The width of a slit is `0.012mm`. Monochromatic light is incident on it. The angular position of first bright line is `5.2^(0)`. The wavelength of the light incident is

Options

A. `6040 A^(0)`
B. `4026 A^(0)`
C. `5890 A^(0)`
D. `7248 A^(0)`

Answer

To find the wavelength of the monochromatic light incident on a slit, we can use the formula for diffraction at a single slit. The formula for the angular position of the first bright fringe is given by:

\[ d \sin \theta = (2n + 1) \frac{\lambda}{2} \]

Where:
- \(d\) is the width of the slit,
- \(\theta\) is the angular position of the bright fringe,
- \(n\) is the order of the bright fringe (for the first bright fringe, \(n = 1\)),
- \(\lambda\) is the wavelength of the light.

### Step 1: Convert the slit width to meters
Given:
- Width of the slit \(d = 0.012 \, \text{mm}\)

Convert to meters:
\[ d = 0.012 \, \text{mm} = 0.012 \times 10^{-3} \, \text{m} = 12 \times 10^{-6} \, \text{m} \]

### Step 2: Identify the angular position and convert to radians if necessary
Given:
- Angular position of the first bright line \(\theta = 5.2^\circ\)

We can use the sine function directly in degrees:
\[ \sin(5.2^\circ) \approx 0.0906 \]

### Step 3: Substitute values into the formula for wavelength
Using the formula:
\[ \lambda = \frac{2d \sin \theta}{2n + 1} \]
Substituting \(n = 1\):
\[ \lambda = \frac{2 \times (12 \times 10^{-6}) \times 0.0906}{2 \times 1 + 1} \]

### Step 4: Calculate the wavelength
Calculate the denominator:
\[ 2n + 1 = 2 \times 1 + 1 = 3 \]

Now substitute into the equation:
\[ \lambda = \frac{24 \times 10^{-6} \times 0.0906}{3} \]

Calculating the numerator:
\[ 24 \times 10^{-6} \times 0.0906 = 2.1744 \times 10^{-6} \]

Now divide by 3:
\[ \lambda = \frac{2.1744 \times 10^{-6}}{3} = 0.7248 \times 10^{-6} \, \text{m} \]

### Step 5: Convert the wavelength to angstroms
1 meter = \(10^{10}\) angstroms, so:
\[ \lambda = 0.7248 \times 10^{-6} \, \text{m} = 7248 \, \text{angstroms} \]

### Final Answer
The wavelength of the light incident is:
\[ \lambda = 7248 \, \text{angstroms} \]