## Problem Restatement **Find the distance in mm between two slits that produces the minimum for 415 nm violet light at an angle of 35°.** --- ## Step 1: Write the-Slit Minimum Condition The condition for the first minimumdark fringe) a double-slit interference pattern is: \d \sin \theta = mlambda \] For the **first minimum**, \( m = 1 \). - λ (wavelength) = 415 nm = \( 415 \times 10^{-9} \) m - θ (angle) = 35.° - m = 1 So, \[ d \sin 35.^\circ = 1 \times 415 \times 10^{-9} \text{ m} \] **Explanation:** This equation relates the slit separation \( d \), the angle to the minimum \( \theta \), the order of the minimum \( m \), and the wavelength \( \lambda \). --- ## Step 2: Solve for Slit Separation \(d\) \[ d = \frac{415 \times 10^{-9} \text{ m}}{\sin 35.^\circ} \] Calculate \(\sin 35.^\circ\): \[ \sin 35.^\circ \approx .5736 \] Plug in the values: \[ d = \frac{415 \times 10^{-9}}{.5736} = 724 \times 10^{-9} \text{ m} = .724 \times 10^{-6} \text{ m} = .724 \text{ μm} \] Convert to mm: \[ .724 \times 10^{-3} \text{ mm} = .000724 \text{ mm} \] **Explanation:** We solved for the slit separation by dividing the wavelength by the sine of the angle, then converted to millimeters. --- ## **Final Answer** \[ \boxed{.000724 \text{ mm}} \]

# Problem Restatement **Find the distance in nm between two slits that produces the first minimum for 415 nm violet light at an angle of 55°.** --- ## Step 1: Write the Slit Minimum Condition The condition for the first minimum (dark fringe) in a double-slit interference pattern is given by: \[ d \sin \theta = m \lambda \] For the **first minimum**, we have \( m = 1 \). - \( \lambda \) (wavelength) = 415 nm = \( 415 \times 10^{-9} \) m - \( \theta \) (angle) = 55° - \( m = 1 \) Thus, the equation becomes: \[ d \sin 55^\circ = 1 \times 415 \times 10^{-9} \text{ m} \] **Explanation:** This equation shows the relationship between the slit separation \( d \), the angle \( \theta \), the order of the minimum \( m \), and the wavelength \( \lambda \). --- ## Step 2: Solve for Slit Separation \( d \) Rearranging the equation gives: \[ d = \frac{415 \times 10^{-9} \text{ m}}{\sin 55^\circ} \] Calculate \( \sin 55^\circ \): \[ \sin 55^\circ \approx 0.8192 \] Substituting this into the equation: \[ d = \frac{415 \times 10^{-9}}{0.8192} \approx 506 \times 10^{-9} \text{ m} \] Convert to nanometers (1 m = \( 10^9 \) nm): \[ d \approx 506 \text{ nm} \] **Explanation:** We calculated the slit separation by dividing the wavelength by the sine of the angle, and then converted the result to nanometers. --- ## **Final Answer** \[ \boxed{506 \text{ nm}} \]