# Diffraction grating problem

1. Sep 14, 2008

### bcjochim07

1. The problem statement, all variables and given/known data
Light passes through a 200 line/mm grating and is observed on a 1.0 m wide screen located 1.0 m behind the grating. Three bright fringes are seen on both sides of the central maximum. What are the minimum and maximum possible values of wavelength?

2. Relevant equations

d= 1mm/200 lines = 5.0e-6 meters

3. The attempt at a solution
I think I got the maximum value:

The maximum value is where the m=3 bright fringe is on the edge of the screen, so the angle is arctan(.5/1) theta= 26.56 degrees. Then dsin(theta)=m*lambda

(5.0e-6)sin(26.56)=3* lambda lambda= 7.45e-7 meters= 745 nm

But for the minimum, I can't figure out what kind of relationship to set up.

2. Sep 14, 2008

### bcjochim07

Does this have something to do with the fact the bands are SEEN on the screen, so the range of visible wavelengths being 380 nm- 750 nm, my guess would be that the minimum wavelength would be 380 nm.

3. Sep 15, 2008

### alphysicist

Hi bcjochim07,

I do not believe that is right. Instead, think of the fact that you cannot see the fourth fringe on the screen. What limit does that put on the wavelength?

4. Sep 15, 2008

### bcjochim07

Is my maximum correct then? Ok, so the fourth fringe will appear on the edge of the screen when:

(5e-6)sin(26.56)=(4)*lambda lambda= 5.59e-7 m. So the wavelength has to be greater than 559 nm.

5. Sep 15, 2008

### alphysicist

That looks right to me.

6. Oct 10, 2008

### polarized hopes

That depends on your source. If youve used a white light source then minimum wavelength should be blue and longest one should be red.

good luck

7. Oct 10, 2008

### alphysicist

Hi polarized hopes,

In this problem the light source has a single frequency, and they are asking for the maximum and minimum values of that frequency that allows only three fringes to appear on the screen.

8. Oct 10, 2008

### polarized hopes

a diffraction grating resolves the spectral and temporal elements of an incident em wave. since no em wave is actually including single frequency, each fringe includes the wavelength distribution information of incident wave. i believe those fringes that you have observed are not the frequency components, they are first, second and third order diffracted light. (m=1, 2, 3)

9. Oct 10, 2008

### alphysicist

The assumption in these problems is that we have a light source that is as monochromatic as possible, so that for purposes of locating fringes we can calculate the positions based on that single frequency.

So in this problem, the question is if there is only three orders visible on a screen that is 1 meter wide (and a set distance away), what are the limits on that frequency? One limit corresponds to having the third order be at the edge of the screen (and just visible), and the other limit corresponds to having the fourth order be at the edge of the screen (as close as possible to the edge without being visible).

10. Oct 10, 2008

### polarized hopes

lets say 3rd order diffracted zone begins y=1m apart from y=0, where you observe m=0, and ends y=1.2m apart from y=0. Which tells us the width of the fringe is 0.2 m. Distance between screen and grating is 1m. so lets calculate

dsin(tetha)=m lambda
(5E-6)(1.2/sqrt(1^2+1.2^2))=3 lambda_max
(5E-6)(1/sqrt(1^2+1^2))=3 lambda_min
lambda_max=1.28 um
lambda_min=1.18 um

in short i mean the departure from the origin is important that gives the wavelength information out of the grating equation. lambda is a function of sin(tetha).

11. Oct 10, 2008

### alphysicist

The width of the fringe is unimportant here; all that matters is its location. (Also, in your denominator on the left hand side of the equation it should be sqrt(1^2+(1/2)^2) since the total length of the screen is 1m, so from center to edge is 0.5m.)

Let me use the answers you gave in your first post to show what's wrong. In your first post you mentioned that the limiting wavelengths were blue and red light; let's assume that we use the textbook values for those wavelengths: 380nm and 750nm.

For a wavelength of 750nm, the third fringe occurs past the edge of the screen, so it does not have three orders visible on the screen, so 750nm is not an answer for the maximum wavelength. (The real answer, 745nm is close, though.)

For the wavelength of 380nm, if you calculate how many orders are visible, you will find that there are more than three orders of 380nm light visible on the screen, so that is not an answer for the minimum wavelength. The real answer (559nm) corresponds to having the fourth order be right at the edge, and so we could say that is the limit for having three order visible. (If the wavelength were any lower, there would not be exactly three orders visible on the screen.)

So the answers are 559nm and 745nm; any wavelength less than this will have more than three fringes visible on the screen, and any wavelength greater will have fewer than three fringes visible on the screen.

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