A slit of width a is illuminiated by white light. The first diffraction minimum for light of `lamda=6500 `Å is formed at `theta=30^(@)`, then the width a of the slit is

Options

A. `3250Å`
B. `6.5xx10^(-4) cm`
C. `1.3 mum`
D. `2.6xx10^(-4) cm`

Answer

To find the width \( a \) of the slit, we can use the formula for the first diffraction minimum in a single slit diffraction pattern. The condition for the first minimum is given by:

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

where:
- \( a \) is the width of the slit,
- \( \theta \) is the angle at which the minimum occurs,
- \( n \) is the order of the minimum (for the first minimum, \( n = 1 \)),
- \( \lambda \) is the wavelength of the light.

### Step 1: Identify the known values
From the problem, we have:
- \( \lambda = 6500 \) Å (angstroms) = \( 6500 \times 10^{-10} \) m (since \( 1 \) Å = \( 10^{-10} \) m)
- \( \theta = 30^\circ \)
- \( n = 1 \) (for the first minimum)

### Step 2: Substitute the known values into the formula
Using the formula for the first minimum:

\[ a \sin(30^\circ) = 1 \cdot \lambda \]

### Step 3: Calculate \( \sin(30^\circ) \)
We know that:

\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 4: Substitute \( \sin(30^\circ) \) into the equation
Now, substituting \( \sin(30^\circ) \) into the equation gives:

\[ a \cdot \frac{1}{2} = \lambda \]

### Step 5: Solve for \( a \)
Rearranging the equation to solve for \( a \):

\[ a = 2 \lambda \]

### Step 6: Substitute the value of \( \lambda \)
Now substituting \( \lambda = 6500 \times 10^{-10} \) m into the equation:

\[ a = 2 \cdot (6500 \times 10^{-10}) = 13000 \times 10^{-10} \text{ m} \]

### Step 7: Convert to micrometers
To convert meters to micrometers, we multiply by \( 10^6 \):

\[ a = 13000 \times 10^{-10} \text{ m} \times 10^6 = 1.3 \text{ micrometers} \]

Thus, the width \( a \) of the slit is:

\[ \boxed{1.3 \text{ micrometers}} \]