# Diffraction grating that gives max intensity

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1. Aug 19, 2015

### toforfiltum

1. The problem statement, all variables and given/known data

2. Relevant equations
dsinθ=nλ
IαA2

3. The attempt at a solution
I chose D because I thought that D gives the max intensity since the amplitudes superimpose over a smaller area. But the answer is C. I can't see why.

2. Aug 19, 2015

### soarce

From diffracting grating one gets 1st order beam (undeflected), 2nd order deflected beam, 3rd order deflected beam and so on. This would rule out the answer A and D. The intensity decerases with the intensity order, in the picture we have no information about the beam intensities represented by arrows, however we have information about the deflection angle, what's the difference between C and D?

3. Aug 19, 2015

### toforfiltum

Why is D ruled out? It has a 1st order beam unlike A.

4. Aug 19, 2015

### soarce

I have made a mistake in my previous post: the undeflected beam should be 0th order.

I was thinking that the answer D limits the number of diffraction orders to one.

LE: Think on the diffraction orders and their diffraction angles, use the formula you have written.

Last edited: Aug 19, 2015
5. Aug 22, 2015

### toforfiltum

I thought that the intensity of the waves falls off as the order increases?

6. Aug 24, 2015

### soarce

The arrows shows the direction of diffracted beams, from the picture I can't distinguish any difference in arrows width (if any). I assume that each arrow stands for a diffraction order.

Did you reach any conclusion examining the diffraction angles of B, C and D figures ?

7. Aug 24, 2015

### toforfiltum

For C and D, angle for 1st order in C is larger than in D. The only logical thing I can think of is that that beam of light from C is directed away from the centre, which means centre is less bright. I just don't understand the answer.

8. Aug 24, 2015

### soarce

From $d\sin\theta_n = n \lambda$ one can calculate the diffraction angles for each order, n=0,1,2,3 etc Use some particular ratio $\lambda/d$, e.g. 1, 0.5, 0.3, 0.25, to see how the diffraction angles relates to the diffractoin orders. You can write your results here.

9. Aug 24, 2015

### toforfiltum

Oh I see why now. If the order is only until 1, then the angle of light would be at 90°, which is impossible to reach screen. Am I right?

10. Aug 24, 2015

### soarce

Right! The ratio $\lambda/d$ gives a cut-off in the diffraction orders. Now compare $\theta_n$ and $\theta_{n+1}$, where $n=0,1,2,3...$

11. Aug 24, 2015

### toforfiltum

The angles from 0th order increase gradually, so B can't be right because it he increase in 1st angle order is greater than increase in 2nd angle order.

12. Aug 24, 2015

### soarce

Right! What happens in D?

13. Aug 24, 2015

### toforfiltum

The first order beam of light should be at 90° to 0th order?

14. Aug 24, 2015

### soarce

In figure D we have only on diffraction order, the second order would have a diffraction angle greater than 90°. Does this situation fit with the relationship between $\theta_n$ and $\theta_{n+1}$ which we established before?

15. Aug 24, 2015

### toforfiltum

No, but isn't figure D wrong in the first place? Like I said in post #9?

Last edited: Aug 24, 2015
16. Aug 24, 2015

### soarce

Post #9 refers to the second order.
In figure D we have only the 1st order. Following your remark in post #11, what angle must have the 1st order so that the 2nd one goes beyond 90°?

17. Aug 24, 2015

### toforfiltum

45°?

18. Aug 24, 2015

### soarce

That's correct, but in figure D we see a 1st order angle much smaller than 45 degrees.

19. Aug 24, 2015

### toforfiltum

Ok, thanks a lot!

20. Aug 24, 2015

### TSny

If the second order is at 90o, at what angle is the first order? Use dsinθ = nλ to find out.