DOC PREVIEW
WVU PHYS 102 - Applications of Electromagnetic Waves
Type Lecture Note
Pages 3

This preview shows page 1 out of 3 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

PHYS 101 1st Edition Lecture 26 Outline of Last Lecture I. Electromagnetic Waves Outline of Current Lecture II. Electromagnetic Waves continued…III. Deriving Refractive Index IV. Practice ProblemsCurrent LectureII. Electromagnetic Waves continued…A. When light (or EM-wave) enters a material, its speed changes, its wavelength also changes, but its frequency does not change. III. Deriving Refractive Index N = c/v ; where n is refractive index, c is the speed of light in the vacuum and v is the speed of light in the material. Prove that n = in vac/air/materialSo 1n: in vac/air = c/f  c = (in vac/air )(f)In the material max = v/f  v= (max)(f)n= c/v = (in vac/air f)/ (max f) The f’s cancel, leaving n = in vac/air / max.IV. Practice Problems #11. A laser beam is incident from air at an angle of 30 degrees to the vertical onto a solution of corn syrup in water. If the beam is reflected to 19.24 degrees to the vertical… A) n2= ? These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.n1sin1 = n2sin2 ; where n1 is air and n2 is syrup 1 sin (30) = n2 sin (19.24)  n2 = 1.52If in vac/air = 632.8 X 10-9 m B) syrup = ? n = in vac/air / syrup1.52 = 632.8 X 10-9 / syrup syrup = 416 X 10-9 mC) frequency in syrup but we who came to class are smart & know that fsyrup= fvac/airSo, let us calculate fvac/air vac/air = c/f 632.8 X 10-9 m = 3 X 108 / f  f = 4.74 X 1014 Hz D) V in syrup N = C/V 1.52 = 3 X 108 / V  V = 1.97 X 108 m/s ~2 X 108 m/s compare to C (3 X 108 m/s) #11. n1sin1 = n2sin2 ; where n1 is air and n2 is water. Remember that n1= 1 and n2=1.333 (given on exam). 1 sin1 = 1.333 sin(45)sin1 = 1.333/21 = sin-1(1.333/2)1 = 70.5 degrees (from vertical) from the horizontal, = 90-70.5 = 19.5


View Full Document

WVU PHYS 102 - Applications of Electromagnetic Waves

Type: Lecture Note
Pages: 3
Download Applications of Electromagnetic Waves
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Applications of Electromagnetic Waves and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Applications of Electromagnetic Waves 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?