Boundary Behavior of Waves and Polarization AP Physics 2 MCQ – Exam Style Questions etc.
Boundary Behavior of Waves and Polarization AP Physics 2 MCQ
Unit 14: Waves , Sound , and Physical Optics
Weightage : 15–18%
Exam Style Practice Questions ,Boundary Behavior of Waves and Polarization AP Physics 2 MCQ
Question
A wide beam of white light is incident normal to the surface of a uniform oil film. An observer looking down at the film sees green light that has maximum intensity at a wavelength of 5.2 x 10–7 m. The index of refraction of the oil is 1.7.
(a) Calculate the speed at which the light travels within the film.
(b) Calculate the wavelength of the green light within the film.
(c) Calculate the minimum possible thickness of the film.
(d) The oil film now rests on a thick slab of glass with index of refraction 1.4, as shown in the figure below. A light ray is incident on the film at the angle shown. On the figure, sketch the path of the refracted light ray that passes through the film and the glass slab and exits into the air. Clearly show any bending of the ray at each interface. You are NOT expected to calculate the sizes of any angles.
Answer/Explanation
Ans:
a) n = c / v … 1.7 = 3×108/ v … v = 1.76×108 m/s
b) n1 λ1 = n2 λ2 … nair λair = noil λoil … (1)(520 nm) = (1.7) λoil … λoil = 306 nm
c) To see the green light max intensity, we need constructive interference in the film for that green light wavelength. As the light travels from air–oil, it undergoes a ½ λ phase shift, but there is no phase shift at the second boundary. In order to produce constructive interference, we need to produce a total extra ½ λ phase shift from traveling in the film thickness. This requires the film thickness be ¼ λoil = ¼ (306) = 76.5 nm
d) Only the refracted ray is shown. We assume total internal reflection does not occur based on problem statements.
Question
The surface of a glass plate (index of refraction n3 = 1.50) is coated with a transparent thin film (index of refraction n2 = 1.25). A beam of monochromatic light of wavelength 6.0 x 10 –7 meter traveling in air (index of refraction n1 = 1.00) is incident normally on surface S1 as shown. The beam is partially transmitted and partially reflected.
a. Calculate the frequency of the light.
b. Calculate the wavelength of the light in the thin film.
The beam of light in the film is then partially reflected and partially transmitted at surface S2
c. Calculate the minimum thickness d1 of the film such that the resultant intensity of the light reflected back into the air is a minimum.
d. Calculate the minimum nonzero thickness d2 of the film such that the resultant intensity of the light reflected back into the air is a maximum.
Answer/Explanation
Ans:
a) c = f λ … 3×108 = f (6×10–7) … f = 5×1014 Hz.
b) n1 λ1 = n2 λ2 … (1) (6×10–7) = (1.25) = 4.8×10–7 m = 480 nm
c) The light is traveling to progressively more dense materials, so undergoes ½ λ phase shifts at both boundaries S1 and S2 essentially canceling out this phase shift (from flips). To get minimum intensity (destructive), the total phase shift from traveling in the film should be ½ λfilm so the film thickness should ¼ λfilm = ¼ 480 nm = 120 nm
d) Based on the same analysis above. To get maximum intensity (constructive) the total phase shift from traveling in the film should be 1 λfilm, so the film thickness should be ½ λfilm = ½ 480 nm = 240 nm