In Young's double slit experiment, distance between two sources is 0.1mm. The distance of screen from the sources is 20cm. Wavelength of light used is `5460 Å`. Then, angular position of first dark fringe is approximately

Options

A. `0.08^@`
B. `0.16^@`
C. `0.20^@`
D. `0.32^@`

Explanation

To find the angular position of the first dark fringe in Young's double slit experiment, we can follow these steps:

### Step 1: Understand the relationship for dark fringes
In Young's double slit experiment, dark fringes occur at positions where the path difference between the light from the two slits is an odd multiple of half the wavelength. The condition for the position of the dark fringes is given by:
\[ \Delta x = (2n - 1) \frac{\lambda}{2} \]
For the first dark fringe, \(n = 1\), so:
\[ \Delta x = \frac{\lambda}{2} \]

### Step 2: Relate path difference to angular position
The path difference \(\Delta x\) can also be expressed in terms of the angle \(\theta\) and the distance \(d\) between the two slits:
\[ d \sin \theta = \Delta x \]
Substituting for \(\Delta x\):
\[ d \sin \theta = \frac{\lambda}{2} \]

### Step 3: Assume small angle approximation
For small angles, \(\sin \theta \approx \theta\) (in radians). Thus, we can rewrite the equation as:
\[ d \theta = \frac{\lambda}{2} \]
From this, we can solve for \(\theta\):
\[ \theta = \frac{\lambda}{2d} \]

### Step 4: Substitute the given values
Now we will substitute the values into the equation. We have:
- Wavelength \(\lambda = 5460 \, \text{Å} = 5460 \times 10^{-10} \, \text{m}\)
- Distance between the slits \(d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m}\)

Substituting these values:
\[ \theta = \frac{5460 \times 10^{-10}}{2 \times 0.1 \times 10^{-3}} \]

### Step 5: Calculate \(\theta\)
Calculating the above expression:
\[ \theta = \frac{5460 \times 10^{-10}}{0.2 \times 10^{-3}} = \frac{5460 \times 10^{-10}}{2 \times 10^{-4}} = \frac{5460}{2} \times 10^{-6} = 2730 \times 10^{-6} \, \text{radians} \]

### Step 6: Convert radians to degrees
To convert radians to degrees, we use the conversion factor \( \frac{180}{\pi} \):
\[ \theta \text{ (in degrees)} = 2730 \times 10^{-6} \times \frac{180}{\pi} \approx 0.156 \, \text{degrees} \]

### Final Result
The angular position of the first dark fringe is approximately:
\[ \theta \approx 0.16^\circ \]