Question on Diffraction Grating

[tigerguy]

Hi - I'm trying a problem on diffraction grating, and I keep on getting stuck. Maybe someone can help me figure out the last step:

Three, and only three, bright fringes can be seen on either side of the central maximum when a grating is illuminated with light ( wavelegnth = 490 nm). What is the maximum number of lines/cm for the grating?

Basically, what I've done is that I'm using the equation d sin(theta) = m lambda / 2
I figure that because its constructive itnerefrence for 3 fringes, m=3, and the wavelength will equal the aforementioned number. I also know that N= 1/d. So, I'm trying to figure out the value of sin(), but I'm not really sure what theta would equal to. How would I figure that out?

Is my reasoning correct, or am I missing something critical to solving this problem? Thanks so much.

 

[Doc Al]

Basically, what I've done is that I'm using the equation d sin(theta) = m lambda / 2

Where did that 2 come from?

So, I'm trying to figure out the value of sin(), but I'm not really sure what theta would equal to.

Hint: What's the maximum value that theta can be?

 

[tigerguy]

Oh, the 2 shouldn't be there, because its constructive interference, not destructive. Does the m=3 make sense, too?

I'm not sure what the maximum value can be - would it be 90 degrees (pi/2), because that's the maximum value that sin can be?

 

[Doc Al]

Oh, the 2 shouldn't be there, because its constructive interference, not destructive.

Right.

Does the m=3 make sense, too?

Makes perfect sense to me.

I'm not sure what the maximum value can be - would it be 90 degrees (pi/2), because that's the maximum value that sin can be?

Not because that angle gives the max value for sin (which it does), but because the greatest displacement from the central maximum will be at 90 degrees. (Any more than that and the light would have to go backwards.)

 

[tigerguy]

Great, I understand it now. Thanks so much!