Let's solve each part step by step with explanations and summary, following your requested format. --- ### **Part (a): Equation for the cosine of half the phase difference** #### **Step 1: Start with the double-slit intensity equation.** The intensity at any point on the screen due to two slits is given by: \[ I = I_0 \cos^2 \left(\frac{\phi}{2}\right) \] **Explanation:** Here, \( I_0 \) is the maximum intensity (when the two waves are in phase), and \( \phi \) is the phase difference between the two waves. The cosine squared term arises from the interference pattern. --- #### **Step 2: Express \(\cos\left(\frac{\phi}{2}\right)\) in terms of \(I\) and \(I_0\).** Rearrange the equation: \[ \cos^2 \left(\frac{\phi}{2}\right) = \frac{I}{I_0} \] Take the square root: \[ \cos \left(\frac{\phi}{2}\right) = \sqrt{\frac{I}{I_0}} \] **Explanation:** This isolates the cosine of half the phase difference in terms of the initial intensity and the resulting intensity. --- **Summary for (a):** \[ \boxed{\cos \left(\frac{\phi}{2}\right) = \sqrt{\frac{I}{I_0}}} \] --- ### **Part (b): Smallest phase angle \(\phi\) for \(I = 75 \, \text{W/m}^2\), \(I_0 = 180 \, \text{W/m}^2\)** #### **Step 1: Substitute values into the equation from part (a).** \[ \cos \left(\frac{\phi}{2}\right) = \sqrt{\frac{75}{180}} \] Calculate the value: \[ \frac{75}{180} = 0.4167 \] \[ \sqrt{0.4167} \approx 0.6455 \] **Explanation:** The right side gives the cosine value needed for the required intensity. --- #### **Step 2: Solve for the phase angle \(\phi\).** Take the arccosine: \[ \frac{\phi}{2} = \arccos(0.6455) \] \[ \frac{\phi}{2} \approx 0.870 \text{ radians} \] \[ \phi \approx 2 \times 0.870 = 1.74 \text{ radians} \] **Explanation:** This step finds the smallest positive phase angle yielding the desired intensity. --- **Summary for (b):** \[ \boxed{\phi \approx 1.74 \text{ radians}} \] --- ### **Part (c): Formula for the diffraction angle \(\theta\)** The phase difference \(\phi\) relates to the path difference \(\delta\): \[ \phi = \frac{2\pi}{\lambda} \delta \] The path difference for two slits separated by \(d\) at angle \(\theta\) is: \[ \delta = d \sin \theta \] So, \[ \phi = \frac{2\pi}{\lambda} d \sin \theta \] From part (a), \(\cos(\phi/2) = \sqrt{I/I_0}\), so: \[ \phi = 2 \arccos\left(\sqrt{\frac{I}{I_0}}\right) \] Set the two expressions for \(\phi\) equal: \[ \frac{2\pi}{\lambda} d \sin\theta = 2 \arccos\left(\sqrt{\frac{I}{I_0}}\right) \] \[ \sin\theta = \frac{\lambda}{\pi d} \arccos\left(\sqrt{\frac{I}{I_0}}\right) \] **Summary for (c):** \[ \boxed{ \sin\theta = \frac{\lambda}{\pi d} \arccos\left(\sqrt{\frac{I}{I_0}}\right) } \] --- ### **Part (d): Smallest angle \(\theta\) for \(I = 75 \, \text{W/m}^2\)** Given: - \(\lambda = 420\, \text{nm} = 420 \times 10^{-9}\, \text{m}\) - \(d = 18\, \mu \text{m} = 18 \times 10^{-6}\, \text{m}\) - Use \(\arccos(\sqrt{75/180}) \approx 0.870\) from earlier. \[ \sin\theta = \frac{420 \times 10^{-9}}{\pi \times 18 \times 10^{-6}} \times 0.870 \] Calculate the denominator: \[ \pi \times 18 \times 10^{-6} \approx 56.5487 \times 10^{-6} \] \[ \frac{420 \times 10^{-9}}{56.5487 \times 10^{-6}} \approx 0.00743 \] \[ \sin\theta = 0.00743 \times 0.870 \approx 0.00647 \] So, \[ \theta \approx \arcsin(0.00647) \approx 0.00647 \text{ radians} \] --- **Summary for (d):** \[ \boxed{\theta \approx 0.00647 \text{ radians}} \] --- ## **Final Answer Summary** - **(a):** \(\cos \left(\frac{\phi}{2}\right) = \sqrt{\frac{I}{I_0}}\) - **(b):** \(\phi \approx 1.74\) radians - **(c):** \(\sin\theta = \frac{\lambda}{\pi d} \arccos\left(\sqrt{\frac{I}{I_0}}\right)\) - **(d):** \(\theta \approx 0.00647\) radians All solutions are presented step-by-step with explanations for each step and boxed final answers.