Tripling the Distance between Bright Fringes

  • Thread starter jst
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    Fringes
But be careful, it's asking for the distance y, not the width of the fringes. So you would need to double your answer to get the full distance from m=-2 to m=+2.
  • #1
jst
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Homework Statement




A board has two small slits separated by a small distance d. A light ray of a certain wavelength is shone through the holes. A screen is placed in front of the holes and bright fringes are displayed on it. If the light wavelength is 440 nm, the distance between the slits is 4000 nm and the distance between the slits and screen is one meter then how wide would the screen have to be so that the third order fringes just hit its left and right sides?

possible answers
128 cm
14.2 cm
17.5 cm
66.0 cm
38.3 cm

Homework Equations



m*LAMBDA = d * y/L

The Attempt at a Solution



The equation that wrote is the one that I think that I should use, but I am stuck.

Homework Statement



A board has two small slits separated by a small distance d. A light ray of a certain wavelength is shone through the holes. A screen is placed in front of the holes and bright fringes are displayed on it. Which of these actions would triple the distance between the zero order and first order bright fringes on the screen?

possible answers
triple the distance between the slits
triple the distance between the slits and screen
double the distance between the slits
double the distance between the slits and screen
double the wavelength of the light

Homework Equations





The Attempt at a Solution



I could figure this one out if I knew the equation used
 
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  • #2
You have the right equation. Why are you stuck?
 
  • #3
In the first question I posted, the part that says: "how wide would the screen have to be so that the third order fringes just hit its left and right sides?"

wavelength = 440nm
m = 2
d = 4000nm
L = 1 m

solve for y?
 
  • #4
Yes, solve for y. But you want m=3, don't you?
 
  • #5
Yep...my mistake :D

On my 2nd question...does it use the same equation? If so, "triple the distance" is that referring to d or L?
 
  • #6
I think you would use the same equation. In the question, triple the distance refers to y, the distance on the screen between the zero order and first order bright fringes.
 
  • #7
Thanks a lot...all your help is appreciated...this is what happens when you take physics as an online course :D
 
  • #8
You're welcome. Good luck with your course. :smile:
 
  • #9
One problem, when I do:

3 * 440 = 4000* (y/1000000000)

y = 3.3E8 nm
= 33cm

Which doesn't match any of the possible solutions, sort of close to one of them, but not close enough to where I would go ahead and select it.

Anything stand out?
 
  • #10
The question asks for the width for the third order fringes, so technically that would be from m=-3 to m=+3. (double your answer, basically)
 
  • #11
:D I didn't see that..thanks!

On the second problem, I'm thinking: "triple the distance between the slits"

My reason is:

m*LAMBDA = d * y/L

m*LAMBDA = d * 3y/L

y = (1/3) * (LAMBDA * m * L)/d

on the right track?
 
  • #12
I have one more similar problem:

A board has two small slits separated by a small distance d. A light ray of a certain wavelength is shone through the holes. A screen is placed in front of the holes and bright fringes are displayed on it. If the light wavelength is 656 nm, the distance between the slits is 4000 nm and the distance between the slits and screen is one meter then what is the distance y in the diagram?

http://glomawr.com/mc006-1.jpg

possible answers

5.20 cm
7.01 cm
32.8 cm
15.7 cm
24.3 cm

-------------------------

I want to put 32.8cm

because of

2*656 = 4000 * (y/1000000000)
y = 32.8cm

but that doesn't seem correct, because of the diagram. Is y (in the diagram) half of a distance or a distance doubled?
 
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  • #17
For some reason my last post posted several times...not sure...I did hit refresh on the forum, but not the posting.
 
  • #18
jst said:
:D I didn't see that..thanks!

On the second problem, I'm thinking: "triple the distance between the slits"

My reason is:

m*LAMBDA = d * y/L

m*LAMBDA = d * 3y/L

y = (1/3) * (LAMBDA * m * L)/d

on the right track?

It wants to know what will triple the distance. What you did makes it one third.
 
  • #19
jst said:
I have one more similar problem:

A board has two small slits separated by a small distance d. A light ray of a certain wavelength is shone through the holes. A screen is placed in front of the holes and bright fringes are displayed on it. If the light wavelength is 656 nm, the distance between the slits is 4000 nm and the distance between the slits and screen is one meter then what is the distance y in the diagram?

http://glomawr.com/mc006-1.jpg

possible answers

5.20 cm
7.01 cm
32.8 cm
15.7 cm
24.3 cm

-------------------------

I want to put 32.8cm

because of

2*656 = 4000 * (y/1000000000)
y = 32.8cm

but that doesn't seem correct, because of the diagram. Is y (in the diagram) half of a distance or a distance doubled?

I interpret y in the diagram as from the zero order (centre) to m=2. So I would say your answer is correct.
 
Last edited by a moderator:

FAQ: Tripling the Distance between Bright Fringes

What does "tripling the distance between bright fringes" mean?

Tripling the distance between bright fringes refers to increasing the distance between adjacent bright fringes in a diffraction pattern by a factor of three. This can be achieved by changing the distance between the light source and the diffraction grating or by changing the wavelength of the incident light.

Why would someone want to triple the distance between bright fringes?

Tripling the distance between bright fringes allows for more precise measurement of the diffraction pattern and can reveal more detailed information about the diffracting object or medium. It can also help reduce overlap between fringes and make it easier to distinguish individual fringes.

How is the distance between bright fringes measured?

The distance between bright fringes is typically measured using a ruler or a caliper. The distance is measured between the centers of two adjacent bright fringes, and this value is then used in the equation dλ = mλ, where d is the distance between fringes, λ is the wavelength of the incident light, and m is the order of the bright fringe.

What factors can affect the distance between bright fringes?

The distance between bright fringes can be affected by the wavelength of the incident light, the distance between the light source and the diffraction grating, and the spacing of the diffraction grating itself. Other factors such as the angle of incidence, the refractive index of the medium, and the size of the diffracting object can also play a role.

What are some practical applications of tripling the distance between bright fringes?

Tripling the distance between bright fringes is commonly used in diffraction experiments to obtain more accurate and detailed data. It is also used in fields such as astronomy, where precise measurement of the diffraction pattern of light from distant stars and galaxies can provide valuable information about their composition and structure.

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