Ch.35 7 11 16 17 18 39 41 55 61
Answers

Chapter 35 Summary: Below is a pictorial summary of the lecture deriving the law of reflection using Huygens' Principle. We stated that the incident wave front shown strikes the surface at point A, and that point (A) becomes a point source of spherical waves according to Huygens.  AD= BC, and the 2 right triangles ADC and CBA share a common side AC. Therefore, the  triangles ADC and CBA  are similar and the two angles with the horizontal are equal :

We can also construct the angles with the  vertical as shown, which represent the angle of incidence and the angle of reflection. Since the 2  horizontal angles are equal , then the two vertical angles are equal , with  the result that   the angle of incidence = angle of reflection.
  
7. See fig. 35.30....n1sinø1 =n2sin45. Find ø1 .  Note:  n 1 = 1 and n 2 = value for water....Note ø1 >450
11. Click  WEB  and scroll down to Chapter 35 and click problem 11 to see how this problem is done on the student web site at http://www.harcourtcollege.com/physics/pse/student/guide/... If a problem is labeled "web" in the textbook, as this one is, you can always  go to this  link and see the solution.....Please bookmark the link for future reference .....
16. d = (2.00 cm/cosø2)sin(30-ø2), where ø2 is found using snell's law: n1sinø1 = n2sinø2   with ø1 = 300,   n1 = 1 and n2 = value for glass. See table 35. 1.
17. time = distance/v   where v = c/n 2 . Note that distance = 2.00 cm/cosø 2
18. This problem is easy once you think about it for a few minutes or less. SEE PROBLEM 16.  Obviously, ø2 = 200 , where n1sinø1 = n2sinø2   and n1 = 1 and n2 = value for the oil....You can find ø1  = ø= initial angle at the air-oil interface.
Finally, find ø3  using:   n2sin200 = n3sinø3 . Find ø3 = ø,   where n3   = value for water.
39. 
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Solve for the critical angle.
Note: n1 = 1.50 and n2 = 1.33
41. 

The above diagram shows the situation at the extreme angle of viewing. In that case there is total internal reflection. The bubble is on a line going through the middle of the coin. d is the vertical distance of the bubble from the surface. Inside the cube, the angle of incidence is equal to the angle of reflection. The ray from the bubble hits the edge of the coin and the angle of refraction is 900 . The refracted ray "skims" along the surface of the Lucite cube indicated above.

Let us solve for the angle of incidence that the ray from the bubble makes with the normal in the case of total internal reflection. This is the critical angle. We know that:
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n1 = 1.59 and n2 = 1.
At the extreme angle of viewing, the angle of refraction is 90 degrees. Thus,
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Solve this equation for the critical angle. Now, let's analyze the depth d if the bubble cannot be seen at any angle, the case of total internal reflection. Clearly, the depth d is given by:

Here, r = the radius of the coin. If the depth d is greater than this value, then you can see the bubble as indicated by the figure below:

In this case the angle of refraction is less than 90 degrees.




On the other hand, you cannot see the bubble if d is less than the value given above:

Think about this for a minute. Draw a sketch on your paper to help visualize the two cases of d being greater than or being less than the ratio of the radius and the tangent.
Thus, since you cannot see the bubble when it is below the penny, we have that:

Since you can see the bubble when it is below the dime we have that:

Thus:

Compute the upper and lower limits of d.
55. Click  WEB  and scroll down to Chapter 35 and click problem 55 to see how this problem is done 
61. Click  WEB  and scroll down to Chapter 35 and click problem 61 to see how this problem is done  (a) See figure P35.61. Geometrically, we see that as the angle of incidence made with the vertical normal shown increases, then the angle of incidence on the vertical side inside the plastic block decreases. That is to say, when the initial angle of incidence measured with the vertical normal increases, the second angle of incidence with the horizontal normal inside the cube decreases.

Now, assume that we have total internal reflection at the vertical surface inside the cube, and assume that the angle of incidence on the vertical surface is the critical angle at which total internal reflection just occurs. Form a right triangle inside the cube whose sides are (1) the refracted ray, (2) the vertical normal and (3) a perpendicular dropped from the vertical side to the vertical normal. Let the angle of incidence made with the horizontal normal inside the cube be given by:

Now, let:

Determine the critical angle using the usual methods:

Note: n1 = 1 and n2 = 1.49
Next solve for the angle of refraction made with the vertical normal using:

Finally, solve for the initial angle of incidence with the vertical normal using Snell's law at the first interface:

Note this angle is the maximum allowed in order for total internal reflection to occur at the vertical interface inside.
(b) Repeat the problem for n1 = 1.33 and n2 = 1.49.
(c) No total internal reflection is possible. Prove this.