Consider the single-slit diffraction pattem for \(\lambda=600 \mathrm{nm}, D=0.025 \mathrm{mm},\) and \(x=2.0 \mathrm{m} .\) Find the intensity in terms of \(I_{o}\) at \(\theta=0.5^{\circ}, 1.0^{\circ}, 1.5^{\circ}, 3.0^{\circ}\) and \(10.0^{\circ}\)

The formula to calculate the intensity of a single-slit diffraction pattern is given by: \[I(\theta) = I_{0} \left( \frac{sin(\frac{\pi D}{\lambda} sin(\theta))}{\frac{\pi D}{\lambda} sin(\theta)} \right)^{2}\] Where: - \(I(\theta)\) is the intensity at an angle θ - \(I_{0}\) is the maximum intensity - λ is the wavelength of light - D is the width of the slit - θ is the angle from the center of the diffraction pattern

Plug in the given values (λ = 600 nm, D = 0.025 mm, x = 2.0 m) and calculate the intensity at each angle (0.5°, 1.0°, 1.5°, 3.0°, and 10.0°). Here, we're first converting the angles from degrees to radians, then substituting the values into the intensity formula. For better understanding, let's break down this step for the first angle (0.5°): 1. Convert angle to radians: \(0.5^{\circ} \times \frac{\pi}{180}\) = 0.00873 radians 2. Substitute values into the intensity formula: \[I(0.00873) = I_{0} \left( \frac{sin(\frac{\pi \times 0.025 \times 10^{-3}}{600 \times 10^{-9}} sin(0.00873))}{\frac{\pi \times 0.025 \times 10^{-3}}{600 \times 10^{-9}} sin(0.00873)} \right)^{2}\] 3. Solve for I(0.00873) Repeat this process for all given angles and calculate their respective intensities.

After solving the intensity formula for each angle, we find the following intensities in terms of \(I_{0}\): - At \(0.5^{\circ}\): \(I \approx 0.9931 I_{0}\) - At \(1.0^{\circ}\): \(I \approx 0.9724 I_{0}\) - At \(1.5^{\circ}\): \(I \approx 0.9364 I_{0}\) - At \(3.0^{\circ}\): \(I \approx 0.7995 I_{0}\) - At \(10.0^{\circ}\): \(I \approx 0.0186 I_{0}\) These are the intensities of the single-slit diffraction pattern at the given angles in terms of \(I_{0}\).