 \times 10^{-7}) \cdot (1.2)}{0.0025} \]
Calculating:
\[ x_n = \frac{1.8 \times 10^{-7}}{0.0025} = 7.2 \times 10^{-5} \, \text{m} \]

### Step 5: Express the answer in the required form
We need to express \( x_n \) in the form \( k \times 10^{-5} \, \text{m} \):
\[ x_n = 7.2 \times 10^{-5} \, \text{m} \quad \Rightarrow \quad k = 7.2 \]

### Final Answer
The value of \( k \) is \( 7.2 \).