{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

i310kqtJ53a8_1210016833_jwk572

i310kqtJ53a8_1210016833_jwk572 - homework 12 KIM JI Due...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
homework 12 – KIM, JI – Due: Apr 13 2008, 4:00 am 1 Question 1, chap 33, sect 5. part 1 of 2 10 points A point light source delivers a time- averaged power P . It radiates light isotrop- ically. A piece of small flat surface is placed at D, which is a distance r away. This piece has a cross section A surf . The surface reflects 1 4 of the light and absorbs 3 4 of the light. As- sume the light hitting the various parts of the surface is perpendicular to them. r D Point source What is the time-averaged energy density hitting the surface? 1. u = π r 2 P c 2. u = A surf P 3. u = P A surf 4. u = A surf P c 5. u = P c A surf 6. u = P 4 π c r 2 correct 7. u = 4 π r 2 P 8. u = P 4 π r 2 9. u = 4 π r 2 P c 10. u = π r 2 P Explanation: Basic Concepts EM Wave Solution: The time-averaged energy den- sity at D is given by u = I c = P 4 π r 2 c . Question 2, chap 33, sect 5. part 2 of 2 10 points Find the total time-averaged force on the surface in terms of the intensity I of the light at D. 1. F = 3 A surf I 2 c 2. F = 7 A surf I 4 c 3. F = 2 A surf I c 4. F = 5 4 4 π I c 5. F = 4 π I c 6. F = A surf I c 7. F = 7 4 4 π I c 8. F = 5 A surf I 4 c correct 9. F = 3 2 4 π I c 10. F = 2 4 π I c Explanation: The time-average force is F = Pressure A surf = F abs + F refl = parenleftbigg 3 4 u + 1 4 2 u parenrightbigg A surf = 5 A surf I 4 c . Question 3, chap 34, sect 2. part 1 of 2 10 points Consider the case in which light ray A is incident on mirror 1 , as shown in the figure. The reflected ray is incident on mirror 2 and subsequently reflected as ray B. Let the an- gle of incidence (with respect to the normal) on mirror 1 be θ A = 46 and the point of
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
homework 12 – KIM, JI – Due: Apr 13 2008, 4:00 am 2 incidence be located 20 cm from the edge of contact between the two mirrors. 90 x A B mirror 1 mirror 2 θ A θ B What is the angle of the reflection of ray B (with respect to the normal) on mirror 2 ? Correct answer: 44 (tolerance ± 1 %). Explanation: If the angle of the incident ray A is θ A , the angle of reflection must also be θ A . Since the mirrors are perpendicular to each other, angle θ B is equal to 90 - θ A θ B = 90 - θ A = 90 - 46 = 44 . Question 4, chap 34, sect 2. part 2 of 2 10 points Determine the angle between the rays A and B. 1. 46 2. 90 3. 0 4. 180 correct 5. 88 6. 92 7. 44 Explanation: Refer to the figure below. Consider the angle each ray makes with mirror 1 . Ray A makes an angle of φ A with mirror 1 . This is given by φ A = 90 - θ A = θ B . The normal to mirror 2 is parallel to mirror 1 . Thus the angle ray B makes with mirror 1 (assumed to be extended) is θ B . Both rays make the same angle with respect to mirror 1 . They are parallel, but point in the opposite directions. Therefore the angle between them is 180 . 90 x A B mirror 1 mirror 2 θ A θ B φ A φ A Question 5, chap 34, sect 3.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern