Index of refraction, angular seperation

[Bradracer18]

I can't seem to figure these out guys...maybe I'm missing a few equations too.


1. If light of wavelength 600nm in air enters a medium with index of refraction 1.50, its wavelength in this new medium is...
A. unchanged
B. 900nm
C. 400nm
D. 602nm
---I know that n=c/v...but I don't know how to relate that to find wavelength.

2. A 200m radio telescope is used to investigate sources emitting a 21cm wavelength. The minimum angular separation resolvable for this system is...
A. 0.154deg
B. 0.073deg
C. 0.0026deg
D. 0.0013deg

I have no clue on this one...we haven't even covered this stuff, and he tested us on it...trying to figure out how to do it, before the final(this week). Again, any help is great!

Thanks again,
Brad

 

[sdekivit]

1) n = v1/v2 = lambda1 / lambda 2

 

[Bradracer18]

sorry, your image didn't load...

 

[Andrew Mason]

I can't seem to figure these out guys...maybe I'm missing a few equations too.


1. If light of wavelength 600nm in air enters a medium with index of refraction 1.50, its wavelength in this new medium is...
A. unchanged
B. 900nm
C. 400nm
D. 602nm
---I know that n=c/v...but I don't know how to relate that to find wavelength.

What is the speed of the light wave in the medium? What is the relationship between wavelength and speed for a wave of a particular frequency?

2. A 200m radio telescope is used to investigate sources emitting a 21cm wavelength. The minimum angular separation resolvable for this system is...
A. 0.154deg
B. 0.073deg
C. 0.0026deg
D. 0.0013deg

I have no clue on this one...we haven't even covered this stuff, and he tested us on it...trying to figure out how to do it, before the final(this week).

This is essentially a single slit diffraction problem: What is the angular separation of the first minimum for a 200 m wide slit? This works out to: \sin\theta = \lambda/D where D is the diamater of the telescope. It should be the same for radio or optical telescopes. For small angles, \sin\theta \approx \theta

AM

 

[Bradracer18]

Ok, so for #1...I still can't get it figured out...might need a little bit more help. V for the light in the medium is 2x10^8m/s. Using v=c/n. Then the relationship you asked for, is it v=wavelength x frequency. If so, I don't understand how to use that to get my answer.

for #2...so basically, I take .21m/200m...which = .00105 I'm not sure why that answer isn't up there...did I do it right?

 

[Andrew Mason]

Ok, so for #1...I still can't get it figured out...might need a little bit more help. V for the light in the medium is 2x10^8m/s. Using v=c/n. Then the relationship you asked for, is it v=wavelength x frequency. If so, I don't understand how to use that to get my answer.

Does the frequency change? So how does wavelength vary with speed?

for #2...so basically, I take .21m/200m...which = .00105 I'm not sure why that answer isn't up there...did I do it right?

They may be using the Raleigh diffraction formula which is sin\theta = 1.22\lambda/D

Also remember: you must put your answer in degrees. Your answer is in radians (or sin\theta).

AM

 

[Bradracer18]

ok andrew...first let me say thanks...

so for #1, I ended up taking frequency=speed of light/my v(2x10^8)...which = 3/2. I took 3/2 x 600nm = 900nm...answer B.

And for #2...I did the math over(and used your new formula)...and came up with answer B.

Are these correct??

Thanks again,
Brad

 

[Andrew Mason]

ok andrew...first let me say thanks...

so for #1, I ended up taking frequency=speed of light/my v(2x10^8)...which = 3/2. I took 3/2 x 600nm = 900nm...answer B.

And for #2...I did the math over(and used your new formula)...and came up with answer B.

Why is v = 3/2? The speed of light in air is 3 x10^8 m/sec. So the speed in the medium is 2/3 of the speed through air.

#2 looks right.

AM