One of the main factors driving improvements in the cost and complexity of integrated circuits (ics) is improvements in photolithography and the resulting ability to print ever smaller features. modern circuits are made using a variety of complicated lithography techniques, with the goal to make electronic traces as small and as close to each other as possible (to reduce the overall size, and thus increase the speed). in the end though, all optical techniques are limited by diffraction. assume we have a scannable laser that draws a line on a circuit board (the light exposes a line of photoresist, which then becomes impervious to a subsequent chemical etch, leaving only the narrow metal line under the exposed photoresist). assume the laser wavelength is 248 nm (krypton fluoride excimer laser), the initial beam diameter is 1 cm, and the focusing lens (diameter = 1.3 cm) is extremely 'fast', with a focal length of only 0.625 cm."a) what is the approximate width w of the line (defined here as the distance between diffraction minima on either side of the central spot/ridge -- see figure)? b) what is the minimum resolvable line separation between adjacent lines? c) if the laser wavelength is instead reduced to 157 nm (as is now potentially available with argon fluoride excimer lasers, though technical challenges remain), what is the new minimum resolvable line separation? d) assuming the size of the traces is the only constraint limiting the 'areal density' (i. e., how many components can be laid out per square centimeter), by what factor does the areal density increase in going from a writing laser with λ = 248 nm to one with λ = 157 nm? e) going back to the lithography system with a laser that produces light (in air) at λ = 248 nm, what minimum line separation can be expected in this case (i. e., what is the new answer to b), if the space between the lens and the photoresist is filled with water [n = 1.33 for water])? hint: what effect does the index have on the wavelength of the light?