Homework #1-solutions

Homework #1-solutions - mehta (snm425) – Homework 1 –...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: mehta (snm425) – Homework 1 – erskine – (58715) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points For 599 nm light, calculate the critical an- gle for the following materials surrounded by water (index of refraction: 1 . 333). diamond. (index of refraction: 2 . 34) Correct answer: 34 . 7264 ◦ . Explanation: Basic Concept: Snell’s law for total internal reflection: n 1 sin θ c = n 2 sin 90 ◦ where θ c is the critical angle for total internal reflection. From Snell’s law, for total internal reflection in diamond, 2 . 34 sin θ 1 = 1 . 333 sin 90 ◦ So, critical angle for for diamond θ diamond = sin − 1 parenleftbigg 1 . 333 2 . 34 parenrightbigg = 34 . 7264 ◦ . 002 (part 2 of 3) 10.0 points flint glass. (index of refraction: 1 . 64) Correct answer: 54 . 3709 ◦ . Explanation: Similarly, critical angle for flint glass θ glass = sin − 1 parenleftbigg 1 . 333 1 . 64 parenrightbigg = 54 . 3709 ◦ . 003 (part 3 of 3) 10.0 points ice. (index of refraction: 1 . 31) 1. sin − 1 parenleftbigg n water n ice parenrightbigg 2. there is no critical angle correct 3. sin − 1 parenleftbigg 1 n ice parenrightbigg 4. sin − 1 parenleftbigg n ice n water parenrightbigg Explanation: For ice, since its refraction index, 1 . 31, is less than that of water, 1 . 333, there is no critical angle! 004 10.0 points Hint: A ray diagram would be helpful. Determine the minimum height of a vertical flat mirror in which a person 71 in . in height can see his or her full image. Correct answer: 35 . 5 in . . Explanation: In the figure, the mirror is labeled AB . A ray from the woman’s foot F strikes the bottom of the mirror at B , with an angle equal to θ and proceeds to the woman’s eye. L θ θ C D A B E T F M The two right triangles EBM and FBM are identical, since they share the common side MB and angle θ . Therefore EM = MF = 1 2 EF which is also the distance BC . Similarly, a ray from the top of the woman’s head T strikes the top of the mirror at A and proceeds to her eye. The same line of reasoning as above leads to the conclusion that DA = 1 2 TE Thus the length AB of the mirror is 1 2 ( TE + EF ), which is one half the woman’s height. Note that the mirror’s bottom edge must be exactly 1 2 EF from the floor for a full-height image to be possible. Note also that the conclusions reached here are valid regardless of how far she stands from the mirror. mehta (snm425) – Homework 1 – erskine – (58715) 2 005 10.0 points An object is placed at a distance of 1.5 f from a converging lens of focal length f , as shown. 2 f f f 2 f 3 f What type of image is formed and what is the size relative to the object?...
View Full Document

This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas.

Page1 / 9

Homework #1-solutions - mehta (snm425) – Homework 1 –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online