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?