(a) Find the angle of the third diffraction minimum for 633 -nm light falling on a slit of width \(20.0 \mu \mathrm{m}\). (b) What slit width would place this minimum at \(85.0^{\circ}\) ?

The formula to find the minima angles in single-slit diffraction is given by: \[ \sin \theta = \frac{m \lambda}{a} \] Where: θ = angle of the dark fringe (minimum) m = order number of the minimum (m = 1, 2, 3, ...) λ = wavelength of the light a = width of the slit We will use this formula to find the angle of the third diffraction minimum and the slit width required to place this minimum at an angle of 85.0 degrees.

Here, we are given the following values: λ = 633 nm = \( 633 \times 10^{-9} \) m (converting to meters) a = 20.0 μm = \( 20.0 \times 10^{-6} \) m (converting to meters) m = 3 (as we are finding the third minimum) We need to plug these values in the formula and solve for θ: \[ \sin \theta = \frac{3 (633 \times 10^{-9})}{20.0 \times 10^{-6}} \] Now, using a calculator, first find the sine value: \[ \sin \theta = 0.09495 \] Next, find the angle θ by taking the inverse sine of the value we got: \[ \theta = \sin^{-1}(0.09495) \] This gives us the angle: \[ \theta \approx 5.44^{\circ} \] So, the angle for the third diffraction minimum is approximately 5.44 degrees.

In this part, we are given the angle of 85.0 degrees and the order number m = 3. We need to find the slit width (a) that corresponds to this angle. We'll use the same formula as before, but now, we are solving for a instead: \[ a = \frac{m \lambda}{\sin \theta} \] Here, we have the following values: θ = 85.0 degrees m = 3 λ = 633 nm = \( 633 \times 10^{-9} \) m (converting to meters) Plugging these values into the formula and solving for a: \[ a = \frac{3 (633 \times 10^{-9})}{\sin(85^{\circ})} \] Calculating this using a calculator gives us the slit width: \[ a \approx 6.19 \times 10^{-6} \, m \] Which is approximately \(6.19 \mu\)m. So, the slit width required to place the third diffraction minimum at an angle of 85.0 degrees is approximately 6.19 μm.