In a double slit experiment ,the separation between the slits is `d=0.25cm` and the distance of the screen `D=100`cm from the slits .if the wavelength of light used in `lambda=6000Å`and `I_(0)`is the intensity of the central bright fringe. The intensity at a distance `x=4xx10^(-5)`in form the central maximum is-

Options

A. `I_(0)`
B. `I_(0)//2`
C. `3I_(0)//4`
D. `I_(0)//3`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given parameters
- The separation between the slits, \( d = 0.25 \, \text{cm} = 0.25 \times 10^{-2} \, \text{m} \)
- The distance from the slits to the screen, \( D = 100 \, \text{cm} = 1 \, \text{m} \)
- The wavelength of light, \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- The distance from the central maximum, \( x = 4 \times 10^{-5} \, \text{m} \)

### Step 2: Calculate the phase difference
The phase difference \( \Phi \) at a distance \( x \) from the central maximum can be calculated using the formula:
\[ \Phi = \frac{2\pi d x}{\lambda D} \]

Substituting the values:
\[ \Phi = \frac{2\pi (0.25 \times 10^{-2}) (4 \times 10^{-5})}{(6000 \times 10^{-10}) (1)} \]

### Step 3: Simplify the expression
Calculating the values step by step:
1. Calculate \( d \cdot x = (0.25 \times 10^{-2}) \cdot (4 \times 10^{-5}) = 1 \times 10^{-6} \)
2. Calculate \( \lambda \cdot D = (6000 \times 10^{-10}) \cdot (1) = 6000 \times 10^{-10} \)
3. Now substituting these values back into the phase difference formula:
\[ \Phi = \frac{2\pi (1 \times 10^{-6})}{6000 \times 10^{-10}} = \frac{2\pi}{6000} \times 10^{4} \]

### Step 4: Calculate the intensity
The intensity \( I \) at distance \( x \) from the central maximum is given by:
\[ I = I_0 \cos^2 \left( \frac{\Phi}{2} \right) \]

Using the phase difference calculated:
\[ I = I_0 \cos^2 \left( \frac{2\pi}{6000} \times 10^{4} \div 2 \right) \]
\[ = I_0 \cos^2 \left( \frac{\pi}{3000} \times 10^{4} \right) \]
\[ = I_0 \cos^2 \left( \frac{10\pi}{3} \right) \]

### Step 5: Evaluate \( \cos^2 \left( \frac{10\pi}{3} \right) \)
Since \( \frac{10\pi}{3} \) is equivalent to \( \frac{4\pi}{3} \) (as \( \frac{10\pi}{3} - 2\pi = \frac{4\pi}{3} \)), we can find:
\[ \cos \left( \frac{4\pi}{3} \right) = -\frac{1}{2} \]
Thus,
\[ \cos^2 \left( \frac{4\pi}{3} \right) = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 6: Final intensity calculation
Now substituting back into the intensity formula:
\[ I = I_0 \cdot \frac{1}{4} = \frac{I_0}{4} \]

### Conclusion
The intensity at a distance \( x = 4 \times 10^{-5} \, \text{m} \) from the central maximum is:
\[ I = \frac{I_0}{4} \]