s81 - Solutions to Homework #8 Chapter 26 48. Picture the...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to Homework #8 Chapter 26 48. Picture the Problem : A kitchen has twin side-by-side sinks. One sink is filled with water, the other is empty. You observe the bottom of each sink and judge the depth of each. Strategy: Consider the refraction light when it leaves the water to answer the conceptual question. Solution: 1. (a) Water causes the apparent depth to decrease (see Figure 26-22). Therefore, the left sink (with water) appears to be shallower than the sink on the right (without water). 2. (b) The best explanation is II . Water bends the light, making an object under the water appear to be closer to the surface. Thus the water-filled sink appears shallower. Statements I and III each ignore the effects of refraction. Insight: If one sink were filled with ethanol ( n = 1.36) it would appear even shallower than the sink filled with water ( n = 1.33). 64. Picture the Problem : The image shows a beam incident upon a horizontal glass surface. This beam refracts into the glass. When the refracted light hits the vertical surface it totally internally reflects. Strategy: From the figure, note that ( 29 c 2 2 sin sin 90 cos . θ = - = Use Snell’s Law (equation 26-11) to write an equation relating the index of refraction and the refracted angle. Use equation 26-12 to write a second equation relating the refracted angle and index of refraction, where the sine of the critical angle is the cosine of the refracted angle. Square both of these equations and sum them to eliminate the angle and solve for the index of refraction. Solution: 1. (a) Write Snell’s Law for the first refraction: 2 sin sin n = 2. Write equation 26-12 in terms of the refracted angle: ( 29 air c 2 2 2 1 sin sin 90 cos cos 1 n n n n = = = - = = 3. Sum the squares of the two equations and solve for n: ( 29 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 sin cos sin 1 sin cos sin 1 sin 1 sin 70 1 1.4 n n n n n + = + + = = + = + = + = 4. (b) Because the minimum index of refraction is related to the incident angle as 2 sin 1 n = + , decreasing θ will cause the minimum value of n to be decreased. Insight: As an example of decreasing the incident angle, set the incident angle equal to 50 ° . Verify for yourself that in this case, the minimum index of refraction drops to 1.3. 66. Picture the Problem : The image shows a light beam incident on a semicircular disk with index of refraction 1.66. After passing through the disk, the light beam hits a wall 5.00 cm above the center of the disk and at a distance 20.0 cm behind the disk. Strategy:
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/10/2012 for the course PHY 102 taught by Professor Korotkova during the Spring '09 term at University of Miami.

Page1 / 4

s81 - Solutions to Homework #8 Chapter 26 48. Picture the...

This preview shows document pages 1 - 2. Sign up to view the full document.

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