Find the distance between the two third order maxima

[Spruance]

Red light with wavelength 713 nm pass through a double split with "split distance" 0.120 mm. The interference pattern gets observed on a screen that is 2.75 m from the double split. Find the distance between the two
third order maxima we seeing on the screen.


(My english is really awkward)

 

[Doc Al]

Show what you've done so far and where you are stuck. If you haven't a clue, read about "double slit interference" in your text or here: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html

 

[Spruance]

d sin 0 = n*lambda

sin 0 = ((n*lambda)/d) = (3 * 713 * 10^-9 m/ 0.120 * 10^-3 m) = 0.017825 = Sin (1.02)

tan 0 = y/r = y/2.75 m

y = tan 1.02 * 2.75 m = 2.80 m

 

[Doc Al]

y = tan 1.02 * 2.75 m = 2.80 m

Recheck that last step. (Note that for small angles, tan(theta) = sin(theta))

You have caculated the position of the 3rd order maximum from the central position (where y = 0), but the problem asks for the distance between the two 3rd order maxima. Where's the other one?

 

[Spruance]

I don't got any answer-book, but I am sure that my previous answer is wrong.

 

[Spruance]

I really don't know what I have to do ...

 

[Doc Al]

You multiplied by 1.02 instead of tan(1.02). (1.02 is degrees, by the way.)

 

[Spruance]

y = tan 1.02 * 2.75 m = 0.04896 m

It cannot be correct

 

[Hootenanny]

y = tan 1.02 * 2.75 m = 0.04896 m

It cannot be correct

I hope Doc Al doesn't mind me jumping in here, but he seems to be else where. You have calculated the distance between the normal (zero order beam) and the third order beam. The question asks you to calculate the distance between the two third order beams.

HINT: You are half way there.

~H

 

[Spruance]

Sin 0 = (n*lambda)/d = ((3 * 713 *10^-9)/(0.120 * 10^-3)) = 0.018

Sin 0.018 = 1.047

Sin 1.047 = (x/2.75)

x = Sin 1.047 * 2.75 m

x = 0.050 m

 

[Hootenanny]

Sin 0 = (n*lambda)/d = ((3 * 713 *10^-9)/(0.120 * 10^-3)) = 0.018

Sin 0.018 = 1.047

Sin 1.047 = (x/2.75)

x = Sin 1.047 * 2.75 m

x = 0.050 m

I'm not sure what your doing here, in any case it is incorrect, \sin\theta only equals zero when n = 0, i.e. the zero order beam. Alarms bells should have started ringing when you abtained sin 0 = a non zero number. You were very close with your previous answer. BIG HINT: The interference pattern is symetrical about the zero order beam (y=0).

~H

 

[Spruance]

sin theta = (n*lambda)/(d)

= (3*713 * 10^-9)/(0.120 * 10^-3)

= Sin(0.017825) = 1.02135136

tan (1.021) = (x/2.75)

x = tan 1.021 * 2.75 = 0.049

0.049 * 2 = 0.098 m

 

[nrqed]

sin theta = (n*lambda)/(d)

= (3*713 * 10^-9)/(0.120 * 10^-3)

= Sin(0.017825) = 1.02135136

?? How do you get from the first line to the second line??
Where do you get your 0.017825 or your 1.02135136??
What do you get for the first line??

 

[Spruance]

I believe that if you just put the numbers into a calculator, it gives you that (3*713 * 10^-9)/(0.120 * 10^-3) is 0.017825

Invers sinus of 0.017825 is the same as Sin(1.02)

 

[nrqed]

I believe that if you just put the numbers into a calculator, it gives you that (3*713 * 10^-9)/(0.120 * 10^-3) is 0.017825

Invers sinus of 0.017825 is the same as Sin(1.02)

Sorry.. I had used 0.1 mm. You are correct. My apologies

 

[nrqed]

sin theta = (n*lambda)/(d)

= (3*713 * 10^-9)/(0.120 * 10^-3)

= Sin(0.017825) = 1.02135136

tan (1.021) = (x/2.75)

x = tan 1.021 * 2.75 = 0.049

0.049 * 2 = 0.098 m

This is correct...

 

[Hootenanny]

y = tan 1.02 * 2.75 m = 0.04896 m

It cannot be correct

Just for reference, the same result can be obtained by simply doubling this number because (as I said earlier) the interference pattern is symmetrical around the zero order beam.

~H