# Diffraction grating, magnification

1. May 1, 2014

### unscientific

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

Part (a):Sketch zeroth and first order rays. Find the angle of first order rays make to normal.
Part (b): Sketch intensity distribution at 7.5mm and 10mm.
Part (c): Explain how this is useful in understanding a microscope.

2. Relevant equations

3. The attempt at a solution

Part (a)

Using $d sin\theta = n\lambda$, $\theta = 2.9^o$. Thus angle to normal is 87.1o.

Part (b)

At 7.5mm

Intensity distribution is given by:
$$I = I_0 \frac{sin \left(\frac{Nkd sin\theta}{2}\right)}{sin \left(\frac{kd sin\theta}{2}\right)}$$

Principal Maximas occur at $\theta_{max} = \frac{n\lambda}{d}$. Separation between a principal maxima and a minima is $\frac{\lambda}{16d}$.

At 10mm

There will be bright spots separated by distances $a$. This is equivalent to a N-slit grating. To find a:

$$a = \theta f = f \frac{\lambda}{d}$$
$$d' = f\frac{\lambda}{a} = d$$

Thus we see that at 10mm, the image of the grating slit is simply reproduced. If we observe at distances greater than 10mm, $d' > d$ is amplified.

Part (c)
By varying the distances of the final plane, we can set the magnification and is given by $\frac{v}{u}$ where $v$ is distance from lens to final plane, $u$ is distance from lens to grating.

Last edited: May 1, 2014
2. May 2, 2014

### BvU

By the way, the drawing for the first order maximum is very strange: do both maxima end up on the same side of the zeroth order maximum on the axis ?

3. May 2, 2014

### unscientific

You're right. It should be something like this:

I'm trying to ascertain whether my answers are right, do you mind taking a look at them??

4. May 2, 2014

### BvU

If this the alternative for the 1st maximum, it's worse than the first attempt. Positive lenses do not diffract light rays away from the axis !

The drawing for the 0th maximum is a lot better

Perhaps a suggestion to make life easier: Start with a grating from x=-w to +w. You can erase the rays from the 0 to -w half later on.

How did you derive the intensity distribution in part b from the relevant equations you should have listed under 2. in the template ? What is N, k ? Are he intensities of the 0th and 1th maxima equal ? What about the widths ?

Part c is not about magnification but about limits to resolving power.

5. May 5, 2014

### BvU

Time to pick up on this one. Any new drawings ? Start with the 0th maximum in post 1. Picture isn't all that clear: Do I see all rays go through one point at 7.5 mm and what do I see at 10 mm ?

Then the +1st maximum. rays from all these 16 slits are parallel, but at an angle with the axis. So they go through "one off-axis point" at 7.5 mm. What about where they end up on the 10 mm plane ?

Finally the -1st maximum. Also off axis but in the other direction. So where is the peak at the 7.5 mm plane ?

On the 10 mm plane we expect a depiction of the grating, right ? If the object is at x > 0, where is the image ?
And: is it infinitely sharp ? Why / why not ?
nice picture I found here -- can't one hide a link or a in a spoiler ?

6. May 5, 2014

### unscientific

I realised my first picture at 7.5mm is correct, and my second one is wrong.

They split again, and form back the grating.

Yeah they form below the central maxima which is halfway along the grating.

If the original grating is infinitely sharp, then the image at 10mm is infinitely sharp.

7. May 6, 2014

### BvU

Only if the lens aperture and the difraction pattern in the focal plane are infinitely big, otherwise higher orders get cut off, causing an unsharper image (the image is the fourier transform of the diffraction pattern).

The f/2w is 0.5, a pretty wide aperture, that picks up a considerable number of orders, but not infinitely many.

This detail might well be within the scope of the exercise, since they give lens height as 2w. I think it is a bit much to drag in rayleigh resolution and such, but who knows. Slit height isn't given, only "narrow"(which is why you draw your 1st order same height as 0th). Your call.