Question

When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink. To do this, the separation of the dots must be less than Raleigh's criterion. Take the pupil of the eye to be 3.8 mm and the distance from the paper to the eye of 47 cm; find the maximum separation (in cm) of two dots such that they cannot be resolved. (Assume the average wavelength of visible light is 555 nm.) 1.78e-4 x cm How many dots per inch (dpi) does this correspond to? 8.370-3* dpi t

Answer 1

Q. Giver wavelength of light Diameter of eyes Reppl a 555x10m 3.8mm h 1=555nm = 3.8x10?m Parfer D = 47 cm = 0.47 m distance between The eye and "Bom Raleigh's exiterion The limit of Resolution dots that do 1.22 argular is given by (minimem distorce betuken -can be seen) dots x se eye I a 0 D = 47 cm do = page 1:22 X 555 X10 3.8X1053 do = 178.184 X10 radian (de) D -6 - M The linear sepration the dots x= X = 178.184X10C X0.47 = 83.75 X10 Xmin = 83.75 X 10m The minimum linear sebra hon detween the dots is Kain The distance(p) that we can not (that we can see the dots is p < Kmin so The mox distance ber = 83.75 Xlo Yem dots ) see - Knin that we can The moximum sebration between the dots not resolve by eye is Poner = 83.75 X 10.9 cm • To find dots per inch 8.75 X 10 em hos one dots

per em so Number hos of dots 1 cm 83.75x10 dot density lem 83.75X164 1cm=0.39370 1inch dot density 83.75X10 X 0.3937 inch dot density 303.28 303 dets ?? inca The dot densito ity is 303 dots linch