Question:

Please convert the below content to rendered equations without any changes and errors. This is a **photolithography resolution** problem. The minimum feature that can be resolved by a projection lithography system is given by the **Rayleigh resolution criterion**: [ R = k_1 \frac{\lambda}{NA} ] Where * (R) = minimum resolvable feature size * (k_1) = process constant (≈ **0.6** for contact holes) * (\lambda) = exposure wavelength * (NA) = numerical aperture of the lens From the **layout**, the **smallest feature that must be resolved is the contact opening**, which is: [ R = 1 ,\mu m ] --- # (a) Maximum allowable wavelength Given [ NA = 0.5 ] [ k_1 = 0.6 ] [ R = 1,\mu m ] Using Rayleigh equation: [ 1 = 0.6 \frac{\lambda}{0.5} ] Solve for (\lambda): [ 1 = 1.2\lambda ] [ \lambda = \frac{1}{1.2} ] [ \lambda = 0.833,\mu m ] [ \lambda = 833,nm ] ✅ **Maximum allowable wavelength** [ \boxed{\lambda_{max} \approx 833\ \text{nm}} ] --- # (b) Required numerical aperture for λ = 435.8 nm Given [ \lambda = 435.8,nm = 0.4358,\mu m ] [ R = 1,\mu m ] [ k_1 = 0.6 ] Rearrange Rayleigh equation: [ NA = \frac{k_1\lambda}{R} ] Substitute: [ NA = \frac{0.6 \times 0.4358}{1} ] [ NA = 0.26148 ] [ NA \approx 0.26 ] ✅ **Required numerical aperture** [ \boxed{NA \approx 0.26} ] --- # Final Answers **(a)** Maximum wavelength: [ \boxed{\lambda_{max} \approx 833,nm} ] **(b)** Required numerical aperture: [ \boxed{NA \approx 0.26} ] --- If you want, I can also show the **exam-style short derivation (3–4 line answer format)** that professors usually expect in **VLSI fabrication exams**.