# Double Slit Experiment without the small angle approximation

1. Jul 23, 2011

### Punkyc7

Two very narrow slits are spaced 1.8 $\mu$m 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 Im only concerned with the distance i was thinking about just finding the distance between the bright fringes. The problem is I dont know how to get theta.

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

2. Jul 23, 2011

### mathfeel

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

3. Jul 23, 2011

### Punkyc7

The problem is we dont know y that is what im trying to find, but im going to need$\theta$ first

4. Jul 23, 2011

### SammyS

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

5. Jul 24, 2011

### 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

6. Jul 24, 2011

### 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

7. Jul 24, 2011

### SammyS

Staff Emeritus
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.

8. Jul 24, 2011

### ehild

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

9. Jul 24, 2011

### 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

10. Jul 24, 2011

### ehild

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

ehild