Thin-Film Interference AP Physics 2 MCQ – Exam Style Questions etc.
Thin-Film Interference AP Physics 2 MCQ
Unit 14: Waves , Sound , and Physical Optics
Weightage : 15–18%
Exam Style Practice Questions ,Thin-Film Interference AP Physics 2 MCQ
Question
In a classroom demonstration of thin films, your physics teacher takes a glass plate and places a thin layer of transparent oil on top of it. The oil film is then illuminated by shining a narrow beam of white light perpendicularly onto the oil’s surface, as shown above. The indices of refraction of air, the oil, and the glass plate are given in the diagram. Standing near the light source, you observe that the film appears green. This corresponds to a wavelength of 520 nm.
(a) Determine each of the following for the green light.
i. The frequency of the light in air
ii. The frequency of the light in the oil film
iii. The wavelength of the light in the oil film
(b) Calculate the minimum thickness of the oil film (other than zero) such that the observed green light is the most intense.
(c) As your teacher changes the angle of the light source, the light you observe from the film changes color. Give an explanation for this phenomenon.
Answer/Explanation
Ans:
a) i. c = f λ … 3×108 = f (520 nm) … f = 5.77×1014 Hz.
ii. frequency in film is the same as in the air, it doesn’t change between mediums = 5.77×1014 Hz
iii. n1 λ1 = n2 λ2 … nair λair = nfilm λfilm … (1)(520 nm) = (1.4)λoil … λoil = 371 nm
b) There will be ½ λ phase shifts at each boundary essentially canceling out each ‘flip’. To make constructive interference, a total of 1 λ of phase shift is needed from traveling in the film and this requires the thickness of the film to be ½ λfilm = ½ (371) = 186 nm
c) As the angle changes, the effective thickness of the film changes as well since the rays travel at angles in the film. The white light source has all frequency colors in it, and different thicknesses traveled will cause different constructive interference with different wavelength colors of light. (This is why a soap bubble has many colors reflected in it. The variation in thickness along the bubble makes different wavelength constructive in different regions)
Question
A student performs an experiment to determine the index of refraction n of a rectangular glass slab in air. She is asked to use a laser beam to measure angles of incidence θi in air and corresponding angles of refraction θr in glass. The measurements of the angles for five trials are given in the table below.
(a) Complete the last two columns in the table by calculating the quantities that need to be graphed to provide a linear relationship from which the index of refraction can be determined. Label the top of each column.
(b) On the grid, plot the quantities calculated in (a) and draw an appropriate graph from which the index of refraction can be determined. Label the axes.
(c) Using the graph, calculate the index of refraction of the glass slab.
The student is also asked to determine the thickness of a film of oil (n = 1.43) on the surface of water (n = 1.33).
Light from a variable wavelength source is incident vertically onto the oil film as shown above. The student measures a maximum in the intensity of the reflected light when the incident light has a wavelength of 600 nm.
(d) At which of the two interfaces does the light undergo a 180º phase change on reflection?
___The air-oil interface only ___The oil-water interface only
___Both interfaces ___Neither interface
(e) Calculate the minimum possible thickness of the oil film.
Answer/Explanation
Ans:
a) 
b) 
c) From snells law. ni sin θi = nr sin θr … nair sin θi = nglass sin θr … nglass = sin θi / sin θr
For the graph we have, this would be the inverse slope → 1 / slope. Slope = 0.67 … 1/0.67 … nglass = 1.5
d) ½ λ phase shifts happen when entering a more optically dense material. So only air–oil does a phase shift happen.
e) First we need to know the wavelength in the film (oil) … n1 λ1 = n2 λ2 … (1)(600) = (1.43)λoil … λoil = 420 nm Since there is already a ½ λ phase shift from the boundary flip, we need a total extra phase shift of ½ λ from traveling in the film to produce constructive interference (maximum). For this, ¼ λoil is required = 105 nm.