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`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the formula for intensity in a double slit experiment
In a double slit experiment, the intensity at a point on the screen can be expressed as:
\[ I = I_0 \cos^2 \phi \]
where \( I_0 \) is the intensity of the central maximum and \( \phi \) is the phase difference.

### Step 2: Determine the phase difference \( \phi \)
The phase difference \( \phi \) can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
where \( \Delta x \) is the path difference, which can be expressed as:
\[ \Delta x = \frac{d \cdot x}{D} \]
Here, \( d \) is the separation between the slits, \( x \) is the distance from the central maximum, and \( D \) is the distance from the slits to the screen.

### Step 3: Substitute the values
Given:
- \( d = 0.25 \, \text{cm} = 0.25 \times 10^{-2} \, \text{m} \)
- \( D = 100 \, \text{cm} = 1 \, \text{m} \)
- \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- \( x = 4 \times 10^{-5} \, \text{m} \)

Now, calculate the path difference:
\[ \Delta x = \frac{d \cdot x}{D} = \frac{(0.25 \times 10^{-2}) \cdot (4 \times 10^{-5})}{1} = 1 \times 10^{-6} \, \text{m} \]

### Step 4: Calculate the phase difference \( \phi \)
Now, substitute \( \Delta x \) into the phase difference formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{6000 \times 10^{-10}} (1 \times 10^{-6}) \]

Calculating this gives:
\[ \phi = \frac{2\pi \times 1 \times 10^{-6}}{6000 \times 10^{-10}} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Step 5: Calculate the intensity \( I \)
Now substitute \( \phi \) into the intensity formula:
\[ I = I_0 \cos^2 \left(\frac{\pi}{3}\right) \]

Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ I = I_0 \left(\frac{1}{2}\right)^2 = I_0 \cdot \frac{1}{4} \]

### Step 6: Final result
Thus, the intensity at a distance \( x = 4 \times 10^{-5} \) m from the central maximum is:
\[ I = \frac{I_0}{4} \]