Trigonometry Demystified, 2nd edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 11 |

1. As you read through this solution, follow along in Fig. 11-6. If you stand along
the mirror axis at a distance r_{o} from the center of the mirror, where r_{o}
significantly exceeds the focal length f, then the distance r_{i} of
your body's real image from the mirror relates approximately to r_{o} and f
according to the equation 1 / In this case, 1 / 1.50 = 1 / Working out the reciprocals to six significant figures to avoid cumulative rounding errors, you get 0.666667 = 1 / You can subtract 0.119760 from both sides to obtain 0.546907 = 1 / Taking the reciprocals of both sides tells you that 1.82846 = Rounding to three significant figures produces the final answer of |

2. As you read through this solution, follow along in Fig. 11-9. In this situation,
you've been told that the upper medium has a refractive index of 1.05. That's r
according to the geometry here, so you know that
The lower medium has a refractive index of 1.75, which corresponds to
The ray
Now you can invoke the formula for Snell's law, which tells you that sin where sin Simplifying, and going to six significant figures, you get sin When you multiply each side by 0.866025, you obtain sin You can take the Arcsine of both sides to get Arcsin (sin which works out to
The choices all appear rounded off to the nearest degree; the correct one is B, which
tells you that |

3. In order to work this problem out, we can do either of two things. First, we can
input the angle 90º for θ and re-work the problem through just as we did when
we solved Question 2. Alternatively, we can calculate the critical angle that we'd observe
if we sent the light beam upward from underneath along the path of ray S, letting
it emerge above the boundary along the path of ray R. Let's use the first scheme,
because this question describes a ray going down (not up) through the boundary. We start
once more with the Snell's law formulasin When we substitute the known values into the formula, we get sin which simplifies to sin so therefore sin When we take the Arcsine of both sides, we obtain Arcsin (sin which works out to
When we round off to the nearest degree, we get
The correct choice is C. |

4. Refer to the illustration below, which is the same as Fig. 11-15 except that five
specific points have been named. Let's begin by finding out the distance of the light beam
from wall W when it hits the mirror. We can do this by examining a right triangle
connecting the flashlight bulb (point B), point P, and point Q.
In this right triangle: - The apex (top) lies at point
*P*where the light beam strikes the mirror - The height equals 7.500 feet (ft), the distance
*PQ* - The base length equals the distance
*BQ*along a horizontal line parallel to the mirrored wall
Let's calculate the base length tan 22.50º = which solves to
Now we know that point Now let's think about the right triangle connecting point - The apex (top) lies at point
*P*where the light beam strikes the mirror - The height equals 13.00 feet (ft), the distance
*PR* - The base length equals the distance
*RS*along the wall opposite the mirrored wall
Let's keep in mind the fact that when the line beam strikes the mirrored wall, the
angle of reflection equals the angle of incidence, which we know is 22.50º. Based on this
knowledge, we can calculate the base length tan 22.50º = which solves to
Point |

5. Even though the solution to the previous problem involved a lot of calculation, the entire process depended on the law of reflection, which states that when a light beam strikes a flat reflective surface, the angle of incidence equals the angle of reflection. The answer is D. |

6. Total internal reflection can occur when, but only when, a light ray encounters a well-defined boundary between a medium having a relatively higher index of refraction and a medium having a relatively lower index of refraction. According to Table 11-1, fresh liquid water has a refractive index of 1.33, and crown glass has a refractive index of 1.52. If a ray of light travels through fresh liquid water and strikes the surface of a sample of crown glass, total internal reflection can't occur regardless of the angle of incidence, because the second medium's refractive index exceeds the first medium's refractive index. The correct choice is D. |

7. In this situation, the light beam travels in the opposite direction compared with
the scenario of Question 6, so we'll see total internal reflection if the incidence angle
gets large enough. The critical angle θ_{c} is the largest angle that
an incident ray can subtend, relative to the normal, without reflecting internally at a
boundary where it encounters a medium having a relatively lower index of refraction. We
can calculate this angle as
where
accurate to three significant figures. The correct choice is C. |

8. Let's use the formula for the critical angle and input the known values. Then we
can solve for the smallest refractive index that the prism in Fig. 11-16 can have, in
order to ensure total internal reflection at its base, and assuming that the ray travels
along the path shown in the figure. The rules of geometry tell us that, as the ray
encounters the base of the prism from inside, the angle of incidence equals exactly 45º.
We want that to equal the critical angle, so we set
accurate to six significant figures. For total internal reflection to be possible, the prism must have a higher index of refraction than the surrounding medium. In this case, the surrounding medium is air with a refractive index of 1.00000. Therefore, we set
The critical angle formula tells us that 45.0000º = Arcsin (1.00000 / where sin 45.0000º = sin [Arcsin (1.00000 / which simplifies to 0.707107 = 1.00000 / and solves to
rounded off to four significant figures. The correct choice is B. |

9. Let's envision the situation as shown in the diagram below. We represent the
incident ray, traveling through a vacuum, as R. We represent the emerging ray,
traveling in the clear solid material, as S. We assign the indices of refraction
as shown. We also set the angle of incidence to θ = 25.0000º. We want to
find the angle φ for red light, and then for violet light. The divergence angle
will equal the difference between the angles of refraction.
For the ray of red light, we have sin Inputting the known values, we get sin which simplifies to sin Multiplying each side by 0.422618, we get sin We take the Arcsine of both sides and use a calculator to find that
Let's leave in the extra decimal places until the very end of the calculation process, so as to avoid cumulative rounding errors. We've just found the angle of refraction for the red ray. Now let's repeat all of the arithmetic for the violet ray, where sin sin sin
The divergence angle between the red and violet refracted rays equals the difference between the red angle of refraction and the violet angle of refraction. That's 16.5919º - 16.1431º, which works out to 0.45º when we round off to the nearest hundredth of a degree. The correct choice is B. |

10. Let's revise the foregoing diagram to show the new situation (below). As before,
we represent the incident ray, traveling through a vacuum, as R. We represent the
emerging ray, traveling in the clear solid material, as S. The refractive indices
remain the same, but the angle of incidence has increased to θ = 85.0000º.
Once again, we want to find the angle φ for red light, and then for violet
light, and then find the difference between them.
Following the same process as we did when we solved the previous problem, we set sin sin sin
Now for the violet ray, for which sin sin sin
In this scenario, the divergence angle between the red and violet refracted rays equals 42.3072º - 40.9494º, which works out to 1.36º when we round off to the nearest hundredth of a degree. The correct choice is D. |