OK another yummy one:
FM radio waves from a distant tower pass through a relatively small gap (width b) between two very large buildings that are opaque to radio waves. (assume that the radio waves are plane-waves by the time they reach the gap, and their direction of travel is exactly perpendicular to b). You walk along a street at distance s from the gap, perpendicular to the direction the wave are traveling. Then suppos that b=8.00m, s=20.0m, and the wavelength of the radio waves is 3.049m. As you walk along the x-axis from -xmax= -30.0m to +xmax= 30.0m while carrying your portable radio, at what position will you discover total destructive interference of the radio waves? Find them all, calculate the frequency of the FM station.

Sorry for the long one. Thank you.

madam, are the answers +8.244m,+23.55 m and their negatives?(because i have taken a really dark shot here)
for frequency , we can apply 3 * 10^8=f lambda and calculate f.
waiting for reply.

First let me thank you for your hard work..Thank you.

So, you choose 3 x 10^8 m/s because its the speed of light?

but I still don't know how you calculated the positions.
n=c/v
v= c/n
f=v/lamda ?

draw a formula triangle, speed is always king (on top) with frequency & wavelength under.

With regard the speed of light, all EM waves travel at this speed in a vacuum.

If you want really want to get picky the refractive index of air is 1.0003.

Also zero diffraction will happen through the gap.

assuming my answers are correct, here is how i did......
first of all, as paul has explained, radio waves are part of electromagnetic spectrum. so i can apply c=frequency* lambda, and use c= 3*10^8 and lambda=wavelength(given) to find frequency.=9.8* 10^7 hertz.
now, since wavelength is comparable to dimensions of gap, there will be diffraction.(feynman explains this very well as the phenomenon in which light wave spreads out while passing through a comparable aperture or edge, ie. whenever we squeeze the size of the aperture, the paths for photons to travel is squeezed and so the probability amplitudes of a photon to reach a point in GEOMETRICAL SHADOW is not completely extinguished. the nullification would have happened if there were many paths available, that is why light behaves rectilinearly in geometrical optics.however, this is not the case here. here light will DIFFRACCT, as lambda is comparable to gap dimension)
now, only calculation. diffraction minima occurs at b sin theta =m lambda where lambda= wavelength, theta= angle from central line passing through middle of gap towards the street. and m is integer-1,2 ,3 ...... or their negatives.
so sin theta=m lambda/b ..............eq1
now a bit of geometry, let waves diffracted at angle theta(for destructive interferance) strike portable radio at a distance x from origin. so tan theta=x/s
so cot theta=s/x
so, cosec ^2theta= 1+cot^2theta=1+s^2/x^2
so sin^2 theta=1/(1+s^2/x^2)
so sin theta=root(1/(1+s^2/x^2)).............eq2
divide eq 1 by eq 2 to remove sin theta and solve the result for x.
u will get x=s root(1/(b^2/m^2 lambda^2 -1)
now put all values and use m=1, m=2 m=3...... to get x.
u will see that m=1, m=2 will give x <30(as required)
corresponding negatives give minima on other sides.
SHORTCUT: i realised that it will be easy to calculate the minima angles first and then simply take tan.
in that case , the required angles are sin inverse(m 3.049/8) gives 22.4 degrees for m=1, 49.66 degrees for m=2, m=3 is impossible and so are the rest.
then, using tan theta=x/20, i get the required x.
was the optical fibre problem answer correct?

Ricky,

Just a question here, you state the wavelength 3.049 m is relative to the gap of 8m, so how can it be relative to create diffraction? no diffraction will happen here.

TIP: 8 sin theta= 3.049m gives reasonable values of theta.
gap reminiscent of single slit.
still, i will refer this to genius physicsquest.

Ricky,

I didn't read the initial question properly i see the person is at an angle, thanks for pointing this out.

I sure wish I had your mind.. This was very very helpful. The optic fiber question works out correct. Drawing the picture seems challanging any suggestions?
Thank you.

i know this is going to be the most horrible drawing in this forum.
picture was modified .the rays must strike the lower interface.

Attached Images

untitled.bmp (576.1 KB, 5 views)

OK!! Love your spunk. The picture did make me laugh and not much does that in physics. I can see the overall point, but I am unclear of how the continued path of the ray goes until it reaches right side of the optical fiber. and which rays for the range of angles for theta will pass through both the core and clad interface and the clad air interface (thereby escaping from the upper edge of the clad into the air.

Thank you. Maybe I can re-draw it and we can work on it together?

Sample Drawing.doc

I hope i attached this OK, any help on the sketch from based on the previous optical question would be very helpful.
thanks