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Help!! Single slit diffraction |
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| Apr27-05, 05:31 PM | #1 |
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Help!! Single slit diffraction
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. |
| Apr27-05, 07:54 PM | #2 |
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anyone HELP me please..
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| Apr27-05, 08:14 PM | #3 |
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[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. ~~ |
| Apr27-05, 08:18 PM | #4 |
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Help!! Single slit diffraction
im so confused... wat is the unknown here?? is it L or Y?
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| Apr27-05, 08:21 PM | #5 |
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| Apr27-05, 08:23 PM | #6 |
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and the value for W is 6.9x10^-5?
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| Apr27-05, 08:27 PM | #7 |
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| Apr27-05, 08:29 PM | #8 |
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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..
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| Apr27-05, 08:42 PM | #9 |
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[tex] 2: \ \ \ \ \mbox{Problem Requirement} \ \, = \ \, \color{red}2Y_{1}(Red) \ \, - \ \, 2Y_{1}(Violet) [/tex] Recalculate your results using above Eq #2. ~~ |
| Apr27-05, 08:54 PM | #10 |
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i just have to multiply my answers to 2 or ...??
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| Apr27-05, 08:56 PM | #11 |
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yeah i got teh right answer...
thanks so much man.. |
| Apr27-05, 08:57 PM | #12 |
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but why do u have to do that .. i mean multiply it to 2??
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| Apr27-05, 08:58 PM | #13 |
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| Apr27-05, 09:00 PM | #14 |
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oh now i see.. thanks again man..
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| Apr27-05, 09:14 PM | #15 |
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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?? |
| Apr27-05, 10:03 PM | #16 |
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[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)}. ~~ |
| Apr28-05, 03:22 PM | #17 |
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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??
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