Homework Assignment Diffraction

11/16/23, 6:33 PM Started on State Completed on Time taken Marks Grade Question 1 Correct Mark 1.00 out of 1.00 A slit 396 um wide is illuminated by light of wavelength 571 nm. A diffraction pattern appears on a screen 3.1 m away. On the screen, what is Wednesday, 2 December 2020, 4:05 PM Finished Wednesday, 2 December 2020, 6:19 PM 2 hours 13 mins 7.80/8.00 97.50 out of 100.00 Homework 11: Diffraction: Attempt review the distance between the first and second diffraction minima on one side of the central maximum? Answer: [ 0.00447 o (= Use a sin(f) = mA for m = 1 and m = 2 to find the angular position of the first and second minima. For each angle, find the distance from the centre of the pattern to the fringe using: ..where D is the distance from the slit to the screen. The difference y5 — y; gives the distance separating the first and second dark fringes. The correct answer is: 0.00447 m Marks for this submission: 1.00/1.00. y = Dtan(6) https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6660007 &cmid=4375438 1/9

11/16/23, 6:33 PM Homework 11: Diffraction: Attempt review Question 2 Correct Mark 0.90 out of 1.00 Light of wavelength 574 nm passes through a slit, creating a diffraction pattern on a screen. The angular separation between the first diffraction minima on either side of the central maximum is 4.3°. Determine the width of the slit. Answer: [ 15300.23 ]\/ nm asin(f) = mA For the first diffraction minima, let m = 1. The angular separation between the two minima on either side of the central maximum is 26. The correct answer is: 1.53e4 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=6660007 &cmid=4375438 2/9

11/16/23, 6:33 PM Homework 11: Diffraction: Attempt review Question 4 Correct Mark 1.00 out of 1.00 In a double-slit experiment, a screen is placed a distance D = 3.00 m from a pair of vertical slits. The slit separation distance is d = 15.6 um. When light with wavelength 589 nm is incident upon the slits, an interference pattern appears on the screen. Assume that the width of each slit is much smaller than the wavelength. Consider a location y on the screen, located 1.31 cm (measured horizontally) from the centre of the entire pattern. What is the ratio of the Enter your answer as a decimal number with three significant figures. Iy Ictr B Central bright fringe 1 Answer: ‘ 0.874 j\/ Correct, well done! We use the formula: 27d sin(6) I = 41, cos? (f) 2 A The intensity at the middle of the central bright fringe is I ;. = 41. where ¢ = Using a small angle approximation, sin(f) ~ tan(f) = %, The intensity ratio is The correct answer is: 0.874 Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6660007&cmid=4375438 4/9

11/16/23, 6:33 PM Homework 11: Diffraction: Attempt review Question 5 Correct Mark 1.00 out of 1.00 A double slit experiment (A) is performed using light of wavelength A. Then a second experiment (B) is performed with the same wavelength on a different pair of slits. Figures A and B show the positions of bright fringes (principal maxima) within the central diffraction envelope for the two experiments. Consider the slit width a and the slit separation distance d in each case. Which of the following statements is/are correct? Caution: There may or may not be more than one correct answer. Select all correct answers and no incorrect answers for full credit. g (U B B B N Select all that apply: The slit-width a in experiment A is greater than that in experiment cross out B. The slit separation distance d in experiment A is greater than that Cross out «/ in experiment B. The ratio d/a in Experiment A is greater than that in experiment Cross out «# B. The width of each slit is much smaller than the wavelength: a << A cross out Your answer is correct. Refer to the diagram below for terminology. Single-slit diffraction envelope Bright fringes due to double slit interference First consider single slit diffraction, where the angular position of the first minimum (m = 1) is given by sin(0) = % This gives the width of the central diffraction envelope. Experiment A has a wider central diffraction maximum, so the ratio % is larger. Since both experiments use the same wavelength, this means that the slit width a in experiment A is smaller than that in experiment B. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6660007 &cmid=4375438 5/9

11/16/23, 6:33 PM Homework 11: Diffraction: Attempt review Question 6 Correct Mark 0.90 out of 1.00 A diffraction grating has 405 lines (rulings) per mm. When white light shines on the grating, all wavelengths experience constructive interference at the centre of the diffraction pattern, so the central maximum (m = 0) is white. However, the higher order maxima (side fringes) occur at different angles for different wavelengths, so the colours are dispersed. How many complete orders of the visible spectrum will be produced in the diffraction pattern? Assume that the range if visible wavelengths is 400-700 nm. Your answer should be an integer. Answer:; [: jv Diffraction maxima can be found using: dsin(f) = mA We want to find the larges m value for the longest wavelength such that the angular position of the fringe is less than or equal to 90 degrees. Take A = 700 nm and 8 = 90", we solve for m: d 103m A (405)(700X107° M) Round down to the next integer. The correct answer is: 3 Marks for this submission: 1.00/1.00. Accounting for previous tries, this gives 0.90/1.00. Question 7 Correct Mark 1.00 out of 1.00 A diffraction grating with a width of 5.0 mm contains 617 lines (rulings) per mm. When it is fully illuminated, what is the smallest wavelength interval this grating can resolve in the 3rd order, if the average wavelength is taken to be 691 nm? Answer: [ 0.07466 }/ nm AN = ——— ..where m = 3 and N = number of rulings = 617 lines per mm x 5.00 mm. The correct answer is: 0.0747 nm Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6660007 &cmid=4375438 7/9

11/16/23, 6:33 PM Homework 11: Diffraction: Attempt review Question 8 Correct Mark 1.00 out of 1.00 You are observing the surface of the Moon with a 1.3 m diameter telescope. Assuming that the telescope resolution is limited only by diffraction effects, what is the diameter of the smallest crater that the telescope can resolve? Assume that the wavelength of light is 574 nm. Hint: In order to "resolve” the crater as a distinct object, consider that the crater-rims on opposite sides of the crater are two points that can just barely be resolved. Take the Earth-Moon distance to be 385,000 km. Answer: [ 207 ]\/ m The angular resolution of the telescope is 0 = 1.22 (%) where d = diameter of telescope aperture. For small angles, O = % where D = the distance between two barely resolvable points (i.e., the crater diameter), and L = distance to the Moon. Solving for D: _ (1L2)(@m)) d D The correct answer is;: 207 m Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6660007 &cmid=4375438 8/9