Register to reply 
Phase difference b/w airwater interface and waterair interface 
Share this thread: 
#1
Apr410, 07:05 AM

P: 67

1. The problem statement, all variables and given/known data
Light of wavelength 418 nm is incident normally on a film of water 1.0 μm thick. The index of refraction of water is 1.33. (a) What is the wavelength of the light in the water? (b) How many wavelengths are contained in the distance 2t, where t is the thickness of the film? (Do not round your answer to a whole number.) (c) What is the phase difference between the wave reflected from the top of the airwater interface and the one reflected from the bottom of the waterair interface in the region where the two reflected waves superpose? 2. Relevant equations 3. The attempt at a solution my attemp at a solution: (a)lambda"water" = lambda/n = 445nm (b)N = 2t/ lambda"water" = 4.5 (c) Light is going from a medium of smaller index of refraction (air n = 1) to a medium with higher one (water n = 1.33) so the reflected wave is out of phase by π respect to the transmitted wave right? So the phase diff = π + 2π(2t/lambda"water") = π +2π(4.5) = ??? My answers i have tried: 25.1 rad, 3.14rad, 31.4 rad, 1 rad, 4.72 rad, 0 rad 31.42rad 1 rad Im really stumped...pls help!!!!!!!!!!!!!!!!!!!!!!! 


#3
Apr410, 07:43 AM

P: 67

This is from my HW: Correct Problem Statement: Light of wavelength 592 nm is incident normally on a film of water 1.0 µm thick. The index of refraction of water is 1.33. then part b and c follow 


#4
Apr410, 08:03 AM

Mentor
P: 41,325

Phase difference b/w airwater interface and waterair interface
What book is this from? 


#5
Apr410, 08:05 AM

Mentor
P: 41,325




#6
Apr410, 08:24 AM

P: 67

this is web based homework. The problem is from tipler 6th ed ch32 prob22 


#7
Apr410, 08:39 AM

Mentor
P: 41,325




#8
Apr410, 09:13 AM

P: 67

π +2π(4.49) = 8.89π subtracting 7π leaves 1.98π = 6.22 radians 


#9
Apr410, 09:13 AM

P: 67

which is correct!



#11
Apr410, 09:46 AM

P: 67

Thanks for your help DOC as usual you come through! 


#12
Apr2811, 04:37 PM

P: 3

Quick question: So CMurda got n + 2n(4.5) because 4.5 is the value of the 2t/lambda(water), and 2n because it travels through water first, but where did he get the 7n that he uses in his calculations? And which n is air and which is water?



#13
Apr2811, 04:38 PM

P: 3




#14
Apr2811, 05:03 PM

Mentor
P: 41,325

The phase difference is 2π(4.49)  π = 7.98π. When you express that mod 2π, you end up subtracting 6π, for a phase difference of 1.98π. (His post #8 was not quite clear and contained a few typos.) 


#15
Apr2811, 06:49 PM

P: 3

Thank you very much  makes WAY more sense now. Regards.



#16
Feb2012, 01:02 AM

P: 2

I don't understand what "Expressing mod 2pi" means. I'm really having a hard time understanding why you subtract that 6pi.



#17
Feb2012, 05:50 AM

Mentor
P: 41,325




#18
Feb2012, 01:12 PM

P: 2

RIGHT RIGHT. Oh my goodness I'm so dumb haha. Thanks for clearing that up haha.



Register to reply 
Related Discussions  
Air oil water interface  Introductory Physics Homework  1  
Why does an IP addreess need to be associated with an interface?  Programming & Computer Science  14  
Interface dielectrics  Introductory Physics Homework  0  
Program interface  Computing & Technology  19  
Building an interface for gtk  Computing & Technology  0 