Wave Optics

jst

1. The problem statement, all variables and given/known data


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

2. Relevant equations

m*LAMBDA = d * y/L

3. The attempt at a solution

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

1. The problem statement, all variables and given/known data

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

2. Relevant equations



3. The attempt at a solution

I could figure this one out if I knew the equation used
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

hage567

You have the right equation. Why are you stuck?

jst

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?

hage567

Yes, solve for y. But you want m=3, don't you?

jst

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?

hage567

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.

jst

Thanks a lot...all your help is appreciated....this is what happens when you take physics as an online course :D

hage567

You're welcome. Good luck with your course.

jst

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?

hage567

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)

jst

: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?

jst

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?



possible answers

5.20 cm
7.01 cm
32.8 cm
15.7 cm
24.3 cm

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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?

jst

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jst

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jst

For some reason my last post posted several times....not sure...I did hit refresh on the forum, but not the posting.