Optics hw

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Some homework q's i don't quite understand. Any help is appreciated.

A double slit arrangement produces fringes on a distant screen. Using the light from a sodium spectral lamp (lamda = 589 nm) the fringes have an angular separation of 0.015 degrees.
a) What is the separation of the slits?
b) What would the new angular separation be if
i) the slit separation were doubled?
ii) the distance to the screen were doubled?
iii) the entire arrangement were immersed in water (n=1.33)?
iv) the sodium lamp were replaced by an argon ion laser (lamda = 488 nm)?
 

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  • #2
siddharth
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How have you attempted to solve this question?

You need to show some work before you get help.
 
  • #3
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i'm not sure if i'm doing it correctly, thus the reason why i posted.
let L = lamda so i don't have to figure out how to type lamda on this.

part a) d = mL/sin @
assume m = 1?
L=589*10^-9
d=2.25*10^-3 metres (using calc in degree mode)
d=2.25mm

bi)double the seperation means d = 4.50mm
@ = arcsin (mL/d) (keeping calc in degree mode)
@ = 0.0075 degrees

bii)i think this has something to do with the equation y = R mL/d but am unsure. (where y is the distance of the fringes on the screen and R is the distance from the slit to the screen.)

biii)putting the apparatus in water shouldn't affect it i don't think.

biv)use @ = arcsin (mL/d) where m = 1, L = 488*10^-9 metres and d 2.25*10^-3 metres gives @ = 0.012 degrees
 
  • #4
siddharth
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i'm not sure if i'm doing it correctly, thus the reason why i posted.
let L = lamda so i don't have to figure out how to type lamda on this.

part a) d = mL/sin @
assume m = 1?
L=589*10^-9
d=2.25*10^-3 metres (using calc in degree mode)
d=2.25mm
Do you realize why to assume m=1? The value of theta is the angular separation for m=1 (ie, the first maximum)

bi)double the seperation means d = 4.50mm
@ = arcsin (mL/d) (keeping calc in degree mode)
@ = 0.0075 degrees
Didn't check the numbers, but your idea is right. (With m=1)

bii)i think this has something to do with the equation y = R mL/d but am unsure. (where y is the distance of the fringes on the screen and R is the distance from the slit to the screen.)
If the distance to the screen is doubled, would theta change? Hint: [tex] \tan \theta = y/R[/tex]

biii)putting the apparatus in water shouldn't affect it i don't think.
No, not right. What property of light, relevant to interference, changes when it travels in water?

biv)use @ = arcsin (mL/d) where m = 1, L = 488*10^-9 metres and d 2.25*10^-3 metres gives @ = 0.012 degrees
Yes.
 
Last edited:
  • #5
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Hey thanks for your help. just some things i want to verify.

If the distance to the screen is doubled, would theta change? Hint: [tex] \tan \theta = y/R[/tex]
QUOTE]
i'm assuming no. because
tan @ = y/R
but y = RmL/d
so tan @ = [RmL/d]/R = mL/d which is independant of R.

and

No, not right. What property of light, relevant to interference, changes when it travels in water?
QUOTE]
(new L)= L/n where (new L) has been reduced so the angle will be smaller.

again thank you so much for helping me with this :)
 

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