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Light-path length difference

by Foxhound101
Tags: distance, light
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Foxhound101
#1
Sep29-08, 03:25 PM
P: 52
1. The problem statement, all variables and given/known data
Two narrow slits are 0.12 mm apart. Light of wavelength 550 nm illuminates the slits, causing an interference pattern on a screen 1.0 m away. Light from each slit travels to the m=1 maximum on the right side of the central maximum.

Part A -
How much farther did the light from the left slit travel than the light from the right slit?
Express your answer using two significant figures.
2. Relevant equations
r=dsin(theta)
(theta)m = m*(lambda/d)
y=L*tan(theta)

ym = (m*lambda*L)/d

3. The attempt at a solution

I don't understand how to do these problems...

thetam = (m*lambda*)/d
thetam = (1*(5.5*10^-7m)/(1m)
thetam = 5.5*10^-7


path length difference = dsin(theta)
so...
r = d*sin(theta)
r = 1m *sin(5.5*10^-7)
r = 9.599^-9m
r = 9.6nm
That doesn't appear to be the correct answer(Unless MasteringPhysics is wrong). Sadly, I don't know if I did the right steps or used the correct equations.

Any help is appreciated.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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alphysicist
#2
Sep29-08, 07:27 PM
HW Helper
P: 2,249
Hi Foxhound101,


Quote Quote by Foxhound101 View Post
1. The problem statement, all variables and given/known data
Two narrow slits are 0.12 mm apart. Light of wavelength 550 nm illuminates the slits, causing an interference pattern on a screen 1.0 m away. Light from each slit travels to the m=1 maximum on the right side of the central maximum.

Part A -
How much farther did the light from the left slit travel than the light from the right slit?
Express your answer using two significant figures.
2. Relevant equations
r=dsin(theta)
(theta)m = m*(lambda/d)
y=L*tan(theta)

ym = (m*lambda*L)/d

3. The attempt at a solution

I don't understand how to do these problems...

thetam = (m*lambda*)/d
thetam = (1*(5.5*10^-7m)/(1m)
thetam = 5.5*10^-7
Remember that this is really:

[tex]
\sin\theta=\frac{m\lambda}{d}
[/tex]

The approximation you are using ([itex]\theta=\frac{m\lambda}{d}[/itex]) is fine since the angle is small enough, but remember that this approximation is true if the angle is measured in radians. So the angle you found is [itex]5.5\times 10^{-7}\mbox{ rad}[/itex].



path length difference = dsin(theta)
so...
r = d*sin(theta)
r = 1m *sin(5.5*10^-7)
r = 9.599^-9m
This number was calculated with the angle measure set to degrees, not radians.


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