Question

If 640-nm light falls on a slit 0.0525 mm wide, what is the angular width of the central diffraction peak? Express your answer using three significant figures.

Answer 1

Then given data,

d = 0.0525mm = 0.0525*10^-3 m

folution given data, d=6youm= 640xlom d = 0.0525= 0.0525 x 10 ms then mal Half width = 0 = sin' () 80, put Valles, O= sin / 1X640x109 0.0525XTOS 0 = sin(0.01219) 0 = 0.698° then full angular width=20 = 2 x 0.698 = 10396 Answer I hope help you!!