Homework Assignment Interference of Light

11/16/23, 6:32 PM Homework 10: Interference of Light: Attempt review Started on Wednesday, 25 November 2020, 8:49 PM State Finished Completed on Thursday, 26 November 2020, 1:31 AM Time taken 4 hours 41 mins Marks 5.93/7.00 Grade 84.76 out of 100.00 Question 1 Correct Mark 0.90 out of 1.00 A soap film (index of refraction 1.33) is surrounded on both sides by air. When white light shines nearly perpendicularly on the film, you see bright interference colours of wavelength = 485 nm in the reflected light. What is the second smallest possible value for the thickness of the film? Answer: [ 273 JV nm Correct, well done! There is a 180 degree (or 7 radian) phase difference due to reflection, so the condition for bright fringes is: (m + %))\ n 2L = The smallest value of the thickness L would be obtained from m = 0. For the second-smallest value, we let m = 1 and solve for L. (1.5)A (2)(1.33) The correct answer is: 273 nm Marks for this submission: 1.00/1.00. Accounting for previous tries, this gives 0.90/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6588054&cmid=4375425#question-7036907-1 1/10

11/16/23, 6:32 PM Homework 10: Interference of Light: Attempt review Question 2 Correct Mark 0.67 out of 1.00 A thin film (n,) is sandwiched between two other transparent media with indices of refraction n, and n3, respectively. Monochromatic light of wavelength A (in air) is normally incident upon the material as shown in the diagram. In this case, assume that n; > nq = n3. For which film-thickness(es) will constructive interference of reflected light occur? Caution: There may (or may not) be more than one correct answer. To get any marks all the correct answers and no incorrect ones must be selected. ny thin s nz film N3 Select all that apply: A cross out 4 —_— A cross out 2 _— A cross out 4n1 A cross out 2n1 A Cross out 4'n,2 A cross out 2’]?,2 3A Cross out ¢ 4’)’12 A cross out 4n3 A cross out Ny E— zero (i.e.,, negligible film thickness) cross out Your answer is correct. In this case there is a phase change for reflection from the 1-2 boundary but not the 2-3 boundary so for constructive interference we want the phase change due to propagation to be a half-integer number of wavelengths. However we need to remember that the wavelength in the film is A/ and so we require: (2m + 1)\ (2m + 1)A 27?,2 4’]?12 where m is an integer. The correct answers are: 4% 2 3\ I 4n2 Marks for this submission: 1.00/1.00. Accounting for previous tries, this gives 0.67/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6588054&cmid=4375425#question-7036907-1 2/10

11/16/23, 6:32 PM Homework 10: Interference of Light: Attempt review Question 4 Correct Mark 1.00 out of 1.00 Two glass microscope slides, of length 10 cm, are in contact at one end and separated by a sheet of paper at the other end, which creates a wedge-shaped film of air. The thickness of the paper is t = 90 um, and the glass has index of refraction n = 1.5. Light with wavelength 0.631 um shines vertically down on the slides. In the resulting interference pattern, what is the distance separating successive dark fringes when viewed from above? Answer: [ 0.35 ]\/ mm Correct, well done! There is a 180 degree (or 7 radian) phase difference due to reflection. The thickness L of the air wedge where dark fringes occur is: B mA 2n ..where n = index of refraction of air = 1.00. (In this case, the thin film is the air between the plates, so the index of refraction of the glass is not relevant). If @ is the angle between the slides and d = position of a dark fringe (measured along the slide from the point of contact): tanf — t _L an_1Ocm_d Solving for d (with n_{air}=1), the position of the mth dark fringe is: = g~ (1) () The distance separating two successive fringes (i.e., for m and m+1) is Ad — 0.631um 10cm 2 90um The correct answer is: 0.0351 cm Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6588054&cmid=4375425#question-7036907-1 4/10

11/16/23, 6:32 PM Homework 10: Interference of Light: Attempt review Question 5 Correct Mark 0.90 out of 1.00 The interference phenomenon known as Newton's rings has many applications in the field of optics. Suppose a convex lens with radius of curvature R is placed (curved side down) on a flat glass surface, and illuminated from above with monochromatic light of wavelength \ = 662 nm. When you measure the ring pattern, you find that the radius of the fiftieth bright ring is 5.9 mm. Use this information to determine the radius of curvature of the lens. Hint: you will want to use an equation derived in the video. Answer: [ 1.07 ]\/ m As derived in the video, the radius of a bright fringe is given by: - ) The first bright fringe corresponds to n = 0, so the 50th bright fringe has n = 49. Solving for the radius of curvature R, with n = 49: 272 (n+3)A The correct answer is: 1.06 m Marks for this submission: 1.00/1.00. Accounting for previous tries, this gives 0.90/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6588054&cmid=4375425#question-7036907-1 5/10

11/16/23, 6:32 PM Homework 10: Interference of Light: Attempt review Question 7 Correct Mark 0.90 out of 1.00 A Michelson interferometer, as shown below, is set up and a series of bright fringes is observed. A thin piece of transparent material with a refractive index of 1.50 is then placed into one of the arms of the interferometer such that the light passes through the full thickness of the sample. This causes the fringe pattern to shift such that where there was a bright fringe there is now a dark fringe. If the wavelength of light used in the interferometer is 442 nm what is the smallest possible value for the thickness of the sample? ',, ) Mirror Compensating Plate Half-silvered I Mirror <« JOMIN I 111 Screen mnswer:| 221 o Correct, well done! To invert the pattern one of the beams must have its phase inverted. If the sample has a thickness h then the phase change in the sample compared to an equivalent thickness of air will be: A = K (2h) — k(2h) = (2m + D)m where m is an integer, k' is the wavenumber in the sample and k the wavenumber in air and we remember that the beam passes through the sample twice. Expanding out we get: 4hm 3 (n—1)=(2m+ )x where 7 is the refractive index of the sample. Rearranging we get: ~ (2m+1)A 4(n — 1) So the smallest possible thickness is when m = 0 and so we get: A 442 x 107° h = = 0.221 pm dn—1) 4x (150 —1) https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6588054&cmid=4375425#question-7036907-1 7/10

11/16/23, 6:32 PM Homework 10: Interference of Light: Attempt review The correct answer is: 0.221 um Marks for this submission: 1.00/1.00. Accounting for previous tries, this gives 0.90/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6588054&cmid=4375425#question-7036907-1 8/10

11/16/23, 6:32 PM Homework 10: Interference of Light: Attempt review Explanation: Water has a higher index of refraction than air, so the wavelength of light would decrease in water. As stated above, the film- thickness needed to form each fringe (via con- or de-structive interference) would be smaller. So the pattern would contract toward the centre; each fringe would shift to a smaller radius where the water film is thinner. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6588054&cmid=4375425#question-7036907-1 10/10