Double Slit Experiment without the small angle approximation

[Punkyc7]

Two very narrow slits are spaced 1.8 \mum apart and are placed 35.0 cm from the screen. What is the distance from the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with \lambda =550 nm

r1-r2=m\lambda
dsin\theta=m\lambda
Rtan\theta=y

Since I am only concerned with the distance i was thinking about just finding the distance between the bright fringes. The problem is I don't know how to get theta.

Ive tried m\lambda=dsin\theta and solved for theta but that didnt work. I am not sure you can find r1 and r2 because the distance between the fringes is missing.

 

[mathfeel]

Two very narrow slits are spaced 1.8 \mum apart and are placed 35.0 cm from the screen. What is the distance from the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with \lambda =550 nm

r1-r2=m\lambda
dsin\theta=m\lambda
Rtan\theta=y

Since I am only concerned with the distance i was thinking about just finding the distance between the bright fringes. The problem is I don't know how to get theta.

Ive tried m\lambda=dsin\theta and solved for theta but that didnt work. I am not sure you can find r1 and r2 because the distance between the fringes is missing.

You can use \theta = \arctan \frac{y}{R}.

 

[Punkyc7]

The problem is we don't know y that is what I am trying to find, but I am going to need\theta first

 

[SammyS]

Don't the "Dark Lines" correspond to destructive interference?

 

[ehild]

SammyS is right, the problem asks about dark fringes. For them, the angle of deviation from the central line is related with d and λ as

d sin(\vartheta)=(2m+1)λ, with m=0, ±1, ±2 ...

y is the distance of a dark spot from the centre of the interference pattern on the screen. If the screen is at D distance from the two slits, tan(\vartheta)=y/D.
You need the distance between the first and second dark lines. For the first one, m=0, and m=1 for the second one.

Calculate \vartheta-s with m=0 and m=1, calculate the corresponding y values and find their difference.

ehild

 

[Punkyc7]

isnt the distance between the dark fringes the same as the distance between the bright fringes?

so i would do

dsin\theta=\lambda
and dsin\theta=3\lambda

solve for \theta in terms of y and then solve for y

 

[SammyS]

The title says "... without the small angle approximation"

λ/d ≈ 3.5

I wouldn't assume the lines in the diffraction pattern are equally spaced.

 

[ehild]

isnt the distance between the dark fringes the same as the distance between the bright fringes?

so i would do

dsin\theta=\lambda
and dsin\theta=3\lambda

solve for \theta in terms of y and then solve for y

The values sin\theta are equally spaced, but the position of the fringes are proportional to tan\theta.
I am sorry, there is a mistake in my last post, I meant

dsin\theta=(2m+1)\lambda/2

So calculate the distances from
dsin\theta=\lambda/2
and dsin\theta=3\lambda/2.

ehild

 

[Punkyc7]

Just to clear something up, If we were using the approximation then the bright and dark fringes would be the same length.

Thank you so much for your help

 

[ehild]

Yes, the approximate method yields equal distances but the approximation can be applied for small angles (less than 10°).

ehild