Interference of Light help?

Question

Light passing through a double slit with separation d=3.33*10^-5 created its first minimum (m=1) at an angle of .551 deg. What is the wavelength of the light in nanometers?

 

I thought I was supposed to use dsinθ=(m-1/2)λ but I keep getting .00003486 and it says the answer is between 400 and 700. I'm not sure where I messed up, but I think it could be putting sin there? My teacher told me it would be sin but I've seen tan being used and I don't know when I'm supposed to use either of them.

Answer 1

For very small θ,

 

sinθ ≅ tanθ

 

In this problem, θ = 0.551°, which is very small, so the approximation may be used.

 

sin(0.551°) ≅ 9.6166 x 10^-3

tan(0.551°) ≅ 9.6171 x 10^-3

 

You can see they are close, so it does not matter which one you use, as long as the angle is very small.  However, if you look at the diagram for the setup of 2-slit diffraction in your physics book, you should see that it really is tanθ.

 

The correct formula is

 

dsinθ=(m + 1/2)λ, for m = 0, ±1, ±2, ...

 

To get to the first dark fringe, you can work with m + 1/2 where m = 0, or m - 1/2 where m = 1.  They will give the same answer.  I prefer to start in the middle (m = 0) and work outward.

 

λ = dsinθ / (m + 1/2) = (3.33 x 10^-5 m) sin(0.551º) / (0 + 1/2) ≅ 6.405 x 10^-7 m = 640.5 nm

 

How did you get 0.00003486 ?