1.) Not sure what type of wave he is talking about - or what precisely he means by motion or source… Typically a wave is a disturbance that involves the continual transfer of energy from one form to another as the disturbance propagates through some medium. For light, energy is transferred between electric and magnetic fields, where for a sound wave, energy is transferred from kinetic energy of a parcel of air, to thermal energy as the next parcel is compressed and so on… For a pendulum, kinetic energy is transferred to potential and back again… But the source is usually some displacement (work) from equilibrium that produces an excess of energy that can then propagate.

2.) I'm guessing that by wave me means a continuous or standing wave - where a pulse is finite in size and in duration.

3.)

T = \frac{1}{\nu}

The period T is inversely proportional to frequency, so if we double the frequency, we halve the period.

4.) As the train moves faster towards you, the sound waves are emitted from locations that are closer and closer to you, which means that each successive wave has a shorter and shorter distance to travel, so while they travel at the same speed through the air, they take less and less time to reach you, so they arrive closer and closer together in time, and the frequency with which the waves arrive is higher.

SInce the speed of sound remains constant, and the frequency increases, the wavelength must decrease…

c = \lambda f

5.)

\lambda = \frac{c}{f} = \frac{3.00 \times 10^8 m/s}{670 nm} \frac{1 \times 10^9 nm}{1 m} \frac{Hz \times s}{1} = 4.47 \times 10^{14} Hz = 447 GHz

6.) The period of a pendulum is given by

T = 2 \pi \sqrt{\frac{L}{g}}

so

L = \frac{g T^2}{4 \pi^2}

L = \frac{\left ( 9.8 m/s^2 \right ) \left (1 s^2 \right )}{4 \pi^2} = .248 m

The clock will not keep the right time on moon because g is different…

If we want to keep T=1s on the moon, we need to adjust the length.

L_{moon}=\frac{g_{moon} T^2}{4 \pi ^2} = \frac{\left ( 1.63 m/s^2 \right ) \left ( 1s^2 \right )}{4 \pi ^2} = 0.0412 m

7.)

d = v t

t = \frac{d}{v} = \frac{1.5 \times 10^8 km}{3.0 \times 10^8 m/s} \frac{1 \times 10^3 m}{1 km} = 500 s = 8.33 \mbox{ minutes}

8.)

c = \lambda f

so

\lambda = \frac{c}{f} = \frac {3.00 \times 10^8 m/s}{2.45 \times 10^9 Hz} = .122 m

9.)

The angle of reflection = the angle of incidence = 25 degrees

10.)

if the light ray approaches at 22 degrees from the mirror surface - then it is 90-22=68 degrees from the surface normal - so the angle of incidence = 68 degrees, so the angle of reflection will also be 68 degrees.

11.) snells law says that

\frac{n_1}{n_2}=\frac{\sin \theta_2}{\sin \theta_1}

where n is the refractive index of each material, and \theta is the angle of incidence or refraction depending on which way the light is headed…

If light is traveling from material 1 into material 2, then \theta_1 is the angle of incidence, and \theta_2 is the angle of refraction

In this problem, the light travels from the unknown substance into air, so we are trying to find n_1

n_1 = n_2 \frac{\sin \theta_2}{\sin \theta_1} = 1.0 \frac{\sin 33.8}{\sin 24.5} = 1.34

12.) Since the angle of incidence = angle of reflection, the path of light bouncing of a mirror is symmetric… So the path from your eye to the truck drivers eye is the same as the path of light from the truck drivers eye to your eye. So if you can see the truck driver's eyes, he can see your eyes.

13.) The refractive index of water is 1.333 which is greater than air, so as light rays from the fish leave the water, they are bent away from the normal, which will make the fish appear to be further away than it is - so you would aim below the fish.

now, if we assume that the refractive index of the water is the same for all wavelengths of light, than the laser light would be bent going towards the fish, in the same way that the light from the fish is bent towards your eye… so you would aim the laser at the apparent location of the fish

14.) When the light exceeds the critical angle, light cannot be refracted and cannot penetrate (escape) the internal surface of the fiber - so it is completely reflected and the signal strength is maintained.