A thin film of acetone n=1.25coats a thick glass platen=1.50White light is incident normal to the film. In the reflections, fully destructive interference occurs at 600nmand fully constructive interference at700nm. Calculate the thickness of the acetone film.

Short Answer

Expert verified

The thickness of the acetone coating is 840nm.

Step by step solution

01

Given Data

  • The refractive index of first medium, that is, air isn1=1.00.
  • The refractive index of the thin film isn2=1.25.
  • The refractive index of the third medium n3=1.50.
  • The maximum intensity occurs at wavelengthλ700=700nm.
  • The minimum intensity occurs at wavelength λ600=600nm.
02

Interference in thin films

Bright colours reflected from thin oil on water and soap bubbles are a consequence of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colours can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

2L=m+12λn2m=0,1,2,...(Maxima—bright film in the air)

where λis the wavelength of the light in air, Lis its thickness, andn2is the film’s refractive index.

Here, in this case, light is travelling in an air medium and incident on the acetone film whose refractive index n2=1.25is higher than air. And then the light gets reflected of the thick glass of refractive index n3=1.50while travelling through acetone and again the refractive index is higher than acetone. As a result, the condition for constructive and destructive interference is

2L=mλ700n2(Constructive)

2L=m+12λ600n2(Destructive)

The constructive interference occurs atλ700=700nm and destructive at λ600=600nm.

03

Determine the thickness of the thin layer

Equating the above two equations,

mλ700n2=m+12λ600n2mλ700-mλ600=λ6002m700nm-600nm=600nm2m=3

Inserting the value of in the constructive condition

L=mλ7002n2=3700nm21.25=840nm

The thickness of the acetone coating is 840nm.

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Most popular questions from this chapter

In Fig. 35-4, assume that two waves of light in air, of wavelength 400nm, are initially in phase. One travels through a glass layer of index of refraction n1=1.60and thickness L. The other travels through an equally thick plastic layer of index of refraction n2=1.50. (a) What is the smallest value Lshould have if the waves are to end up with a phase difference of 5.65 rad? (b) If the waves arrive at some common point with the same amplitude, is their interference fully constructive, fully destructive, intermediate but closer to fully constructive, or intermediate but closer to fully destructive?

Figure 35-46a shows a lens with radius of curvature lying on a flat glass plate and illuminated from above by light with wavelength l. Figure 35-46b (a photograph taken from above the lens) shows that circular interference fringes (known as Newton’s rings) appear, associated with the variable thickness d of the air film between the lens and the plate. Find the radii r of the interference maxima assumingr/R1.

In Fig. 35-23, three pulses of light— a, b, and c—of the same wavelength are sent through layers of plastic having the given indexes of refraction and along the paths indicated. Rank the pulses according to their travel time through the plastic layers, greatest first.

In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a)What is the angular separation in radians between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen 50 cm from the slits?

Find the slit separation of a double-slit arrangement that will produce interference fringes0.018radapart on a distant screen when the light has wavelengthλ=589nm.

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