Question

please answer the following questions

3. Many cell phones broadcast their signals at 800 MHz. Imagine you are behind 2 large, radio-wave-absorbing building with only a 13 m space between them. What is the angular distance, in degrees, of the radio wave after it emerges from between the buildings? 4. You want to recreate the circular diffraction pattern you saw in lab to show some friends. You borrowed a helium-neon laser (lambda = 633 nm) and made a 0.15 mm diameter pinhole. If you want the central maximum to have a diameter of 0.80 cm, how far behind the pinhole should you place the viewing screen?

Answer 1

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so order is , 276.08 nm , 497 nm , 828.25 nm

apply V = Lf

so wavelength L = 3e8/800 e6

L = 0.375 m

apply d sin theta = mL

sin theta = 0.375/13

sin theta = 0.0288

theta = 1.652 deg
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apply Y = 1.22 LR/D

R = YD/1.22L

here Y = 0.8/2 = 0.4 cm ( fron centre to one end)

R = 0.4 cm *0.15 mm /(1.22* 633nm)

R = 0.776m or 77.6 cm