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 step by step, we will use the concepts of Young's double-slit experiment and the relationship between intensity and phase difference.

### Step 1: Understanding the Intensity Relation
In Young's double-slit experiment, the intensity \( I \) at a point on the screen is given by:
\[ I = I_0 \cos^2 \left( \frac{\delta}{2} \right) \]
where \( I_0 \) is the intensity of the central maximum and \( \delta \) is the phase difference between the two waves arriving at that point.

### Step 2: Setting Up the Equation for Given Intensity
We need to find the position where the intensity is \( \frac{I_0}{2} \):
\[ \frac{I_0}{2} = I_0 \cos^2 \left( \frac{\delta}{2} \right) \]
Dividing both sides by \( I_0 \):
\[ \frac{1}{2} = \cos^2 \left( \frac{\delta}{2} \right) \]
Taking the square root:
\[ \cos \left( \frac{\delta}{2} \right) = \frac{1}{\sqrt{2}} \]

### Step 3: Finding the Phase Difference
The angle \( \theta \) corresponding to \( \cos \left( \frac{\delta}{2} \right) = \frac{1}{\sqrt{2}} \) is:
\[ \frac{\delta}{2} = \frac{\pi}{4} \implies \delta = \frac{\pi}{2} \]

### Step 4: Relating Phase Difference to Path Difference
The phase difference \( \delta \) is related to the path difference \( \Delta x \) by:
\[ \delta = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \delta = \frac{\pi}{2} \):
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x \]
Solving for \( \Delta x \):
\[ \Delta x = \frac{\lambda}{4} \]

### Step 5: Finding the Position on the Screen
The position \( y \) on the screen corresponding to the path difference \( \Delta x \) can be calculated using:
\[ y = \frac{\Delta x \cdot D}{d} \]
where \( D \) is the distance from the slits to the screen and \( d \) is the distance between the slits.

### Step 6: Substituting Values
Given:
- \( d = 0.25 \, \text{cm} = 0.0025 \, \text{m} \)
- \( D = 120 \, \text{cm} = 1.2 \, \text{m} \)
- \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)

Substituting these values into the equation:
\[ \Delta x = \frac{6 \times 10^{-7}}{4} = 1.5 \times 10^{-7} \, \text{m} \]
Now substituting into the position equation:
\[ y = \frac{(1.5 \times 10^{-7}) \cdot (1.2)}{0.0025} \]
Calculating:
\[ y = \frac{1.8 \times 10^{-7}}{0.0025} = 7.2 \times 10^{-5} \, \text{m} \]

### Step 7: Expressing in Required Form
We need to express \( y \) in the form \( k \times 10^{-5} \, \text{m} \):
\[ y = 7.2 \times 10^{-5} \implies k = 7.2 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{7.2} \]