Asked by Leiah_

Hi! I need help with this question...
A double slit experiment, using light of wavelength 600 nm, results in fringes being produced on a screen. The fringe separation is found to be 1.0 mm.

When the distance between the double slits and the viewing screen is increased by 2.0 m, the fringe separation increases to 3.0 mm. What is the separation of the double slits producing the fringes?
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Supposedly the answer is 0.6 mm, but I've been getting 1.2 mm. Where did I go wrong?

x=(wavelength*distance)/fringe separation

Eq 1 => 1*10^-3 = 6*10^-9 (D) /A
Eq 2 => 3*10^-3 = 6*10^-9 (D+2)/A

Help is much appreciated!!! :D
Answered by Leiah_
Oh my gosh! I forgot nano is 10^-7. Oops! :C
Answered by bobpursley
600 nm= 600E-9=.6E-9 meters=6E-7meters

Is that your error?

Answered by bobpursley
nano is -9....
Answered by Leiah_
Hello Bob! Thanks for aswering :)
Oh my gosh, nano is ^-9, what am I saying haha, so confused :C

I think I've been using 6E-9 but I just can't seem to get it...
Answered by bobpursley
using light of wavelength 600 nm, " IS NOT 6E-9, it is 600 E-9, or 6E-7
Answered by Leiah_
OH! Ok, I redid my working using 6E-7 this time :)

Somehow, I managed to get a 6 in my answer haha, but not 0.6 mm.

Eq 1 -> 3*10^-3 =6*10^-7(D+2)
Eq 2 -> 1*10^-3 =6*10^-7(D)

A=6*10^-4D
A=2*10^-4(D+2)

6*10^-4D=2*10^-4(D+2)
D=1.99997
A=2*10^-4(-1.9997+2)
=6*10^-9 m
=6*10^-6 mm

Sorry for the messy working :C
Still can't get it!!
Answered by bobpursley
UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
General Certificate of Education
Advanced Subsidiary Level and Advanced Level
PHYSICS
9702/12
Paper 1 Multiple Choice
October/November 2010
1 hour
Additional Materials: Multiple Choice Answer Sheet
Soft clean eraser
Soft pencil (type B or HB is recommended)
Answered by Leiah_
I took a look at the paper but they only give the answer and they don't show the working method :C
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