Question

A thin flake of mica (n ! 1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe (m ! 7).If l ! 550 nm,what is the thickness of the mica

Answer 1

Answer: 6.64 × 10^ -6 m

Step-by-step explanation:

for a constructive interference;

ρ= 2πm

remember that we are looking for the seventh bright side fringe.

so, m=7

ρ=14π

We now have to find the phase different

ρ`=2πx/ρ - 2πx/ρ

p` is the wavelenght in the mica. x= width of the mica wavelength in the mica.

ρ= ρ/n.

n= refractive index.

The phase will now be

ρ`= 2πx/ρ - 2πx = 2πx(n-1)/ρ.

ρ= ρ`

14π = 2πx (n-1)/ ρ

we now solve for x; x = 7ρ/n-1

x = 7(550× 20^ 9)/ 1.58 - 1

= 6.64 × 10^-6