course Phy 232
July 6 at 6:10 p.m.
*********************************************
Question: `q**** query univ phy problem 33.38 (34.28 10th edition) 3 mm plate, n = 1.5, in 3 cm separation between 450 nm source and screen. How many wavelengths are there between the source and the screen?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The distance between the source and the screen is 1.8 cm but the glass inserted between themis only .25 thick. So the wavelengths going through the glass are going to be different than those that exist in the aire between the source and the screen.
Number of wavelengths everywhere but glass:
#wavelengths = d/lambda = 1.55 cm/(540*10^-7 cm) = 28703.7 wavelengths
Number of wavelengths in glass:
#wavelengths = d/lambda,
lambda = lambda_0/n = 540/1.4 = 385.7 nm
#wavelengths = d/lambda = 0.25 cm/(385.7*10^-7 cm) = 6481.7 wavelengths
Summing the two #wavelengths together gives us the total amount of wavelengths between the source and the screen.
Total number of wavlengths = 35185
.............................................
Given Solution:
`a** The separation consists of 1.5 cm = 1.5 * 10^7 nm of air, index of refraction very close to 1, and 1.5 mm = 1.5 * 10^6 nm of glass, index of refraction 1.5.
The wavelength in the glass is 450 nm / 1.5 = 300 nm, approx..
So there are 1.5 * 10^7 nm / (450 nm/wavelength) = 3.3 * 10^4 wavelengths in the air and 1.5 * 10^6 nm / (300 nm/wavelength) = 5.0 * 10^3 wavelengths in the glass. **
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
For some reason my original wavelength and distance between the source and the screen are different than yours. However, I think we both solved the problem through the same process.
I agree. Good work.