Light of wavelength `6,000 Å` falls on a single slit and diffracts. It is observed that the first maximum in diffraction pattern subtends an angle of `45^@`. Calculate the slit width.

Answer

To solve the problem of finding the slit width when light of wavelength \(6000 \, \text{Å}\) falls on a single slit and the first maximum subtends an angle of \(45^\circ\), we can follow these steps:

### Step 1: Understand the Condition for Maxima
In a single slit diffraction pattern, the condition for the position of the maxima is given by:
\[ \sin \theta = \frac{m \lambda}{b} \]
where \(m\) is the order of the maximum (for the first maximum, \(m = 1\)), \(\lambda\) is the wavelength of the light, \(b\) is the slit width, and \(\theta\) is the angle at which the maximum occurs.

### Step 2: Convert Wavelength to Meters
The given wavelength is \(6000 \, \text{Å}\). We need to convert this to meters:
\[ \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

### Step 3: Substitute the Known Values
We know that the angle \(\theta\) for the first maximum is \(45^\circ\) and we can use the sine function:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
Now substituting \(m = 1\), \(\lambda = 6 \times 10^{-7} \, \text{m}\), and \(\sin(45^\circ)\) into the maxima condition:
\[ \sin(45^\circ) = \frac{1 \cdot (6 \times 10^{-7})}{b} \]

### Step 4: Rearranging the Equation
Rearranging the equation to solve for \(b\):
\[ b = \frac{6 \times 10^{-7}}{\sin(45^\circ)} = \frac{6 \times 10^{-7}}{\frac{1}{\sqrt{2}}} \]

### Step 5: Calculate the Slit Width
Now we can calculate \(b\):
\[ b = 6 \times 10^{-7} \times \sqrt{2} = 6 \times 10^{-7} \times 1.414 \approx 8.49 \times 10^{-7} \, \text{m} \]

### Step 6: Final Result
Thus, the slit width \(b\) is approximately:
\[ b \approx 8.49 \times 10^{-7} \, \text{m} \text{ or } 849 \, \text{μm} \]