Diffraction grating sum, 1 mark mcq question (a level)

[mutineer123]

Homework Statement

Im having a bit trouble with diffraction gratings lately, with sums like this:
A parallel beam of light of wavelength 450 nm falls normally on a diffraction grating which has
300 lines /mm.
What is the total number of transmitted maxima?
A 7
B 8
C 14
D 15

Homework Equations

dsin∅=nλ

The Attempt at a Solution

Well, my answer has no reasoning, so you might as well not read it:
I took d as 3.33X10^-6, lambda as 450 X 10^-9, and ∅as 90(to be honest I don't really know why I took it as 90!)


Can anyone redirect me to a good diffraction grating video/website if they know of any? I really neeed to understand how to plug in units into the formulae.

 

[Emilyjoint]

Your numbers are correct. Put them in the equation to get n

 

[mutineer123]

Your numbers are correct. Put them in the equation to get n

I did I am getting 7, but its wrong(can u also tell me why we take sin 90?)

 

[gneill]

I did I am getting 7, but its wrong(can u also tell me why we take sin 90?)

sin(90°) = 1 is the largest real value that sin(θ) can have. In other words 90° is the limit for possible diffraction angle magnitude. Rearrange the diffraction grating formula and you have

$sin(\theta) = n\frac{\lambda}{d}$

Clearly n can only take values that result in the R.H.S. being less than or equal to one.

Note that the system is symmetric about the center line from the diffraction grating to the projected image. That means maxima will occur on both sides of the center line with the same angular spacing (and there is also a central maximum on the center line).

 

[Emilyjoint]

I also got 7 (7.4 which must be
Deleted something
Sorry. I was going to give an answer

 

[Emilyjoint]

some more ! n is number of orders so 7.4 must be rounded down to 7 (can't have 0.4 of an order)

 

[mutineer123]

some more ! n is number of orders so 7.4 must be rounded down to 7 (can't have 0.4 of an order)

The answers not 7, and it embarrassing for me, but i forgot the right answer(its def not 7 because i got that, which was not right). But i think the guy above you may be right when he talked about symmetry, it could be 15 i think. because 7X2=14 + 1 for the central maxima

 

[mutineer123]

sin(90°) = 1 is the largest real value that sin(θ) can have. In other words 90° is the limit for possible diffraction angle magnitude. Rearrange the diffraction grating formula and you have

$sin(\theta) = n\frac{\lambda}{d}$

Clearly n can only take values that result in the R.H.S. being less than or equal to one.

Note that the system is symmetric about the center line from the diffraction grating to the projected image. That means maxima will occur on both sides of the center line with the same angular spacing (and there is also a central maximum on the center line).

Thnks

 

[mutineer123]

sin(90°) = 1 is the largest real value that sin(θ) can have. In other words 90° is the limit for possible diffraction angle magnitude. Rearrange the diffraction grating formula and you have

$sin(\theta) = n\frac{\lambda}{d}$

Clearly n can only take values that result in the R.H.S. being less than or equal to one.

Note that the system is symmetric about the center line from the diffraction grating to the projected image. That means maxima will occur on both sides of the center line with the same angular spacing (and there is also a central maximum on the center line).

gneill, can you help me in two other diffraction sums i am having problems with?



https://www.physicsforums.com/showthread.php?p=3855002&posted=1#post3855002


https://www.physicsforums.com/showthread.php?p=3855015#post3855015