Help!! Single slit diffraction

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.

six789

anyone HELP me please..

xanthym

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.


~~

six789

im so confused... wat is the unknown here?? is it L or Y?

xanthym

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).


~~

six789

and the value for W is 6.9x10^-5?

xanthym

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}.


~~

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..

xanthym

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.


~~

six789

i just have to multiply my answers to 2 or ...??

six789

yeah i got teh right answer...
thanks so much man..

six789

but why do u have to do that .. i mean multiply it to 2??

xanthym

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.


~~

six789

oh now i see.. thanks again man..

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??

xanthym

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)}.


~~

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??