Homework Assignment Geometric Optics

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University of Alberta *

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Physics

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Dec 6, 2023

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11/16/23, 6:27 PM Started on State Completed on Time taken Marks Grade Question 1 Correct Mark 1.00 out of 1.00 A man stands in front of a plane mirror. He wishes to see a full-length image of himself, from eye-level all the way down to his feet. If the man is 165 cm tall, what minimum length of mirror is required? Assume that the mirror length is measured vertically. Wednesday, 28 October 2020, 9:23 PM Finished Thursday, 29 October 2020, 6:16 PM 20 hours 53 mins 9.33/11.00 84.85 out of 100.00 Answer: 82.5 Correct, well done! JV cm Homework 7: Geometric Optics: Attempt review The minimum mirror-length is half the person's height, as illustrated in the ray diagram below. A E The correct answer is: 82.5 cm Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 1/12
11/16/23, 6:27 PM Homework 7: Geometric Optics: Attempt review Question 2 Correct Mark 1.00 out of 1.00 A glass hemisphere is completely submerged in a fluid. When a beam of light is incident upon the flat glass surface, the light is refracted with the angles shown. The angles are: ¢1 = 49.0° ¢2 = 600 If the speed of light in the fluid is 0.693c, what is the index of refraction in the glass? Answer: 1.89 j\/ The angles of incidence and refraction are measured relative to the normal, so 8; = 90° ¢; and 85 = 90° ¢,. The law of refraction then requires that: nysin(f1) n cos(¢r) sin(6;) cos(¢a) Now we also need to know that the refractive index of the fluid is: n, = Ur where c is the speed of light in vacuum and vy is the speed of light in the fluid. Now remembering also that: cos ¢y = cos 60° = 1/2 we get: ng = —Cos ¢ v Putting in the numbers this gives: Ny x c0s49.0° = 1.893 B 2 "~ 0.693 The correct answer is; 1.89 Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 2/12
11/16/23, 6:27 PM Homework 7: Geometric Optics: Attempt review Question 3 Correct Mark 0.00 out of 1.00 Refer to the previous problem about the glass hemisphere in a fluid. The beam of light enters the flat surface at the exact centre of the glass hemisphere, and exits the glass at the point P. What is the angle of refraction as the light enters the fluid at P? Select one: D cross out 2 cross out 90° ¢ cross out 90° ¢, cross out 0.0° Cross out ¢ 90° cross out total internal reflection occurs at point P, so the light does not enter cross out the fluid Your answer is correct. Because the light strikes the centre of the hemisphere, it travels along a radius of the sphere to the point P. Any radial path is perpendicular to the surface of the sphere. Therefore the ray of light is normal to the glass-fluid interface at point P. The angle of incidence and angle of refraction are both zero. In other words, the light is not refracted at P. The correct answer is: 0.0° Marks for this submission: 1.00/1.00. Accounting for previous tries, this gives 0.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 3/12
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11/16/23, 6:27 PM Homework 7: Geometric Optics: Attempt review Question 4 Correct Mark 1.00 out of 1.00 iceCube tab The Ice Cube Neutrino Observatory at the south pole looks for tiny flashes of light from electrons, muons and tau particles produced deep in the ice, ~2km below the surface, by neutrinos. In order to be able to reconstruct these events the speed of light in the ice is required. This is measured to be 2.19x108 m/s. What is the refractive index of ice? [c=3.0x108 m/s] Answer: | 1.3699 v Deep Core Correct, well done! Refractive index is defined as: © 19 n=—. 2 So all that is needed is to divide the speed of light in vacuum by the speed of light in the ice. The correct answer is: 1.37 Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 4/12
11/16/23, 6:27 PM Homework 7: Geometric Optics: Attempt review Question 5 Correct Mark 1.00 out of 1.00 A glass, right-angled prism with triangular cross-section is immersed in a liquid with index of refraction n = 1.55. A horizontal ray of light passes through the glass and strikes the glass/water interface at the point P. Total internal reflection occurs at P, and the reflected light travels vertically downward as shown. What is the minimum index of refraction of the glass such that total internal reflection would occur at P? Answer: 2.19 :JV Correct, well done! From the geometry of the situation we see that the angle of incidence at P is 45 degrees. If the index of refraction of the glass is the minimum required for total internal reflection to occur, then the angle of incidence must be equal to the critical angle. n . ny f e ng 9 sin45° f v2 Putting in the numbers this gives: ng = 1.8 X V2 =2.192 The correct answer is: 2.19 Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 s glass liquid 5/12
11/16/23, 6:27 PM Question 6 Correct Mark 1.00 out of 1.00 Homework 7: Geometric Optics: Attempt review The diagram shows three layers of transparent media. Light originating inside medium 2 strikes the upper interface (between 1 and 2) and undergoes total internal reflection. The reflected ray strikes the lower interface (between 2 and 3) where the ray is partially reflected and partially refracted. Rank the indices of refraction for the three media (if possible), from smallest to largest. 1 Select one: 5] ns no n na ng n n < N9y < < ng < ng <mn < Ny = ng < ng < ng < Ny <mn < Ny < ng < mn < Ny and ng < ng but we cannot rank nj vs. ng There is insufficient information to rank any of them Your answer is correct. At the upper interface, total internal reflection occurs so we know that n; < ns. At the lower interface, the transmitted ray is refracted away from the normal, so we know that ng3 < na,. cross out cross out Ccross out Cross out cross out cross out cross out Ccross out cross out But how do we rank the indices of refraction for the first and third layers? We know that (1) the angle of reflection equals the angle of incidence, and (2) the two interfaces are parallel surfaces, therefore the two angles of incidence (see diagram below) must be equal. We also know that total internal reflection occurs at the upper interface, so 8; exceeds the critical angle for the upper interface, but not the lower one. Therefore; 2o sin(6;) < n2 ny < ng https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 ns ny 6/12
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11/16/23, 6:27 PM Homework 7: Geometric Optics: Attempt review The correct answer is: ny < ng < N9 Marks for this submission: 1.00/1.00. Question 7 Correct Mark 0.33 out of 1.00 Drag the correct words into the blanks to complete the following sentence correctly: When an object is placed on the optical axis 40 cm from a convex, spherical mirror which has a focal length of 20 cm it produces a v image which is | rightway up [ and on the « side of the mirror compared to the object. Note: You only get one attempt at this question because there are only two options for each blank! real inverted Your answer is correct. When the object is placed in front of a convex mirror the image produced is always virtual, right way up and on the opposite side of the mirror to the object. The correct answer is: Drag the correct words into the blanks to complete the following sentence correctly: When an object is placed on the optical axis 40 cm from a convex, spherical mirror which has a focal length of 20 cm it produces a [virtual] image which is [rightway up] and on the [opposite] side of the mirror compared to the object. Note: You only get one attempt at this question because there are only two options for each blank! Marks for this submission: 1.00/1.00. Accounting for previous tries, this gives 0.33/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 7/12
11/16/23, 6:27 PM Homework 7: Geometric Optics: Attempt review Question 8 Correct Mark 1.00 out of 1.00 Drag the correct words into the blanks to complete the following sentence correctly: When an object is placed on the optical axis 20 cm from a diverging lens which has a focal length of 10 ¢cm it produces a v image which is | rightway up and on the « side of the lens compared to the object. Note: You only get one attempt at this question because there are only two options for each blank! Your answer is correct. When an object is placed in front of a diverging lens the image produced is virtual, rightway up and on the same side of the lens as the object regardless of the focal length of the lens. The correct answer is: Drag the correct words into the blanks to complete the following sentence correctly: When an object is placed on the optical axis 20 cm from a diverging lens which has a focal length of 10 cm it produces a [virtual] image which is [rightway up] and on the [same] side of the lens compared to the object. Note: You only get one attempt at this question because there are only two options for each blank! Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 8/12
11/16/23, 6:27 PM Homework 7: Geometric Optics: Attempt review Question 9 Correct Mark 1.00 out of 1.00 Drag the correct words into the blanks to complete the following sentence correctly: When an object is placed on the optical axis 20 cm from a convex, spherical mirror which has a focal length of 40 cm it produces a v image which is [ rightway up | ¥ and on the v side of the mirror compared to the object. Note: You only get one attempt at this question because there are only two options for each blank! real inverted Your answer is correct. When the object is placed in front of a convex mirror the image produced is always virtual, right way up and on the opposite side of the mirror to the object. The correct answer is: Drag the correct words into the blanks to complete the following sentence correctly: When an object is placed on the optical axis 20 cm from a convex, spherical mirror which has a focal length of 40 cm it produces a [virtual] image which is [rightway up] and on the [opposite] side of the mirror compared to the object. Note: You only get one attempt at this question because there are only two options for each blank! Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 9/12
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11/16/23, 6:27 PM Homework 7: Geometric Optics: Attempt review Question 10 Correct Mark 1.00 out of 1.00 An object is placed on the central axis of a thin converging lens, at a distance of 24 cm from the lens. The magnitude of its focal length is [f|= 11 cm. What is the linear magnification, M, of the resulting image? Answer:| -0.846 v Correct, well done! This is a converging lens so we know that the sign of f is positive. Starting with the lens equation: 1 4 1 1 u v f and so the position of the image is just: 1 1 1 uf v f u u—f and the magnification is v 1 M = —= —— U 1u f The correct answer is: -0.846 Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 10/12
11/16/23, 6:27 PM Homework 7: Geometric Optics: Attempt review Question 11 Correct Mark 1.00 out of 1.00 An object is placed on the central axis of a thin lens, at a distance of 19 cm from the lens. The linear magnification of the resulting image is 0.69. What is the focal length, f, for this lens? Answer: [ -42.29 ]\/ cm Correct, well done! From the formula for magnification the image distance is v=—Mu The focal length can be calculated from the lens equation: Rearranging this gives: The correct answer is: -42.3 cm Marks for this submission: 1.00/1.00. https://eclass.srv.ualberta.ca/mod/quiz/review.php?attempt=6312116&cmid=4375382 11/12