In a double experiment D=1m, d=0.2 cm and `lambda = 6000 Å`. The distance of the point from the central maximum where intensity is 75% of that at the centre will be:

Options

A. 0.01 mm
B. 0.03 mm
C. 0.05 mm
D. 0.1 mm

Answer

To solve the problem of finding the distance from the central maximum where the intensity is 75% of that at the center in a double-slit experiment, we can follow these steps:

### Step 1: Understand the relationship between intensity and phase difference
In a double-slit experiment, the intensity \( I \) at a point on the screen can be expressed as:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( I_0 \) is the maximum intensity (at the central maximum) and \( \phi \) is the phase difference.

### Step 2: Set up the equation for 75% intensity
Given that the intensity \( I \) is 75% of \( I_0 \):
\[ I = 0.75 I_0 \]
Substituting this into the intensity equation gives:
\[ 0.75 I_0 = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ 0.75 = \cos^2\left(\frac{\phi}{2}\right) \]

### Step 3: Solve for the phase difference
Taking the square root of both sides:
\[ \cos\left(\frac{\phi}{2}\right) = \sqrt{0.75} = \frac{\sqrt{3}}{2} \]
This implies:
\[ \frac{\phi}{2} = \frac{\pi}{6} \quad \text{or} \quad \frac{\phi}{2} = \frac{11\pi}{6} \]
Thus,
\[ \phi = \frac{\pi}{3} \quad \text{or} \quad \phi = \frac{11\pi}{3} \]

### Step 4: Relate phase difference to path difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \phi = \frac{\pi}{3} \):
\[ \frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x \]
Solving for \( \Delta x \):
\[ \Delta x = \frac{\lambda}{6} \]

### Step 5: Calculate the distance \( y \) on the screen
The distance \( y \) from the central maximum where this path difference occurs can be calculated using:
\[ y = \frac{D}{d} \Delta x \]
where \( D \) is the distance from the slits to the screen and \( d \) is the separation between the slits.

### Step 6: Substitute the values
Given:
- \( D = 1 \, \text{m} \)
- \( d = 0.2 \, \text{cm} = 0.002 \, \text{m} \)
- \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)

Substituting these values:
\[ \Delta x = \frac{6 \times 10^{-7}}{6} = 1 \times 10^{-7} \, \text{m} \]
Now calculate \( y \):
\[ y = \frac{1}{0.002} \times 1 \times 10^{-7} = 5 \times 10^{-5} \, \text{m} = 0.05 \, \text{mm} \]

### Final Answer
The distance from the central maximum where the intensity is 75% of that at the center is:
\[ \boxed{0.05 \, \text{mm}} \]