Two slits spaced 0.25 mm apart are placed 0.75 m from a screen and illuminated by coherent light with a wavelength of 650 nm. The intensity at the center of the central maximum `(theta = 0^(@))` is `I_(0)`. The distance on the screen from the center of the central maximum to the point where the intensity has fallen to `I_(0) // 2` is nearly

Options

A. 0.1 mm
B. .25 mm
C. 0.4 mm
D. 0.5 mm

Explanation

To solve the problem, we need to determine the distance on the screen from the center of the central maximum to the point where the intensity has fallen to half of its maximum value. Let's break down the solution step by step.

### Step 1: Understanding the Problem
We have two slits separated by a distance \( d = 0.25 \, \text{mm} = 0.25 \times 10^{-3} \, \text{m} \) and a screen placed at a distance \( D = 0.75 \, \text{m} \). The wavelength of the light used is \( \lambda = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m} \).

### Step 2: Maximum Intensity
The intensity at the center of the central maximum is denoted as \( I_0 \). We need to find the distance \( x \) from the center where the intensity falls to \( \frac{I_0}{2} \).

### Step 3: Intensity Formula
The intensity at a point on the screen due to two coherent sources can be expressed as:
\[ I = I_0 \cos^2 \left( \frac{\phi}{2} \right) \]
where \( \phi \) is the phase difference between the two waves arriving at that point.

### Step 4: Finding the Phase Difference
To find the point where the intensity is half, we set:
\[ I = \frac{I_0}{2} \]
Substituting into the intensity formula gives:
\[ \frac{I_0}{2} = I_0 \cos^2 \left( \frac{\phi}{2} \right) \]
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 5: Relating Phase Difference to Path Difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the equation:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \phi = \frac{\pi}{2} \):
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x \]
Solving for \( \Delta x \):
\[ \Delta x = \frac{\lambda}{4} \]

### Step 6: Calculating the Path Difference
Substituting the value of \( \lambda \):
\[ \Delta x = \frac{650 \times 10^{-9}}{4} = 162.5 \times 10^{-9} \, \text{m} = 162.5 \, \text{nm} \]

### Step 7: Using Geometry to Find \( x \)
The path difference can also be expressed in terms of the geometry of the setup:
\[ \Delta x = \frac{x d}{D} \]
Substituting the known values:
\[ \frac{650 \times 10^{-9}}{4} = \frac{x \cdot 0.25 \times 10^{-3}}{0.75} \]
Rearranging gives:
\[ x = \frac{650 \times 10^{-9} \cdot 0.75}{4 \cdot 0.25 \times 10^{-3}} \]
Calculating \( x \):
\[ x = \frac{650 \times 10^{-9} \cdot 0.75}{0.25 \times 10^{-3}} = \frac{650 \times 0.75}{0.25} \times 10^{-6} \]
\[ x = 1950 \times 10^{-6} \, \text{m} = 1.95 \, \text{mm} \]

### Step 8: Final Result
Thus, the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \( \frac{I_0}{2} \) is approximately:
\[ \boxed{1.95 \, \text{mm}} \]