# Finding the angle of phase difference - two slit model

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1. Mar 17, 2016

### vetgirl1990

1. The problem statement, all variables and given/known data
Light of a wavelength 548nm illuminates two slits separated by 0.25mm. At what angle would one find the phase difference between the waves from two slits to be 2 rads?

2. Relevant equations
σ / λ = ΔΦ / 2π

σ: path difference
λ: wavelength
Δφ: phase difference

3. The attempt at a solution
Based on geometry of a two slit model, σ = dsinθ.
I thought this would be a simple substitution problem, but I'm not getting the correct answer (Answer = 0.04°), so I'm worried that I'm overlooking something.

σ / λ = ΔΦ / 2π
(dsinθ) / λ = ΔΦ / 2π
sinθ = (ΔΦ / 2π)(λ / d)
θ = sin-1(ΔΦ / 2π)(λ / d)

θ = 0.000697°

I also tried making the small angle approximation (sinθ ≈ θ) , but still didn't get the right answer: θ = 0.0000122°

2. Mar 17, 2016

### blue_leaf77

The unit of the above value should be in radians, not degree.

3. Mar 17, 2016

### vetgirl1990

The final answer is reported in degrees (0.04°)

4. Mar 17, 2016

### blue_leaf77

Yes, I know, but the value of 0.000697 for $\theta$ should be in radians? Then convert it to degrees.
What's the value you got when calculating θ = sin-1(ΔΦ / 2π)(λ / d)?

5. Mar 17, 2016

### vetgirl1990

θ = sin-1(ΔΦ / 2π)(λ / d)
= sin-1 (2rads / 2π)(5.48x10-7m / 2.5x10-4m)
= sin-1 (0.31831 * 0.002192)

0.03998 rads * 180/π = 2.29

The answer is still wrong even when I leave it as radians until the end.

6. Mar 17, 2016

### blue_leaf77

Check again the output of your calculator if it is in radians or degrees. If it's in degrees already, then you are getting the correct answer after rounding up to two figures behind the decimal.

7. Mar 17, 2016

### vetgirl1990

Ah I see! sin-1 is automatically calculating my answer in degrees already. Thank you for your help and patience!

8. Mar 17, 2016

### blue_leaf77

You are welcome.