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
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!!
Oh my gosh! I forgot nano is 10^-7. Oops! :C
600 nm= 600E-9=.6E-9 meters=6E-7meters
Is that your error?
nano is -9....
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...
using light of wavelength 600 nm, " IS NOT 6E-9, it is 600 E-9, or 6E-7
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)
I took a look at the paper but they only give the answer and they don't show the working method :C
To solve this problem, we need to use the equation for the fringe separation in a double-slit experiment:
λ = (wavelength * distance) / fringe separation
where:
λ is the wavelength of light,
distance is the distance between the double slits and the screen, and
fringe separation is the distance between adjacent bright fringes.
Let's start by using the given information for the first scenario (Eq 1):
λ = 600 nm = 6 * 10^-7 m
fringe separation = 1.0 mm = 1.0 * 10^-3 m
Plugging in these values into Eq 1, we get:
6 * 10^-7 = (6 * 10^-9 * D) / A
Now let's move on to the second scenario (Eq 2):
distance = distance in Eq 1 + 2.0 m = A + 2.0 m
fringe separation = 3.0 mm = 3.0 * 10^-3 m
Plugging in these values into Eq 2, we get:
6 * 10^-7 = (6 * 10^-9 * (D + 2.0)) / (A + 2.0)
Now, we have two equations (Eq 1 and Eq 2) with two unknowns (D and A). We can use these equations to solve for D, the separation of the double slits producing the fringes.
Let's rearrange Eq 1 to solve for A:
A = (6 * 10^-9 * D) / (6 * 10^-7)
Now, substitute this expression for A in Eq 2 and solve for D:
6 * 10^-7 = (6 * 10^-9 * (D + 2.0)) / ((6 * 10^-9 * D) / (6 * 10^-7) + 2.0)
Now simplify this equation by multiplying both sides by ((6 * 10^-9 * D) / (6 * 10^-7) + 2.0):
6 * 10^-7 * ((6 * 10^-9 * D) / (6 * 10^-7) + 2.0) = 6 * 10^-9 * (D + 2.0)
Simplifying further, we have:
(6 * 10^-9 * D) + 12 * 10^-7 = 6 * 10^-9 * D + 12 * 10^-9
The D terms on the left side will cancel, leaving:
12 * 10^-7 = 12 * 10^-9
Now, divide both sides by 12 * 10^-9:
1.0 = D
Therefore, the separation of the double slits producing the fringes is 1.0 mm or 1.0 * 10^-3 m, not 0.6 mm.
It appears there may have been a mistake in the given answer of 0.6 mm. Your calculation of 1.2 mm is the correct result.