# Problem with wave optics - diffraction grating?

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1. Feb 22, 2015

### winsticknova

1. The problem statement, all variables and given/known data
For your science fair project, you need to design a diffraction grating that will disperse the visible spectrum (400-700nm) over 30 degrees in first order.
How many lines per mm does your grating need?

2. Relevant equations
sin(Θ) = mλ/d
y = Ltan(Θ)

I'm pretty sure the only equation I need is the first one.

3. The attempt at a solution
I'm not really sure where to start. I tried plugging in 400 nm and 700 nm in the first equation with m =1 and Θ=30°. But I don't get the right answer with either. If I use 400 nm, I get 1250 lines/mm which is close to the answer but not exact. The answer is 1230 lines/mm but I don't know how they get that answer.

2. Feb 23, 2015

### BvU

Hello Win, welcome to PF

As usual, you start with reading the question carefully, and trying to find out what it is they want from you.
From your attempt at a solution, I see that what you do is to have the first maximum for 400 nm light at 30 degrees.
That is not what is asked for. I think you misinterpret the word 'disperse' (english: spread -- I think)
If you look at the top picture here you see two peaks with m=1. Any idea what angle the exercise wants to be 30 degrees for the line density you have to calculate ?

3. Mar 1, 2015

### orsanyuksek2013

The grating formula is

m λ = d (sin β + sin α)

where

m: Diffraction order

λ: Wavelength of the incoming light

d: Grating constant

β: Angle between the incoming light and the grating normal

α: Angle between the outgoing light (dispersed light) and the grating normal (exit angle)

In the question above β = 0° and m = 1. So

λ = d sin 30°

and the d is asked.

Note that, for each wavelength the exit angle is different.

Therefore you have to consider the wavelength, 550 nm

(400 nm + 700 nm) / 2

as a reference wavelength. In this case d = 1,1 x 10^(-6) m.

That is, the grating must have

1 / d = 909 l / mm.

----------------------------
Örsan Yüksek

4. Mar 1, 2015

### BvU

Dear Örsan,

That's not what the problem statement says. I think you also misinterpret the word 'disperse' in

"disperse the visible spectrum (400-700nm) over 30 degrees in first order"​

The value for d the exercise wants as an answer is the d that causes a difference of 30 degrees between the first order maximum for 400 nm light and the first order maximum for 700 nm light. In other words$$400 \; {\rm nm} = 1\;{10^{-3} \; {\rm m}\over N}\;\sin\phi\\ 700 \; {\rm nm} = 1\;{10^{-3} \;{\rm m}\over N}\;\sin(\phi+{\pi\over 6})$$

5. Mar 8, 2015

### orsanyuksek2013

Hi BvU,

than for the clarification.

In this case:

400 = d sin a

700 = d sin b

b - a = 30°

4 sin a
--- = --------------------
7 sin b

4 sin b = 7 sin a

a = 29.5°
b= 59.5°

400 = d sin 29.5°

1 / d = 1230 l / mm

----------------------------
Örsan Yüksek

6. Mar 9, 2015

### BvU

Well done ! -- But now Win hasn't had the opportunity to learn and gather experience from this exercise. It IS against PF rules to give away a full solution, but I don't think that's a big problem ( in view of the fact that the first post is two weeks old.

The difficult part in this exercise was "reading the question carefully, and trying to find out what it is they want"