In single slit diffraction `a= 0.14mm, D= 2m` and distance of second dark band from central maxima is `1.6 cm`. The wavelength of light is

Options

A. `6500 A^(0)`
B. `7500 A^(0)`
C. `5600 A^(0)`
D. `8500 A^(0)`

Answer

To find the wavelength of light in a single slit diffraction experiment, we can use the formula for the position of dark bands. The formula for the position of the m-th dark band in single slit diffraction is given by:

\[ a \sin \theta = m \lambda \]

Where:
- \( a \) is the width of the slit,
- \( \theta \) is the angle of diffraction,
- \( m \) is the order of the dark band (for the second dark band, \( m = 2 \)),
- \( \lambda \) is the wavelength of light.

Given:
- \( a = 0.14 \, \text{mm} = 0.14 \times 10^{-3} \, \text{m} \)
- \( D = 2 \, \text{m} \)
- Distance of the second dark band from the central maximum \( y = 1.6 \, \text{cm} = 1.6 \times 10^{-2} \, \text{m} \)

### Step 1: Small Angle Approximation
For small angles, we can use the approximation:
\[ \sin \theta \approx \tan \theta \approx \frac{y}{D} \]

### Step 2: Substitute into the Dark Band Equation
Substituting the small angle approximation into the diffraction formula:
\[ a \left( \frac{y}{D} \right) = m \lambda \]

### Step 3: Rearranging for Wavelength
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = \frac{a y}{m D} \]

### Step 4: Substitute Known Values
Now, substituting the known values into the equation:
- \( a = 0.14 \times 10^{-3} \, \text{m} \)
- \( y = 1.6 \times 10^{-2} \, \text{m} \)
- \( m = 2 \)
- \( D = 2 \, \text{m} \)

\[ \lambda = \frac{(0.14 \times 10^{-3}) (1.6 \times 10^{-2})}{2 \times 2} \]

### Step 5: Calculate the Wavelength
Calculating the values:
\[ \lambda = \frac{0.14 \times 1.6 \times 10^{-5}}{4} \]
\[ \lambda = \frac{0.224 \times 10^{-5}}{4} = 0.056 \times 10^{-5} \, \text{m} \]
\[ \lambda = 5.6 \times 10^{-7} \, \text{m} = 5600 \, \text{Å} \]

### Final Answer
The wavelength of light is \( \lambda = 5600 \, \text{Å} \).

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