Calculating Distance Between Double Slits Using the Double-Slit Equation

[pinkyjoshi65]

[SOLVED] Double slit equations

In an experiment, blue light with a wavelength of 645 nm is shone through a double-slit and lands on a screen that is located 1.35 m from the slits. If the distance from the centre maximum to the 8th order bright fringe is 2.6 cm, calculate the distance between the two slits.

Here is what I did:
wavelength=6.45*10^-7 m
L=1.35 m
m=8
X=0.026m

I used X/L = m*wavelength/d to find d.
Is this right?

 

[pinkyjoshi65]

got another question:

Two slits are separated by a distance of 2.00 10-5 m. They are illuminated by light of wavelength 5.60 10-7 m. If the distance from the slits to the screen is 6.00 m, what is the separation between the central bright fringe and the third dark fringe?

Have no idea to solve it since it involves both a bright and a dark fringe..

 

[pinkyjoshi65]

Anyone??

 

[pinkyjoshi65]

i would really appreciate it if someone could help me with this..

 

[rl.bhat]

I used X/L = m*wavelength/d to find d. This formula is correct.
The position of the dark fringe is given by X/L = (2m + 1)*wavelength/2d .
m = 0 gives the first dark fringe

 

[pinkyjoshi65]

ok but I am asked to find the distane between the middle bright fringe and the 3rd dark fringe.

 

[pinkyjoshi65]

Oh wait, so you are saying I sub in m=2, and the given values to find X?

 

[rl.bhat]

All the distances are measured from the central bright fringe. You want to find the distance of third dark fringe. So put m = 3

 

[pinkyjoshi65]

umm..ok..got it..!..thnkz..got another question..this one is on single slit phenonmenon:
Light of wavelength 625 nm shines through a single slit of width 0.32 mm and forms a diffraction pattern on a flat screen located 8.0 m away. Determine the distance between the middle of the central bright fringe and the first dark fringe?

So what i did was using wavelength/width= X/L, i found X. this is the distance from one dark to another dark fringe (on the other side of the middle light fringe) then i divided tht by 2. tht right?

 

[rl.bhat]

So what i did was using wavelength/width= X/L, i found X. this is the distance of the first dark fringe from the central bright fringe.

 

[pinkyjoshi65]

um..ok..so this is the dist. of the 1st dark fringe to the dark fringe after the central bright fringe yea?

 

[pinkyjoshi65]

i am still unclear on how to find the distance from the 1st dark fringe to the MIDDLE of the central bright fringe..

 

[rl.bhat]

The distance of the first dark fringe is half the width of the central bright fringe.

 

[pinkyjoshi65]

so my answer is write yea..?..i mean the method..find findind the distance, and then divinding it by 2 gives the right answer..

 

[rl.bhat]

No need to divid by 2. X itself is the distance of the first dark fringe from the central maximum.

 

[pinkyjoshi65]

Oh..ok..thankz..:)..One more question..Light of wavelength 600 nm is incident upon a single slit with width 4.0 × 10-4 m. The figure shows the pattern observed on a screen positioned 2.0 m from the slits.Determine the distance s. see attached pic..
So here we have the wavelenght, w and L. Now since both the bright and the dark fringe is invloved do we have to cobine their equations??
like: (m+0.5)wave/w= X/L...Dont know how to proceed with this problem..

 

[rl.bhat]

I can't see the picture. Any way, the distance of the nth secondary maximum from the mid point of the central maximum = (2n + 1)lambda*L/2d and the distance of the nth secondary minimum from the mid point of the central maximum = n*lambda*L/d

 

[pinkyjoshi65]

umm..but in the pic..we are supposed to find the distance between the middle of the central max...to the middle of the 2nd dark fringe from the central max..

 

[rl.bhat]

All the fringes are of finite width. In the case of diffraction the fringe widths are not same. As you move away from the central bright fringe the width decreases. So always you have to take the reading from center to center of any fringe.

 

[pinkyjoshi65]

so then, for solving this problem we would use X=Lnlamda/w. I have attached the pic again for you to see.

 

[pinkyjoshi65]

And here are solutions to some of the questions. Could you please just check my solutions.
1)Calculate the angle that will form between the centre maximum and the third-order dark fringe if the wavelength of the light that strikes two slits is 650 nm and the distance between the slits is 2.2 10-6 m.
1. n=3, λ=6.5* m, d=2.2* m
(n-0.5)λ= d*Sinθ
Sinθ= (n-0.5)*λ/d
= (3-0.5)6.5* /2.2*
= 2.5*6.5* /2.2
= 7.3863*0.1= 0.73863
Sinθ= 0.73863
θ= (0.73863)= 47.61 degrees
2)In an experiment, blue light with a wavelength of 645 nm is shone through a double-slit and lands on a screen that is located 1.35 m from the slits. If the distance from the centre maximum to the 8th order bright fringe is 2.6 cm, calculate the distance between the two slits.

2. m=8, λ=6.45* m, L =1.35m, X=0.026m
X/L=mλ/d
d=Lmλ/X
d= 1.35*6.45* *8/0.026
=2.68* m
3)In an experiment, the distance from one slit to the third dark fringe is found to be 2.200 046 8 m. If the wavelength of the light being shone through the two slits is 590 nm, calculate the distance from the second slit to this same dark fringe.
3. n=3, PS1= 2.2000468m,λ=5.9* m
(n-0.5)λ=lPS1-PS2l
+/- (3-0.5)5.9* = 2.2000468-PS2
+/- 14.75* = 2.2000468-PS2
PS2= 2.2000468-14.75* OR 2.2000468+14.75*
PS2= 2.200045325m or 2.200048275m

4)Two slits are separated by a distance of 2.00 10-5 m. They are illuminated by light of wavelength 5.60 10-7 m. If the distance from the slits to the screen is 6.00 m, what is the separation between the central bright fringe and the third dark fringe?

4. X/L= (2n+1)λ/2d
X= (2n+1)λL/2d= 42*5.60*10-6/4*10-5
= 58.8*10-2= 0.60m

 

[pinkyjoshi65]

anyone?

 

[rl.bhat]

650 nm = 650*10^-9 = 6.5*10-7m
Sinθ= 0.073863
Recheck all the problems

 

[pinkyjoshi65]

i did change nm to m..i don't know y its not showing that in the messege. I have sent a file to you as a private msg. Those are the same solutions, i pasted in here. Take a look at them please. And for the previous question (with the pic), we have to use X=Lnlamda/w yes?

 

[pinkyjoshi65]

I am unable to send it to you: here is the file. Plese check it.

 

[pinkyjoshi65]

Light of wavelength 625 nm shines through a single slit of width 0.32 mm and forms a diffraction pattern on a flat screen located 8.0 m away. Determine the distance between the middle of the central bright fringe and the first dark fringe

λ=6.25*〖10〗^(-7)m, w=0.32*〖10〗^(-3)m, L=8.0m
λ/w=X/L
X=Lλ/w= 8*6.25*〖10〗^(-7)/0.32*〖10〗^(-3)
= 156.25*10000
1.56*〖10〗^6m

Light of 600.0 nm is incident on a single slit of width 6.5 mm. The resulting diffraction pattern is observed on a nearby screen and has a central maximum of width 3.5 m. What is the distance between the screen and the slit?

λ=6.00*〖10〗^(-7)m, w=6.5*〖10〗^(-3)m, 2X=3.5;X=1.75m
λ=wX/L
L=wX/λ= 6.5*1.75*〖10〗^(-3)/6*〖10〗^(-7)= 11.375*〖10〗^4/6 =2.00*〖10〗^4m

A monochromatic beam of microwaves with a wavelength of 0.052 m is directed at a rectangular opening of width 0.35 m. The resulting diffraction pattern is measured along a wall 8.0 m from the opening. What is the distance between the first- and second-order dark fringes?

. λ=0.052m, w=0.35m, L=8m
λ/w=X/L
X= Lλ/w= 8*0.052/0.35
= 1.18 m
Light from a red laser passes through a single slit to form a diffraction pattern. If the width of the slit is increased by a factor of two, what happens to the width of the central maximum?

λ1=wy1/L, y1= Lλ/w
λ2=2wy/L, y2= Lλ/2w
y2/y1= λ2L*w/2w*Lλ1
y2/y1= 0.5
y2=0.5y1

And for the previous question (with the pic), we have to use X=Lnlamda/w yes?

 

[pinkyjoshi65]

please could someone go through this?

 

[pinkyjoshi65]

Hello..Anyone?

 

[pinkyjoshi65]

could someone please check my solutions..?

 

[pinkyjoshi65]

Hello..??

 

[pinkyjoshi65]

Anyone, please someone just look through my solutions. Its urgent..

 

[rl.bhat]

X/L= (2n+1)λ/2d
X= (2n+1)λL/2d= 42*5.60*10-6/4*10-5
= 58.8*10-2= 0.60m
For third dark fringe you have to use (2n - 1).

λ=6.25*〖10〗^(-7)m, w=0.32*〖10〗^(-3)m, L=8.0m
λ/w=X/L
X=Lλ/w= 8*6.25*〖10〗^(-7)/0.32*〖10〗^(-3)
= 156.25*10000
1.56*〖10〗^6mCheck this calculation.

 

[pinkyjoshi65]

X/L= (2n+1)λ/2d
X= (2n+1)λL/2d= 42*5.60*10-6/4*10-5
= 58.8*10-2= 0.60m
For third dark fringe you have to use (2n - 1).

This is for the 4th question with the picture? And the next half is for the 4th question (double slit equation)?

 

[rl.bhat]

λ=6.25*〖10〗^(-7)m, w=0.32*〖10〗^(-3)m, L=8.0m
λ/w=X/L
X=Lλ/w= 8*6.25*〖10〗^(-7)/0.32*〖10〗^(-3)
= 156.25*10000
1.56*〖10〗^6mCheck this calculation.. The answer is 1.56*10^-2m.

 

[pinkyjoshi65]

X/L= (2n+1)λ/2d
X= (2n+1)λL/2d= 42*5.60*10-6/4*10-5
= 58.8*10-2= 0.60m
For third dark fringe you have to use (2n - 1)

Which one is this for..:S..?

 

[pinkyjoshi65]

Hello..??

 

[pinkyjoshi65]

ok so I used 2n+1 and got 0.60m, then i use (2n-1), and whteve value i get, i add 0.60 to it?

 

[Doc Al]

single slit diffraction

Oh..ok..thankz..:)..One more question..Light of wavelength 600 nm is incident upon a single slit with width 4.0 × 10-4 m. The figure shows the pattern observed on a screen positioned 2.0 m from the slits.Determine the distance s. see attached pic..
So here we have the wavelenght, w and L. Now since both the bright and the dark fringe is invloved do we have to cobine their equations??
like: (m+0.5)wave/w= X/L...Dont know how to proceed with this problem..

You just need to apply the condition for single slit diffraction. There's a simple formula to find the location of any dark fringe from the central maximum:
y = \frac{m\lambda D}{w}

where D is the distance to the screen and w is the width of the slit.

 

[pinkyjoshi65]

um..ok..and how did you derive this equation? Also could you also go through the solutions in posts 21 and 26..

 

[Doc Al]

um..ok..and how did you derive this equation?

Read up on diffraction here: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html#c1"

Also could you also go through the solutions in posts 21 and 26..

Unfortunately, I don't have much time right now. (I'll get to them as soon as I can.)

 

[pinkyjoshi65]

Ok, thnkz for your help..really appreciate it. It would be nice if I could an answer by tomorrow..thnkz again..:)

 

[pinkyjoshi65]

Hi, sorry got a question about the equation you gave me. Since we are supposed to find the distance between the centre of the max fringe to the DARK fringe, (m=2), so arent we supposed to use this equation:
(m+0.5)wavelength= wX/L, and find X?

 

[Doc Al]

Hi, sorry got a question about the equation you gave me. Since we are supposed to find the distance between the centre of the max fringe to the DARK fringe, (m=2), so arent we supposed to use this equation:
(m+0.5)wavelength= wX/L, and find X?

Don't confuse that equation (which is for the double slit pattern) with the formula that I gave you for single slit diffraction. The formulas are different.

 

[pinkyjoshi65]

umm..but that equation is for single slit equations (constructive) The ones for the double slit equations are (n-0.5)wave= dX/L (destuctive)..

 

[Doc Al]

umm..but that equation is for single slit equations (constructive)

I guess you could use it for that, but it's a bit bogus. There's no simple equation for finding the bright spots of a single slit pattern. There's an equation for finding the dark spots, which is the one I gave. You can estimate the positions of the bright spots by saying they are in the middle between two dark spots.

Note that you're being asked to find the position of a dark fringe, which is destructive interference.

 

[pinkyjoshi65]

However in the image I have attached, it seems like they are asking us to find the distance from the centre to the CENTRE of the 2nd dark fringe..How can I use X=m*wave*L/w to find the distance till the CENTRE od the dark fringe?

 

[Doc Al]

However in the image I have attached, it seems like they are asking us to find the distance from the centre to the CENTRE of the 2nd dark fringe..How can I use X=m*wave*L/w to find the distance till the CENTRE od the dark fringe?

I'm not sure what else I can tell you: I gave you the formula to use; I gave you a link that discusses how the formula is derived; I told you why the other formula that you wanted to use is incorrect.

Don't get distracted by talk of the center of the fringe--whether it's a dark or a bright fringe, those simple formulas always give the position of the center of the fringe. (Don't confuse this with the position of the central maximum, which is at the center of every fringe pattern.)

 

[pinkyjoshi65]

umm..ok..so i'll use X=m*wavelength*L/w..Thankz a lot..Will be waiting for your reply on the other posts.

 

[pinkyjoshi65]

Sorry one more question, m is equal to 2 yes?

 

[pinkyjoshi65]

um..