Light of wavelength 633 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is 1.20 What is the width of the slit?

Answer

To solve the problem of determining the width of a slit given the wavelength of light and the angle between the first diffraction minima, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Wavelength of light, \( \lambda = 633 \, \text{nm} = 633 \times 10^{-9} \, \text{m} \)
- Angle between the first diffraction minima on either side of the central maximum, \( \theta_{total} = 1.20^\circ \)

2. **Calculate the Angle for One Side**:
- Since the angle given is for both sides of the central maximum, we need to find the angle for one side:
\[ \theta = \frac{\theta_{total}}{2} = \frac{1.20^\circ}{2} = 0.60^\circ \]

3. **Use the Formula for Diffraction Minima**:
- The condition for the first diffraction minimum in a single slit is given by:
\[ a \sin(\theta) = m \lambda \]
where \( m = 1 \) for the first minimum, and \( a \) is the width of the slit.

4. **Rearranging the Formula**:
- We can rearrange the formula to solve for \( a \):
\[ a = \frac{\lambda}{\sin(\theta)} \]

5. **Calculate \( \sin(0.60^\circ) \)**:
- Using a calculator or trigonometric tables, find:
\[ \sin(0.60^\circ) \approx 0.0105 \]

6. **Substituting Values**:
- Now substitute the values of \( \lambda \) and \( \sin(\theta) \) into the equation:
\[ a = \frac{633 \times 10^{-9} \, \text{m}}{0.0105} \]

7. **Perform the Calculation**:
- Calculate \( a \):
\[ a \approx \frac{633 \times 10^{-9}}{0.0105} \approx 6.04 \times 10^{-5} \, \text{m} \]

8. **Final Result**:
- The width of the slit is approximately:
\[ a \approx 6.04 \times 10^{-5} \, \text{m} = 60.4 \, \mu\text{m} \]

### Final Answer:
The width of the slit is approximately \( 60.4 \, \mu\text{m} \).

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