Question

10. A soap film (n = 1.33) is formed in a loop of wire that is mounted vertically is not exactly uniform in thickness because of its weight, but wedge-shaped in cross section. When the film is illuminated with green light of wavelength 520 nm, one sees that a distance of 2.40 cm separates two bright fringes that are 4 orders apart. By how much does the film differ in thickness over this distance?

Answer 1

answer) to find the thickness we need to use the formula

t=d...............1)

here we have the fringe order =4

so we have

x4=2(m+4)+1)/4n

x4=(2m+9)/4n

the distance between two bright fringes is

x4-xo

xo=(2m+1)/4n

so we have

d=(2m+9)/4n-(2m+1)/4n

d=8/4n

now putting all the values in this eqn we have

d=8*520*10^-9/4*1.33*0.024=32.58145*10^-6m

now using the eqn 1) we find t

t=0.024*32.58145=0.782 m

so the answer is 0.782 m or 0.78m or 0.78195m