How Does Changing Variables Affect Light Diffraction Patterns?

[student07]

Homework Statement

A laser emitting light with a wavelength of 560 nm is directed at a single slit, producing an interference pattern on a screen that is 3.0 m away. The central maximum is 5.0 cm wide.
a) Determine the width of the slit and the distance between adjacent maxima.
b) What would the effect on this pattern be, if
i) the width of the slit was smaller?
ii) the screen was moved further away?
iii) a larger wavelength of light was used?
c) How would this interference pattern differ if the light was shone through a
i) double slit?
ii) diffraction grating?

Homework Equations

λ = WΔy / L
λ = 560 nm = 5.60 x 10^-7 m
L = 3 m
Δy = 5 cm = 0.05 m
W=?

The Attempt at a Solution

a) 5.60 x 10^-7 m = W(0.05 m) / 3 m
W = (5.60 x 10^-7 )(3 m) / 0.05 m = 3.36 x 10^-5 m
I'm thinking that the distance between adjacent maxima and width of slit is the same in this case.

b) i) If the width of the slit is smaller then the angle with the horizontal must increase.
ii) Ifthe screen is moved further away then the width of the slit is decreased.
iii) If a larger wavelength is used the the distance between minima and maxima is increased (not sure about this one).
c) i) On a double slit the interference pattern is much clearer.
ii) On a diffraction grating destructive interference might occur.
I need some hints about b and c please help.

 

[BvU]

Dear student, you should make clear what the variables in your relevant equations stand for. e.g in the first one: λ = WΔy / L, where Δy is the distance from the axis to the first minimum. Which is not the width of the central maximum.

I'm thinking that the distance between adjacent maxima and width of slit is the same in this case

Are you saying the distance between adjacent maxima is equal to 3.36 x 10^-5 m (or whatever you get when you have fixed your mistake) ?

b. ii) the screen is moved. The slit stays the same.

b iii) what's the reasoning ? I can't read your thoughts

c i) and why is that / Are there any other differences ?

c ii) destructive interference occurs in 1, 2, or many slits.

Study the subject. Preferably in your notes or textook. Or else at hyperphysics.

--

 

[student07]

Dear student, you should make clear what the variables in your relevant equations stand for. e.g in the first one: λ = WΔy / L, where Δy is the distance from the axis to the first minimum. Which is not the width of the central maximum.

Are you saying the distance between adjacent maxima is equal to 3.36 x 10^-5 m (or whatever you get when you have fixed your mistake) ?

b. ii) the screen is moved. The slit stays the same.

b iii) what's the reasoning ? I can't read your thoughts

c i) and why is that / Are there any other differences ?

c ii) destructive interference occurs in 1, 2, or many slits.

Study the subject. Preferably in your notes or textook. Or else at hyperphysics.

--

Yes that's what I calculated for the adjacent maxima to be 3.36 x 10^-5 m which is also 33600 nm.
Now for b) and c) I sort of guessed but no worries I'll find more info about it.

 

[student07]

then central maximum is W? and I'm looking for Δy?

 

[BvU]

No the central maximum is not W. At least not in your notation (I think: after all you set Δy = 5 cm in post #1 and W = ?). You could have avoided this kind of confusion by adhering more religiously to the template (a list of "all variables and given/known data" -- with dimensions ( which you have) and meaning/description).

Now we have to sort that out in an advanced stage of the thread -- cumbersome, but all is not lost :

You were given that the width of the central maximum is 5 cm. The first equation you mobilize (λ = WΔy / L) is the condition for Δy, the position of the first minimum on the screen wrt the axis. Which is not the width of the central maximum. (I have deja vu now...). W is the slit width, which you calculate to be 34 μm. I think that's not correct, but to fix it is easy.

My compliments for your "no worries I'll find more info about it" ! Happy to answer specific questions; the subject as a whole is too extensive to treat and explain comprehensively in a forum. It is very rich and very important in all kinds of areas in science and technology: any investment is well worth the effort.

single slit (with a lot of further links and a calculator to check your result)

intensity (showing a formula for the pattern)

I have a comment here: your exercise asks for the distance between adjacent maxima. Taken literally, that would lead to a complicated process of differentiation and solving trigonometric equations. I am convinced that what is wanted here is actually the distance between consecutive minima. (which is almost but not exactly the same. It is the same as the distance between positions where the sine = 1)​

 

[student07]

So after doing some research I found that I need the distance from the center to the first minimum which is half of the central maximum that is 2.5 cm or 0.025 m.
Then with that I need to find an angle: θ = tan^-1 = (0.025m / 3m) = 0.477 deg.
Then to find the width of the slit: d = λ / sinθ = (5.60 x 10^-7m) / (sin 0.477) = 6.73 x 10^-5 m
Now once the width is found I can use it to find the adjacent maxima:
Δy = λL / w = (5.60 x 10^-7m)(3m) / (6.73 x 10^-5 m) = 0.025 m

Now for part b and c you said I can find some answers at hyperphysics website?
Thanks for your help

 

[BvU]

Your problem statement didn't mention angles.
Usually the small angle approximation $\tan x \approx x \approx \sin x$ holds nicely (better than 1% up to 22.5 degrees)

 

[S. M. I. Wahidi]

What about part b and c?

 

[BvU]

What about part b and c?

What about it ? Do you have a specific question or comment ?

 

[Heitorthehandyman]

I am a physics student and I had to answer this question and this is what I came up with. I just figured that when answering if then statements in physics it is always key to look at equations and how our variables relate.
I am still trying to gt used to answering these questions fully ( i might have the right answer but they expect me to write more about it)
do you guys in the physics forum community think its enough??

b)i. if the width of the pattern was smaller then the distance between fringes would also be smaller since they are directly related. [PLAIN]http://file///C:/Users/Redd/AppData/Local/Temp/msohtmlclip1/01/clip_image002.png

ii. if the length between slits and screen increase then the distance between fringes would decrease since they are inversely related. [PLAIN]http://file///C:/Users/Redd/AppData/Local/Temp/msohtmlclip1/01/clip_image002.png

iii. if the wavelength of light emitting from laser increases then the distance between fringes would decrease since they are inversely related. http://file///C:/Users/Redd/AppData/Local/Temp/msohtmlclip1/01/clip_image004.png
the images that are not showing are just the following equation: lambda=w*delta y/L