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, we need to find the distance from the central maximum where the intensity is 75% of the maximum intensity in a double-slit experiment. We are given the following parameters:

- Distance to the screen (D) = 1 m
- Slit separation (d) = 0.2 cm = 0.002 m
- Wavelength (λ) = 6000 Å = 6000 x 10^-10 m = 6 x 10^-7 m

### Step-by-step solution:

1. **Understanding the intensity condition**:
The intensity at a point on the screen in a double-slit experiment is given by:
\[ I = I_{max} \cos^2\left(\frac{\delta}{2}\right) \]
where \(I_{max}\) is the maximum intensity, and \(\delta\) is the phase difference.

Given that the intensity is 75% of the maximum intensity:
\[ I = 0.75 I_{max} \]
This leads to:
\[ 0.75 I_{max} = I_{max} \cos^2\left(\frac{\delta}{2}\right) \]
Dividing both sides by \(I_{max}\):
\[ 0.75 = \cos^2\left(\frac{\delta}{2}\right) \]

2. **Finding the phase difference**:
Taking the square root:
\[ \cos\left(\frac{\delta}{2}\right) = \sqrt{0.75} = \frac{\sqrt{3}}{2} \]
Therefore:
\[ \frac{\delta}{2} = \frac{\pi}{6} \quad \Rightarrow \quad \delta = \frac{\pi}{3} \]

3. **Calculating the path difference**:
The path difference (\(\Delta x\)) corresponding to the phase difference is given by:
\[ \Delta x = \frac{\delta}{2\pi} \lambda \]
Substituting \(\delta = \frac{\pi}{3}\):
\[ \Delta x = \frac{\frac{\pi}{3}}{2\pi} \lambda = \frac{\lambda}{6} \]
Now substituting \(\lambda = 6 \times 10^{-7} \, \text{m}\):
\[ \Delta x = \frac{6 \times 10^{-7}}{6} = 10^{-7} \, \text{m} \]

4. **Finding the position on the screen (y)**:
The position \(y\) on the screen can be calculated using the formula:
\[ y = \frac{D \Delta x}{d} \]
Substituting the values:
\[ y = \frac{1 \times 10^{-7}}{0.002} = \frac{10^{-7}}{2 \times 10^{-3}} = 5 \times 10^{-5} \, \text{m} \]

5. **Converting to millimeters**:
To convert meters to millimeters:
\[ y = 5 \times 10^{-5} \, \text{m} = 0.05 \, \text{mm} \]

### Final Answer:
The distance from the central maximum where the intensity is 75% of that at the center is **0.05 mm**.