Microwave Emitting Star, and detector with diffraction

[Carid]

Well done, you spotted my "deliberate mistake"...gulp...So what is the proper value for angle theta?

 

[TFM]

It would be:

$$\theta = asin(10.3/100)$$

Which gives the angle to be: 5.9 degrees.


Does this look okay?

 

[Carid]

10.3 ? I'm not the only one making "deliberate mistakes" this morning...

 

[TFM]

Should be 6.02 degrees.

Silly Me!

TFM

 

[TFM]

Have we made an error - I think we have found the minima, but the question asks:

At what angle above the horizon is the star when the first maximum is received?

?

TFM

 

[Carid]

Why do you think we have found a minimum?

 

[TFM]

I'm not sure, I suddenly saw the question and panicked. I think I remembered putting destructive interference in a previous post.

I also wrote down how destructive interference is when the waves out half wavelength out of phase written done in my work, despite the fact that the question just above said maxima. Suddenly gt very worried and panicked. - I get worried far too easily

Sorry,

TFM

 

[Carid]

Panic is good.
It means "let's go back to the fundamentals and see if we have understood and be sure we haven't neglected something along the way".

 

[TFM]

So can I just check something, just to see. I wrote down

destructive interference is when the waves out half wavelength out of phase

But we made the calculations with half a wavelength added on. Would this not make the final wave a minima

Just Checking

Also, Panic is good, unless the homework is due in in less then 1 hr...

TFM

 

[Carid]

OK let's examine the fundamentals.

We will get our maximum if the path length between the two rays differs by a single wavelength. Now we found an extra half-wavelength in the lower ray path. We need another half-wavelength. Where's that coming from? It comes from the reflection which advances the phase of the lower ray by half a wavelength. Now we've got two half wavelengths adding up to one whole wavelength. We should get that first maximum.

I hope you get to see the diagram I sent earlier; very simply the 1 metre height is NOT arbitrary. It's twice the 50cm height and is a question of symmetry.

 

[TFM]

Yeah I worked the 1m from symmetry after thinking about. So the water reflection from the water makes the wave half out of phase. See, I didn't realize that, should have though, since the the link showed the reflection wave going from a peak, reflected into a trough.

That makes sense. Thanks,

Out of interest, not part of the question, but I am intrigued, how would you find, say a second or third maxima? Would you just add on a whole wavelength onto the half in the triangle, eg:

$$\theta_n = (n-1)\lambda + \frac{1}{2}{\lambda}$$

where n is the nth maxima?

TFM

 

[Carid]

As the star rises in the sky we'd arrive at a moment when the extra path length is now 3/2 of a wavelength (31.5cm) and the angle would be about 17.5degrees.

That equation you quoted looks like something you'd use in optics where the wavelength is extremely short and the angle very small so that trig functions are discarded.

 

[TFM]

I could have sworn I put the asin in. Should have been:

$$\theta_n = asin\left((n-1)\lambda + \frac{1}{2}{\lambda}\right)$$

Thanks for all your assistance.

TFM

 

[turin]

So, I see Carid has got you covered, TFM. Great discussions, too. I just thought you might benefit from an alternative perspective. Can you see how to identify this with a Young's double slit phenomenon? In your problem, you have a detector, a water surface, and a single light source. In Young's experiment, there is a screen and two slits. Can you identify the analogies? Can you identify the crucial difference?

 

[TFM]

I can definatley see the similarities. The Two slits would be where the wave is reflected, straight up, on Carid's Diagram, the isosceles triangle. The crucial difference is that in Young's, the two waves would be in phase, whereas this version has the two waves out of phase.

TFM