580 nm light shines on a double slit with d = 0.000125 m. What is the
angle of the second bright
interference maximum (m = 2)?
(Remember, nano means 10-9.)
(Unit = deg)

0.532

Explanation:

Your equation to find the second bright interference maximum is gonna be this: d sin (Θ) = m λ

First, find your variables.

λ = 580 · 10^-9

d = 0.000125

m = 2

Next, fill in the equation.

d sin (θ) = m λ

(0.000125) sin (θ) = (2) (580·10^-9)

Then isolate your variable.

θ = arcsin ( (2)(580·10^-9) / (0.000125) )

Run your equation and you will end up with 0.53171246 , which rounds to 0.532.

The main thing you have to watch out for is make sure you are calculating for the bright interference and not the dark interference, as well as checking you're calculating for the maximum, not the minimum.

I hope this helps :D

where is the circuit diagram ?