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Help Single slit diffraction

by six789
Tags: diffraction, single, slit
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six789
#1
Apr27-05, 05:31 PM
P: 127
heya all... this is the problem..
Predict whether violet light(lambda=404nm) or red light (lambda=702nm) will have a wider central maximum when used to generate a single-slit diffraction pattern. Calculate the difference if the light is incident on a 6.9x10^-5 wide slit falling onot a screen 85cm away..

i know that the red light will have a wider central maximum, but i dont know the unknown, is it Ym from the formula Ym=(m*lambda*L)/W? and i dont know the value for m, is it 5 since the gap between the red light and the violet light is 5... im not sure... and the value of red light in my book is 7.3x10^-3m... HELP me please...im soo confused..

THANKS SO MUCH in advance for the big help.
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six789
#2
Apr27-05, 07:54 PM
P: 127
anyone HELP me please..
xanthym
#3
Apr27-05, 08:14 PM
Sci Advisor
P: 412
Quote Quote by six789
heya all... this is the problem..
Predict whether violet light(lambda=404nm) or red light (lambda=702nm) will have a wider central maximum when used to generate a single-slit diffraction pattern. Calculate the difference if the light is incident on a 6.9x10^-5 wide slit falling onot a screen 85cm away..

i know that the red light will have a wider central maximum, but i dont know the unknown, is it Ym from the formula Ym=(m*lambda*L)/W? and i dont know the value for m, is it 5 since the gap between the red light and the violet light is 5... im not sure... and the value of red light in my book is 7.3x10^-3m... HELP me please...im soo confused..

THANKS SO MUCH in advance for the big help.
For single-slit diffraction, the first MINIMUM will occur at (m=1) of your formula:

[tex] 1: \ \ \ \ Y_{1} \ \, = \ \, \frac{\color{red}\mathbf{(1)}\color{black}\lambda L}{W} [/tex]

where "L" is the distance to screen, and "W" the slit width. The "length" of the Central Maximum measured from the first minimum below the Central Peak to the first minimum above the Central Peak is (2*Y1). The above formula applies to each different "λ" you wish to evaluate.


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six789
#4
Apr27-05, 08:18 PM
P: 127
Help Single slit diffraction

im so confused... wat is the unknown here?? is it L or Y?
xanthym
#5
Apr27-05, 08:21 PM
Sci Advisor
P: 412
Quote Quote by six789
im so confused... wat is the unknown here?? is it L or Y?
The unknown is "Y". You know the value of "L", which is the distance from the slit to the screen, given in the problem to be (L = 0.85 meters).


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six789
#6
Apr27-05, 08:23 PM
P: 127
and the value for W is 6.9x10^-5?
xanthym
#7
Apr27-05, 08:27 PM
Sci Advisor
P: 412
Quote Quote by six789
and the value for W is 6.9x10^-5?
You did not include units for the above value. Assuming the units to be "meters", then yes you're correct that {W = 6.9x10^(-5) meters}.


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six789
#8
Apr27-05, 08:29 PM
P: 127
yes, thats what i did.. but why is my answer incorrect to the answer in the book... it should be 7.3x10^-3m for the red light..
xanthym
#9
Apr27-05, 08:42 PM
Sci Advisor
P: 412
Quote Quote by six789
yes, thats what i did.. but why is my answer incorrect to the answer in the book... it should be 7.3x10^-3m for the red light..
According to your msg, problem requires calculation of the DIFFERENCE in Central Maxima "lengths" for Red & Violet light:

[tex] 2: \ \ \ \ \mbox{Problem Requirement} \ \, = \ \, \color{red}2Y_{1}(Red) \ \, - \ \, 2Y_{1}(Violet) [/tex]

Recalculate your results using above Eq #2.


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six789
#10
Apr27-05, 08:54 PM
P: 127
i just have to multiply my answers to 2 or ...??
six789
#11
Apr27-05, 08:56 PM
P: 127
yeah i got teh right answer...
thanks so much man..
six789
#12
Apr27-05, 08:57 PM
P: 127
but why do u have to do that .. i mean multiply it to 2??
xanthym
#13
Apr27-05, 08:58 PM
Sci Advisor
P: 412
Quote Quote by six789
i just have to multiply my answers to 2 or ...??
Use Eq #2 given in Msg #9. The value for each Y1 must be multiplied by 2 to obtain the FULL "length" of the Central Maximum. The Central Maximum "length" must be calculated for BOTH red and violet light, and then placed into Eq #2 given in Msg #9.


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six789
#14
Apr27-05, 09:00 PM
P: 127
oh now i see.. thanks again man..
six789
#15
Apr27-05, 09:14 PM
P: 127
how about for this problem...
Commercial satellites are able to resolve objects separated by only 1.0m. If those satellites orbit the earth at an altitude of 650km, determine the sixe of the satellites' circular imaging aperture. Use 455nm light for the light in the lenses of the satellites..
can u help me again... i dont know waht formula to use here??
xanthym
#16
Apr27-05, 10:03 PM
Sci Advisor
P: 412
Quote Quote by six789
how about for this problem...
Commercial satellites are able to resolve objects separated by only 1.0m. If those satellites orbit the earth at an altitude of 650km, determine the sixe of the satellites' circular imaging aperture. Use 455nm light for the light in the lenses of the satellites..
can u help me again... i dont know waht formula to use here??
The formula for circular aperture diffraction is different from that for single-slit diffraction. The minimum resolvable distance "Y" at distance "L" from a circular aperture of diameter "W" using light of wavelength "λ" is given by:

[tex] 3: \ \ \ \ Y \ \, = \ \, \frac{\color{red}\mathbf{(1.22)}\color{black}\lambda L}{W} [/tex]

Solve for aperture diameter "W" (in meters) using other variable values given by problem {Y=(1 meter), L=(650e3 meters), λ=(455e(-9) meters)}.


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six789
#17
Apr28-05, 03:22 PM
P: 127
where did u get the formula for this?? i just know the formula... tetha = 1.22(lambda)/D... can u tell me how u derived from that formula??
xanthym
#18
Apr29-05, 04:38 AM
Sci Advisor
P: 412
Quote Quote by six789
where did u get the formula for this?? i just know the formula... tetha = 1.22(lambda)/D... can u tell me how u derived from that formula??
[tex]
\setlength{\unitlength}{.001cm}
\begin{picture}(6000,5000)(500,-7000)
\color{red}
\thicklines
\linethickness{1pt}
\put(6001,-2161){\line(-1,-1){600}}
\put(5401,-2761){\line( 1,-1){600}}
\put(6001,-661){\line( 0,-1){4500}}
\put(1201,-2761){\line( 1, 0){4800}}
\put(4201,-661){\line( 0,-1){1800}}
\put(6001,-3361){\line(-3, 1){4815}}
\put(4201,-3061){\line( 0,-1){2100}}
\put(1201,-3350){A}
\put(1201,-1561){B}
\put(6151,-2950){C}
\put(6151,-2300){\color{blue} \large Triangle Represents Point A's Airy Diffraction Central Max}
\put(7500,-3300){\color{blue}$\theta \, = \, \angle CED \, = \, \angle AEB$}
\put(7500,-4400){\color{blue}$ \frac{Length(AB)}{Length(AE)} \, \approx \, \theta \, \approx \, \frac{(1.22)\lambda}{(Aperture \ Diameter)} $}
\put(6151,-3600){\mbox{D}}
\put(4351,-2611){\mbox{E}}
\put(0,-7000){\mbox{.}}
\end{picture}
[/tex]
You are certainly correct that several approximations are involved. At large distances, the approximations shown above are usually valid. The Triangle represents Point "A"s Airy Diffraction resulting from the Circular Aperture at Point "E". Point "B" can be resolved from Point "A" by the Rayleigh Criteria when Point "B"s Airy Diffraction Central MAX falls on Point "A"s First MIN. This occurs at angle "θ" shown above. The other relationships can be determined from the formulas shown.

Derivation and discussion of Airy Diffraction from a circular aperture (a special case of Fraunhofer Diffraction) can be found here:
http://scienceworld.wolfram.com/phys...rAperture.html
A brief overview discussion can be found here:
http://hyperphysics.phy-astr.gsu.edu...pt/cirapp.html
(For this latter ref, follow the indicated links for additional info.)

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