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 find the width of the slit (denoted as 'a') when light of wavelength 633 nm is incident on it, and the angle between the first diffraction minima on either side of the central maximum is 1.20 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Condition for Minima**:
The condition for the minima in a single slit diffraction pattern is given by the formula:
\[ a \sin \theta = m \lambda \]
where:
- \( a \) = width of the slit
- \( \theta \) = angle of the minima
- \( m \) = order of the minima (for the first minimum, \( m = 1 \))
- \( \lambda \) = wavelength of the light

2. **Determine the Angle for the First Minimum**:
Since the total angle between the first minima on both sides of the central maximum is 1.20 degrees, the angle for one side (first minimum) will be half of this:
\[ \theta = \frac{1.20}{2} = 0.60 \text{ degrees} \]

3. **Convert the Angle to Radians**:
To use the sine function, we need to convert the angle from degrees to radians:
\[ \theta = 0.60 \times \frac{\pi}{180} \approx 0.0105 \text{ radians} \]

4. **Substitute the Known Values into the Formula**:
We know the wavelength \( \lambda = 633 \text{ nm} = 633 \times 10^{-9} \text{ m} \) and \( m = 1 \). Plugging these values into the minima condition:
\[ a \sin(0.0105) = 1 \times (633 \times 10^{-9}) \]

5. **Calculate \( \sin(0.0105) \)**:
Using a calculator, we find:
\[ \sin(0.0105) \approx 0.0105 \]

6. **Rearranging the Formula to Solve for 'a'**:
Now we can rearrange the formula to solve for \( a \):
\[ a = \frac{633 \times 10^{-9}}{\sin(0.0105)} \]

7. **Plugging in the Values**:
\[ a = \frac{633 \times 10^{-9}}{0.0105} \approx 6.04 \times 10^{-5} \text{ m} \]

8. **Convert to Micrometers**:
To express the width in micrometers:
\[ a \approx 60.4 \text{ micrometers} \]

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

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