Useful equations

 

1. If an object or a wave moves with constant speed, there is a simple relationship between the distance that it moves, the speed, and the elapsed time. It is possible to find any one of these parameters knowing the other two.

 

If:      D=distance traveled

          S=speed

          T=elapsed time

 

Then:

                   D=S * T                calculate distance given speed and elapsed time

                   S=D/T                  calculate speed given distance and elapsed time

                   T=D/S                  calculate elapsed time given distance an speed

 

Important: All three quantities must be expressed in a consistent set of units.

 

In general, it is easiest to use one of the following two choices:

 

          1. T in seconds (s), D in kilometers (km), S in kilometers per second (km/s)

          2. T in seconds (s), D in meters (m), S in meters per second (m/s)

 

2. For any wave, there is a simple relationship between the frequency, the wavelength and the speed of propagation. It is possible to find any one of these parameters given the other two.

 

If:      c=velocity of propagation

          l=wavelength

          n=frequency

 

          c=n * l       find velocity given frequency and wavelength

          l=c/n          find wavelength given velocity and frequency

          n=c/l          find frequency given velocity and wavelength

 

Important: All three quantities must be expressed in a consistent set of units.

In general the easiest solution is to express c in meters per second (m/s), l in meters (m) and n in Hertz. Other combinations are possible, but can lead to confusion. Remember to convert any metric prefix to the equivalent power of ten during the calculation and convert it back afterwards (if necessary).

 

Example:  If the velocity of light is 300,000 km/s, find the frequency of a wave whose wavelength is 575 nanometers (nm).

 

Remove metric prefixes:  575 nm = 575 x 10-9 m

                                                300,000 km/s= 300,000 x 103 m/s

 

Then:           n=c/l= (300,000 x 103)/(575 x 10-9) = 521.74 x 1012 Hz = 521.74 THz

 

 

3. The magnification of an optical system is the ratio of the size of the image to the size of  the object that produced it. If Li and Lo are the size of the image and the object, respectively, then the magnification is

 

M=Li/Lo

 

The magnification is the ratio of two lengths and has no units. The magnification for a pinhole camera is given by the ratio of the distances of the image and the object from the pinhole. See Topic 9. The pinhole camera for details.

 

4. The ratio of the velocity of light in vacuum to the velocity in any medium is the index of refraction. If c is the velocity of light in vacuum and v is the velocity in the medium, then the index of refraction, n, is given by

 

n= c/v

 

or

 

v= c/n

 

The index of refraction is a ratio of two velocities and has no units. The index of the atmosphere is about 1.0003 and the index of glass is about 1.5.

 

 

 

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