A parallel beam of monochromatic light is incident on a narrow rectangular slit of width `1mm`. When the diffraction pattern is seen on a screen placed at a distance of `2m`, the width of principal maxima is found to be `2.5mm`. The wavelength of light is

Options

A. `6250Å`
B. `6200Å`
C. `5890Å`
D. `6000Å`

Explanation

To find the wavelength of light used in the given diffraction pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Width of the slit (A) = 1 mm = \(1 \times 10^{-3}\) m
- Distance to the screen (D) = 2 m
- Width of the principal maxima (β) = 2.5 mm = \(2.5 \times 10^{-3}\) m

2. **Understanding the Diffraction Pattern:**
- In single-slit diffraction, the width of the principal maxima is related to the wavelength (λ) of the light and the width of the slit (A) using the formula:
\[ \beta = \frac{2\lambda D}{A} \]
- Here, β is the width of the principal maxima, D is the distance from the slit to the screen, and A is the width of the slit.

3. **Rearranging the Formula:**
- We need to find the wavelength (λ), so we rearrange the formula:
\[ \lambda = \frac{\beta A}{2D} \]

4. **Substituting the Values:**
- Now substitute the known values into the equation:
\[ \lambda = \frac{(2.5 \times 10^{-3} \, \text{m})(1 \times 10^{-3} \, \text{m})}{2 \times 2 \, \text{m}} \]

5. **Calculating the Wavelength:**
- Calculate the numerator:
\[ 2.5 \times 10^{-3} \times 1 \times 10^{-3} = 2.5 \times 10^{-6} \, \text{m}^2 \]
- Calculate the denominator:
\[ 2 \times 2 = 4 \, \text{m} \]
- Now divide the numerator by the denominator:
\[ \lambda = \frac{2.5 \times 10^{-6}}{4} = 0.625 \times 10^{-6} \, \text{m} = 6.25 \times 10^{-7} \, \text{m} \]

6. **Converting to Angstroms:**
- To convert meters to angstroms (1 m = \(10^{10}\) angstroms):
\[ \lambda = 6.25 \times 10^{-7} \, \text{m} \times 10^{10} \, \text{angstrom/m} = 6250 \, \text{angstrom} \]

### Final Answer:
The wavelength of the light used is \(6250 \, \text{angstrom}\).

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