Changes between Version 2 and Version 3 of u/johannjc/scratchpad12

Timestamp:

02/03/16 23:23:32 (9 years ago)

Author:

Jonathan

Comment:

—

u/johannjc/scratchpad12

@@ -13 +13 @@
 SInce the speed of sound remains constant, and the frequency increases, the wavelength must decrease...
 
-$c = \lambda \nu $
+$c = \lambda f $
 
 
 5.)  
 
-$ \lambda = \frac{c}{\nu} = \frac{2.99792458 \times 10^8 m/s}{670 nm} {1 \times 10^9 nm}{1 m} \frac{Hz s}{1} = 2.0 \times 10^20 Hz $
+$ \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.