University:
Massachusetts Institute of Technology
Course:
2.71 | Optics
Academic year:

2024
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174

Pages:

8
Author:

zaprjekujl

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.71/2.710 Optics Practice Problems for the Final Exam Spring ’13 Posted Monday, May 13, 2013 1. (Pedrotti 13-21) A glass plate is sprayed with uniform opaque particles. When a distant point source of light is observed looking through the plate, a diffuse halo is seen whose angular width is about 2! . Estimate the size of the particles. (Hint: consider Fraunhoffer diffraction through random gratings, and use Babinet’s principle) Answer: The diffraction pattern of an opaque circular particle is complementary to that due to circular apertures of the same size in an otherwise opaque screen. ! ! ! ! !! ! Under the Fraunhofer condition ( 𝐸 𝑥!, 𝑦! ≈ 1 𝑧 ! ≪ 1, !! ! ! ! !! ! !! ≪ 1) exp (−𝑖𝑘(𝜃! ! 𝑥 + 𝜃! ! 𝑦))𝑡 𝑥, 𝑦 𝐸 𝑥, 𝑦 𝑑𝑥𝑑𝑦 !! !! Where 𝜃! ! ≈ , 𝜃! ! ≈ ! ! For the given problem, we may further assume E(x, y) is a plane wave at normal incidence, and the transmission function t(x, y) for a single particle can be expressed as: 𝑥! + 𝑦! 𝑡 𝑥, 𝑦 = 1 − 𝑐𝑖𝑟𝑐( ) 𝑅 Where 𝑅 is the radius of the opaque particles. 1 𝑥! + 𝑦! ! ! ! ! 𝐸 𝑥 ,𝑦 ≈ exp (−𝑖𝑘(𝜃! 𝑥 + 𝜃! 𝑦)) 1 − 𝑐𝑖𝑟𝑐( ) 𝑑𝑥𝑑𝑦 𝑧 𝑅 𝑥! + 𝑦! 𝑅 1 𝐸 𝑥 ! , 𝑦 ! ≈ ℱ 1 − 𝑐𝑖𝑟𝑐 𝑧 ! ! ! ! With 𝑥 ! = 𝑘! , 𝑦 ! = 𝑘! 𝐸 𝑘! , 𝑘! 1 ≈ 𝛿( 𝑘𝑥 2 + 𝑘𝑦 2 ) − 𝑅 𝑧 The halo is similar to an Airy disc! 1 2 2𝜋𝐽1 𝑅 𝑘𝑥 2 + 𝑘𝑦 2 2 𝑅 𝑘𝑥 + 𝑘𝑦 2 We can evaluate the width of the halo (a second peak) based on the table on Figure 11_08 provided by Pedrotti: © Pearson Prentice Hall. All rights reserved. Where 𝛾 = 𝑅 𝑘𝑥 2 + 𝑘𝑦 2 = !! ! 𝑅𝜃. From the above table, 2𝜋 𝑅∆𝜃 = 7.106 − 3.832 = 3.274 𝜆 Taking central wavelength at visible frequency, λ = 500 nm and given ∆𝜃 = 2! , we find the radius of the particle: 𝑅=𝜆 3.274 3.274 = 500𝑛𝑚× = 7463𝑛𝑚 = 7.4𝜇𝑚 2 ∆𝜃 (2𝜋)! (360) (2𝜋)! 360 2 2. (Adapted from Pedrotti 16-1 and 16-12) © Pearson Prentice Hall. All rights reserved. Figure A. Recording (Left) and Reconstruction (Right) of a Gabor Hologram a) Use the superposition of two beams to show that the recorded intensity pattern on a Gabor zone-‐plate (the hologram of a point source) is given approximately by 𝐼 = 𝐴 + 𝐵𝑐𝑜𝑠 ! (𝑎𝑟 ! ) Where 𝐴 = 𝐼! + 𝐼! − 2 𝐼! 𝐼! , 𝐵 = 4 𝐼! 𝐼! , and 𝑎 = 𝜋/(2𝑠𝜆). Here I1 and I2 are the intensity due to the reference and signal beams, respectively, s is the distance of the object point from the film, and λ is the wavelength of the light. For the approximation, assume the path difference between the two beams is much smaller than s, so we are looking at the inner zones of the hologram. Solution: in this problem, two beams are interfering at the zone plate: a reference plane wave with intensity I1, and a spherical wave with intensity I2. At a distance r from the symmetric axis, the path difference of the two beams can be written as: 𝑟! ! ! 𝛿 = 𝑠 +𝑟 −𝑠 ≈ 2𝑠 Therefore the intensity of the interference pattern can be written as: 𝐼 = 𝐼! + 𝐼! + 2 𝐼! 𝐼! 𝑐𝑜𝑠(𝑘𝛿) And 𝑘𝛿 𝑐𝑜𝑠 𝑘𝛿 = 2𝑐𝑜𝑠 ! −1 2 So we can rewrite the intensity into the following form: 𝑘𝛿 𝐼 = 𝐼! + 𝐼! − 2 𝐼! 𝐼! + 4 𝐼! 𝐼! 𝑐𝑜𝑠 ! 2 𝑘𝛿 𝜋𝑟 ! = 2 2𝜆𝑠 3 b) (2.710 only) Show that the phase delay of the diverging subject beam, at a point on the film at distance r from the axis, is given by 𝜋𝑟 ! /𝑙𝑠. This results follows when r< 𝜆 ! ! ! Combined with part b), we obtain: 𝑑𝜆 = ± 𝑑𝑥 > ±𝜆 ! 8 !