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

[DottZakapa]

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.

Relevant 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]

Homework Statement: In a double slit experiment let d=5.00 D=30.0λ. Estimate...

d is 5.00 what? Meters, inches, light years, ...

 

[DottZakapa]

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]

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 $\beta$.

 

[DottZakapa]

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

 

[DottZakapa]

no one knows?

 

[TSny]

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 $\cos^2(\delta/2)$.

 

[DottZakapa]

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