Monochromatic light of wavelength 441 nm is incident on a narrow slit. On a screen 2.00 m away, the distance between the second diffraction minimum and the central maximum is 1.80 cm.(a) Calculate the angle of diffraction `theta` of the second minimum. (b) Find the width of the slit.

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Calculate the angle of diffraction `theta` of the second minimum.

1. **Identify the given values**:
- Wavelength of light, \( \lambda = 441 \, \text{nm} = 441 \times 10^{-9} \, \text{m} \)
- Distance from the slit to the screen, \( L = 2.00 \, \text{m} \)
- Distance from the central maximum to the second minimum, \( y = 1.80 \, \text{cm} = 0.0180 \, \text{m} \)
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