Visible light is incident perpendicularly on a diffraction grating of 200 rulings/mm. What are the (a) longest, (b) second longest, and (c) third longest wavelengths that can be associ-ated with an intensity maximum at theta=30.0 degrees?

Question

anonymous · Answer

well we know that the if we use the equation dsin(theta)=m(lamda) that d = 5.0*10^-6

anonymous · Answer

How did you get 5.0*10^-6?

anonymous · Answer

It says that its 200 rulling (or lines) per millimeter and that isnt slit spacing, so in order for us to find the slit spacing we need to do 1/200

anonymous · Answer

which would give the slit spacing in millimeters, and my answer is in meters for the slit spacing

anonymous · Answer

Then what should be the value of m in part (a), (b), (c)?

anonymous · Answer

I'm still trying to figure that out right now since i know of 2 different equations to use for a diffraction grating and one of them requires the length from the grating to the screen, thus i dont thing i can use that one

anonymous · Answer

Okay. The answers from the book are 625 nm, 500 nm, 416 nm for parts (a), (b) and (c).

anonymous · Answer

Hm....lets see if i can backtrack to see how they got it then