Homework 3 Solution

# Homework 3 Solution - Version PREVIEW – Homework 3 –...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version PREVIEW – Homework 3 – Savrasov – (39825) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Air to Diamond 001 10.0 points Light is incident at an angle of 40 ◦ on the surface of a diamond. Find the angle of refraction. Correct answer: 15 . 4035 ◦ . Explanation: By Snell’s Law, for an angle of incidence α and an angle of refraction β , n air sin α = n sin β sin β = sin α n , since n air = 1. β = arcsin parenleftbigg sin α n parenrightbigg . Drinking Glass Bottom 04 002 10.0 points An cylindrical opaque drinking glass has a diameter 6 cm and height h , as shown in the figure. An observer’s eye is placed as shown (the observer is just barely looking over the rim of the glass). When empty, the observer can just barely see the edge of the bottom of the glass. When filled (with a transparent liquid with an index of refraction of 1 . 26) to the brim, the observer can just barely see the center of the bottom of the glass. θ i h 6 cm θ r e y e Calculate the angle θ r . Correct answer: 63 . 7321 degrees. Explanation: Looking at the figure below, R r R i θ i h r d θ r e y e After filling the glass with liquid, we know from Snell’s law that n liquid sin θ i = n air sin θ r . The radius r is one-half the diameter d , there- fore sin θ i ≡ r R i = r √ r 2 + h 2 . Now we can solve for h . Substituting our expressions for the sines into Snell’s Law, we get n liquid r √ r 2 + h 2 = n air d √ d 2 + h 2 , so n 2 liquid r 2 ( d 2 + h 2 ) = n 2 air d 2 ( r 2 + h 2 ) . We can simplify this by using n air = 1 . 0, n liquid = n , and d = 2 r h 2 ( d 2- n 2 r 2 ) = d 2 r 2 ( n 2- 1) h 2 bracketleftbig (2 r ) 2- n 2 r 2 bracketrightbig = d 2 r 2 ( n 2- 1) h 2 r 2 (4- n 2 ) = d 2 r 2 ( n 2- 1) h 2 (4- n 2 ) = d 2 ( n 2- 1) , so d h = radicalBigg (2 r ) 2- n 2 r 2 r 2 ( n 2- 1) = radicalBigg 4- n 2 n 2- 1 . Version PREVIEW – Homework 3 – Savrasov – (39825) 2 we see from the geometry that tan θ r ≡ d h , therefore θ r = arctan parenleftbigg d h parenrightbigg = arctan radicalBigg 4- n 2 n 2- 1 = arctan radicalBigg 4- (1 . 26) 2 (1 . 26) 2- 1 = 63 . 7321 ◦ ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

Homework 3 Solution - Version PREVIEW – Homework 3 –...

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

View Full Document
Ask a homework question - tutors are online