Question

A thin film of acetone (n = 1.25) coats a thick glass plate (n = 1.50). White light is incident normal to the film. In the reflections, fully destructive interference occurs at 480 nm and fully constructive interference at 560 nm. Calculate the thickness of the acetone film.

Answer 1

Given

acetone refractive index n = 1.25 , glass refractive index n1 = 1.5

for destructive interference wavelength lambda L1 = 480 nm

for constructive interference wavelength lambda L2 = 560 nm

Thickness of the acetone

we know that the condition for constructive inteference in thin films is

   2t = (m+0.5) lambda2/n
   t = (m+0.5) lambda 2/ 2*n

substitutitng the values

for m=0,   t = (0+0.5)560/(2*1.25) nm = 112 nm
  

for m=1,   t = (1+0.5)560/(2*1.25) nm = 336 nm

for m=2,   t = (2+0.5)560/(2*1.25) nm = 560 nm

for m=3,   t = (3+0.5)560/(2*1.25) nm = 784 nm

for m=4,   t = (4+0.5)560/(2*1.25) nm = 1008 nm  

for m=5,   t = (5+0.5)560/(2*1.25) nm = 1232 nm

for destructive interference condition is 2t = m*lambda1/n

for m=1,   t = (1)560/(2*1.25) nm = 224 nm

for m=2,    t = (2*560)/(2*1.25) nm = 448 nm

for m=3,    t = (3*560)/(2*1.25) nm = 672 nm

for m=4,    t = (4*560)/(2*1.25) nm = 896nm

for m=5,    t = (5*560)/(2*1.25) nm = 1120 nm

for m=2,    t = (2*560)/(2*1.25) nm = 1344 nm

the minimum thickness is 1120 nm - 1008 nm =112 nm

the thickness of the acetone film. ia 112 nm