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Double slit: ratio of intensity of 3rd- and 0th-order maxima

  • Thread starter DottZakapa
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65
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Homework Statement
In a double slit experiment let d=5.00 D=30.0λ. Estimate the ratio of the intensity of the third order maximum with that of the zero-order maximum.
Homework Equations
interference diffraction
Homework Statement: In a double slit experiment let d=5.00 D=30.0λ. Estimate the ratio of the intensity of the third order maximum with that of the zero-order maximum.
Homework Equations: interference diffraction

i guess the goal is this equation

##I_{(\theta)}=I_0 \times(cos^2\beta)\times \left ( \frac {sin\alpha} \alpha \right)^2##

then i do

## D\sin \theta = 3\lambda##

##\sin\theta= \frac {3\lambda} D, \space \theta=5.74^0##

##\beta= \frac {\pi d} \lambda \sin \theta##

##\alpha = \frac {\pi D} \lambda \sin \theta \space##

substituting the data

##\alpha=\frac {\pi 30.0\lambda} \lambda \frac {3\lambda} {30.0\lambda}\space##
next

##\beta= \frac {\pi 5.00} \lambda \frac {3\lambda} {30.0\lambda}## i don't know how to solve this one, and solve the rest of the problem, how do i get rid of ##\space\lambda\space## at denominator?

any help please?
 

kuruman

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Homework Statement: In a double slit experiment let d=5.00 D=30.0λ. Estimate...
d is 5.00 what? Meters, inches, light years, ....
 
65
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d is 5.00 what? Meters, inches, light years, ....
That is what the text says. There is no other data informato. But i guess that d is the distance between the two slits
And D is the slits width.
 

kuruman

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That is what the text says. There is no other data informato. But i guess that d is the distance between the two slits
And D is the slits width.
I'm sure you are right about the meaning of d and D. Nevertheless, units are needed for d. Note that if you can express d as a multiple of the wavelength λ, then you will be able to get rid of the extra λ in the denominator in the expression for ##\beta##.
 
65
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I'm sure you are right about the meaning of d and D. Nevertheless, units are needed for d. Note that if you can express d as a multiple of the wavelength λ, then you will be able to get rid of the extra λ in the denominator in the expression for ##\beta##.
this is the solution but... i don't understand from where those result come from
Screen Shot 2019-08-30 at 21.53.10.png
 
65
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no one knows?
 

TSny

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Looks like the solution is using D = 6λ and d = 30λ. (These are different values from what you gave in the statement of the problem.) Also, it appears to me that there is a mistake in the formula for ##I## given in the solution. I think the factor of ##\sin^2(\delta/2)## should be ##\cos^2(\delta/2)##.
 
65
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I think the factor of ##\sin^2(\delta/2)## should be ##\cos^2(\delta/2)##.
you right, unfortunately i did not have the chance to meet the teacher and ask
 

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