Given the distance between the two towers = 40 km We know that for distances smaller than Fresnels distance(ZF) , spreading due to diffraction is smaller as compared to the size of the waves. But the hill is half way between the tower so distance upto which much of diffraction will not be observable should be 40/2 = 20km = 20,000m We know that Fresnels disatnce is the minimum distance a beam of light can travel before its deviation from straight line path due to diffraction becomes significant. Hence ZF = 20,000 m For the hill not to obstruct the spreading radio beam, the radial spread of the beam over the hill 20 km away must not exceed 50 m. / lambda= 50*50/ 20, 000 = 0.125 m. Therefore the longest wavelength of radiowaves which can be sent between the towers without appreciable diffraction = 0.125 m
113) Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?
12.5 cm
Related Optics MCQ with Answers
0.2 nm
use x=n ?1D/d
39.4?
Join the point of incidence (call it A) to the mid point of the second mirror (call it B) Let the corner point be called C. Once you join A and B you know that: tan ?BAC = 14/11.5 = 1.2174 ? ?BAC = tan ?? (1.2174) = 50.6 ? Angle of incidence is angle made with the normal at the point of incidence = (90 - 50.6) = 39.4 ? Angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0cm long) after reflecting from the first mirror = 39.4 ?
9I and I
By I(max) = I1 + I2 + 2?I1 x I2 =>I(max) = 0.I + 0.4I + 2?(0.I x 0.4I) =>I(max) = 0.9I By I(min) = I1 + I2 - 2?I1 x I2 =>I(min) = 0.I + 0.4I - 2?(0.I x 0.4I) =>I(min) = 0.I
I0/4
the positions of minimum intensity will not be completely dark is I0/4